Category Archives: Teachers’ Resources

Transport in plants
This is the movement of water and mineral salts from the soil to other parts of the plant and food materials from the leaves to the rest of the plant.
Most single –celled organisms are very small hence have a very high S.A to volume ratio hence substance rapidly get in and out of the cell by simple diffusion.eg mosses and liverworts
Multicellular organisms are big hence have a small S.A to volume ratio thus they need a special transport system to efficiently move substances into and out of the cells.
In higher plants, the transport function is carried out by a specialized transport system known as the vascular bundle.
It comprises of ;
- (i)Xylem –transports water and mineral salts from the soil.
- (ii)Phloem –transports dissolved food substances such as sugars from the leaves.
- ROOTS
- Functions of roots
- (i)For anchorage-hold the plant firmly in the soil
- (ii)Absorption of water and mineral salts from the soil
- (iii)As storage organs of some plants e.g. carrots
- (iv)As breathing roots ( gaseous exchange) in some plants
- Internal structure of a root
- Piliferous layer
This is a special epidermis of young roots whose cells give rise to root hairs.
Its cells are thin walled to allow passage of water and mineral salts
As the root tissues mature a less permeable suberised epidermis replaces the piliferous layer
- Root cap
It covers the apex of the root.
It consists of simple parenchyma cells that protect the growing part ( apical meristem) of the root tip as it is pushed past soil particles.
Cells of root cap are relatively impermeable to water and solutes.
- NB Roots of aquatic plants lack root caps because they grow suspended in water.
- Epidermis
It’s the outermost layer of cells that protect the inner tissues.
The layer is one cell thick to reduce the distance over which water and mineral salts pass.
Some epidermal cells are modified to form root hairs.
- Cortex
Tissue found between epidermis and endodermis
Made up of loosely packed, thin walled parenchyma cells
- Endodermis
It’s a layer of surrounding the vascular bundles.
Its characterised by;
Rectangular shaped cells
Starch grains – hydrolysed to release energy
Many mitochondria
Root pressure is thought to develop within the endodermis.
Casparian strip – has an impervious deposit on the radial and cross walls.
Endodermis controls the amount of water and mineral salts entering into the vascular bundles.
- Pericycle
It’s a layer of cells found between endodermis and vascular bundles.
Gives rise to lateral roots
- Xylem
Comprises of vessel and tracheid elements.
It transports water and mineral salts from the soil.
Phloem –transports dissolved food substances such as sugars from the leaves to the rest of the plant.
Comparison between monocotyledonous root and dicotyledonous root
- Dicotyledonous root
xylem occupies the centre
Xylem forms a star shape
phloem is found between the two rays of the star
Pith absent
- Monocotyledonous root
Xylem and phloem are arranged to form a ring
Xylem tissue alternates with the phloem tissue
Pith present

Root hairs

they are modified outgrowth of epidermal cells
They are numerous, long and slender to provide a large surface area through which absorption of water and mineral salts take place
They have numerous mitochondria to supply energy for active transport
They have a thin cell wall which ensures rapid movement of materials
Cell vacuole has high solute concentration to maintain a high osmotic pressure to absorb water
Cell vacuole is large to store absorbed water and salts
Have short life span but are continuously replaced by new ones that develop nearer to the tip.
- Stem
- Functions
To support and expose the leaves and flowers to the environment
To conduct water and mineral salts from the roots to the rest of the plant
To conduct manufactured foods from the leaves to the rest of the plant
Other functions
Storage of food and water-in potato stem tubers
For gaseous exchange e.g. lenticels
Perennation –survival of perennial and biennial plants from one year to the next by vegetative means
Substance in the stem are transported within the vascular tissue
- Comparison between monocotyledonous stem and dicotyledonous stem
- Dicotyledonous stem
Vascular bundles arranged to form a ring
Have a central pith
Xylem located on the inside while the phloem on the outside
Cambium present between the xylem and phloem
Monocotyledonous stem
Vascular bundles arranged randomly/scattered
Cambium absent

NB most tissues in the root and stem are similar because these tissues are continuous from the root into the stem
Common tissues to both root and stem are;
- Epidermis
Cells are elongated
Outer walls are covered by a waxy cuticle that;
Prevents excessive loss of water through evaporation
Protects inner tissues from infection and mechanical injury
- Cortex
Made up of various types of cells i.e.
- (i)Collenchyma cells
They are angular
They have thick cellulose cell walls to provide support to the root
- (ii)Parenchyma cells
Spherical in shape
Have thin cell walls
Cells are loosely packed creating intercellular spaces filled with air
Cortex act as storage tissue for water and food
They may have chloroplasts to carry out photosynthesis hence called chlorenchyma
(iii) Sclerenchyma
Their walls are thickened by deposition of lignin in a process known as lignifications
It serves as a strengthening tissue
- Pith
It’s the central part of the stem. It consists of the parenchyma cells that store water and food substances. In some stems the pith may be hollow.
- Absorption of water
The soil particles are usually surrounded by a film of water
Root hair cells absorb water from the soil by osmosis. The cell sap in the vacuole of the root hair cell has high concentration of salts and sugars hence it’s hypertonic to the water found between the soil particles.
Due to this concentration gradient, water molecules move by osmosis from the soil through the semi-permeable membrane of the root hair cells into the cell sap.
The root hair cell sap becomes more dilute relative to the adjacent cortex cells. As a result water moves by osmosis into the adjacent cortex cells and their osmotic pressure is lowered relative to the inner cortex cells, which draw water by osmosis
Water passes through successive cortex cells and through the endodermis to the xylem by osmosis.
The endodermis actively pumps ions into the xylem vessels thus enhancing uptake of water into the xylem vessels by osmosis and creating root presure
The root hair cells will take up water as long as their concentration of salts is higher than that in the soil
- Active uptake of mineral salts
The soil water contains dissolved mineral salts which plants require for their growth and proper functioning.
The concentration of cell sap in root hairs is greater than that in the soil hence enter the root hairs against the concentration gradient. This process requires the use of energy therefore referred to as active transport.
Active transport involves substances known as carriers which combine with mineral ions and then carry them across the plasma membrane into the cell.
After absorption, the mineral salts move into the xylem vessels at the centre of the root then carried up the stem into the leaves.
- Factors that affect the absorption of mineral salts
(i) Metabolic inhibitors-these are chemical substances that prevent metabolic activities from taking place.
they prevent the release of energy thus active transport does not take place e.g. cyanide
(ii) Temperature-low temperature reduces the rate at which active transport takes place. Increase in temperature up to the optimum speeds up the rate of chemical reaction. High temperatures denature the enzymes.
(iii) Oxygen concentration- Oxygen is used in in oxidation of substrates that yield energy for use in the active uptake of mineral salts
- Transpiration
It’s the process by which plants lose water in the form of water vapour in the atmosphere.
Loss of droplets of water from the plant is called guttation.
Guttation occurs through special glands found mostly at points where the vein is in contact with the edge of the leaf. The glands are called hydathodes. They are located on the leaf margin or apex.
Guttation usually occurs in plants that grow in wet habitats.
- Role of leaf in transpiration
Water gets into the leaves through the xylem tissue. Water leaves the xylem and enters the cells of spongy mesophyll by osmosis.
Water diffuses into the sub-stomatal air spaces in the form of vapour.
The concentration of water molecules is higher in the air spaces than in the atmosphere. Water diffuses out through the stomata into the atmosphere
Movement of water through a leaf

Types of transpiration

(a) Stomatal transpiration
This is the loss of water in the form of water vapour through the stomata.
It accounts for 80-90% of the total transpiration in plants.
Most stomata occur on the leaves but may also occur on the epidermis of young herbaceous stems.
- (b) Cuticular transpiration
This is the loss of water in the form of water vapour through the cuticle.
In plants with thick cuticles the loss is negligible.
- (c) Lenticular transpiration
This is the loss of water in the form of water vapour through the lenticels.
Lenticels are areas with loosely fitted cells on woody stems.
The loss of water is negligible.
Forces involved in transportation of water and mineral salts
- Transpiration pull
Process by which water moves up the xylem due to evaporation of water in the leaf.
It enables a stream of water to move from the roots up the leaves
Energy from the sun causes evaporation of water increasing the diffusion gradient between the atmosphere and the mesophyll cells which leads to water vapour diffusing into the atmosphere.
The mesophyll cells draw water from the xylem. The water from the xylem is replaced by a continuous column of water known as transpiration stream moving up the roots.

Cohesion and adhesion forces

Water molecules attract one another in such a way that they always stick together. The forces that keep them together are referred to as cohesion force.
Also water molecules are attracted to the walls of the container in which the water is contained by a force referred to as adhesion force.
The cohesive and adhesive forces in very thin columns can be very high and not easily broken.
These forces maintain a continuous and an uninterrupted water column in the xylem vessels up the trees.
- Capillarity
It’s the tendency of water to rise in very narrow tubes.
The lumen of xylem tracheids and vessels is very narrow and this enables water to rise by capillarity.
- Root pressure
It’s the force that pushes water absorbed from the soil to move up the stem from the root.
The energy used to develop root pressure originates from the endodermal cells.
Cells of endodermis actively secrete mineral salts into the xylem. The osmotic pressure of the xylem content is increased thereby encouraging water movement.
When the stem of a plant is cut, water oozes out from the cut stem.
Root pressure can only raise water to a height of about 1 metre hence if a plant is growing in soil with little water the maximum height that the root pressure will raise water will be less than 1 metre.
- Importance of transpiration
Replace water lost through the leaves.
Aid in transportation of water and mineral salts
Cools the plant.
Helps in the removal of excess water especially in aquatic plants
Causes wilting- this is beneficial when a plant cannot obtain enough water to replace that lost by the plant through transpiration
Responsible for turgor in plants
- Factors affecting transpiration rate
- Structural factors
They are related to the morphology of the plant e.g.
Roots
Plants with extensive root system have a high rate of transpiration than those with few roots.
Extensive roots absorb more water hence more is available in the sub-stomatal spaces.
Leaf size
Large leaves have a large surface area over which transpiration takes place hence high rate of transpiration
Leaf structure
Cuticle
A thick cuticle reduces the rate of transpiration
The cuticle in most case is waxy-wax reflect away the sunlight hence lower temperatures in the leaf
Wax is also water proof hence reducing rate of transpiration
Stomata
- -Number of stomata-the fewer the number of stomata the lower the rate of transpiration
- -Position of stomata-the sun shines directly on the upper surface of leaves hence increasing the rate of vapourization thus high water loss
Stomata on the lower surface are sheltered from the suns rays hence lower water loss
-Sunken stomata-when the stomata are sunken water vapour accumulates in the sub-stomatal air spaces thus its not exposed to moving air hence reducing the rate of transpiration
Leaf fall
During periods of drought, some plants such as broad-leafed deciduous trees shed their leaves to reduce the surface area for water loss.
In some species of grass the aerial shoot dries up to ground level.
(e) Hairy leaves
In some plant, leaves are covered with hairs or scales. These trap a layer of still moist air on the surface of the leaves thus reducing transpiration
- Environmental factors
Temperature

High temperature increases the capacity of the atmospheric air to hold more water vapour.
High temperature increases the internal temperature of the leaf which in turn increases the latent heat of vapourization therefore enhancing evaporation from the leaf cells.
- Humidity
It’s the amount of water vapour in the atmosphere.
The humidity difference between the inside and the outside of the leaf is known as saturation deficit and it determines the rate of water loss from the leaf.
In dry weather, the saturation deficit if high hence increasing rate of transpiration
In high humidity, the saturation deficit if low hence decreasing rate of transpiration. Under such conditions some plants secrete droplets of water through specialized pores called hydathodes .
This process of water loss is called guttation and is common in hydrophytes ( plants growing in wet habitats)
- Wind
Wind carries water vapour as fast as it diffuses out of the leaves through the stomata.
This prevents the air around the leaves from being saturated with water vapour. This helps to maintain a high diffusion gradient between the inside and the outside of the leaf
When the air is still, the area around the leaf soon becomes saturated with water vapour. Diffusion of water vapour from the leaf surface is low leading to low rate of transpiration
- Light intensity
The stomata of most plants open fully during daylight hours when the light intensity is high
This brings the sub-stomatal air into direct contact with external environment.
The water vapour therefore diffuses out at a higher rate than in dim light when the stomata are partially closed.
- Atmospheric pressure
The lower the atmospheric pressure the higher the rate of evaporation
At high altitudes the atmospheric pressure is very low hence plants growing there lose a lot of water due to high rate of transpiration
Most of them have adaptations to prevent excessive water loss
- Availability of water
When there is adequate amount of water in the soil, water is absorbed and conducted to all the cells.
The mesophyll cells in the leaves become moist thus more water will diffuse into the inter-cellular spaces increasing the diffusion gradient. More water is lost to the atmosphere through transpiration
- Structure and function of xylem
Xylem comprises of
-xylem vessels
-tracheids
Xylem vessels
They are hollow tubes
They are made of dead cells placed end to end
Walls thickened with lignin to prevent them from collapsing as water is being transported up the plant.
Patterns of thickening
- The hollow part (lumen) provides passage for substances
- Xylem walls have perforations which form simple pits
- The pits on the xylem vessels permit the passage of water in and out of the lumen into the neighbouring cells
- Tracheid elements
- Have tapering or chisel-shaped ends
- Walls thickened with lignin
- Have tiny pores known as pits or perforations
- The pits on the side walls allow lateral water to the cells surrounding the xylem. This makes tracheids less efficient in conducting water than vessels
- NB Xylem vessels are more efficient in transport of water than tracheids because ;xylem vessels cross walls between their cells have dissolved forming a continuous hollow tube while tracheids have tapering ends whose cross walls remain perforated and this increases resistance
- Translocation
- It’s the movement of manufactured food substances from where they are manufactured in the leaves to the rest of the plant. It takes place in the phloem tissue in plants
  - Phloem tissue

Phloem tissue is made up of sieve tubes and companion cells.
Sieve tube – its long with perforated end walls which are called sieve plates
Cytoplasmic strands / Filaments run through sieve plates connecting adjacent cells.
At maturing sieve tube cells lack nuclei and ribosome
Mature sieve tube cells have few mitochondria
Companion cells – these cells have dense cytoplasm and a prominent nucleus and other cell organelles.
Companion cell generate the energy needed in the sieve elements because it has mitochondria
Plasmodesmata – these are passages found on the lateral walls. Substances move through them from the companion cells to the sieve tube cells.
Function of phloem
Materials move from one sieve tube element to another through the sieve pores in the sieve plates between adjacent elements. These materials are transported in solution form in the cytoplasm of the sieve elements
The organic products translocated are; sugar, amino acids and vitamins. They are translocated to;
(i) Growing and developing regions of the plants such as young shoots, leaves, flowers, fruits and roots
Storage organs or tissues such as tubers, corms, bulbs, rhizomes and seeds
Secretory organs such as nectar glands in some insect pollinated plants e.g. bananas
Experiment; Ringing experiment
Make a ring through the bark around the stem of a young tree using a sharp knife
Make a second ring 5cm above the first ring and peel off the bark between the two rings
X
Observe the experiment over the next two months
Discussion
When the ring of bark is removed, the phloem beneath it is also removed. After several weeks swelling above the ring is noted eg
X
This swelling is due to accumulation of food substances that were being transported from the leaves but could not get across the debarked part of the stem. As a result, there is no swelling on the lower part of the ring.

Transport in animals

Circulatory system
A circulatory system transports the substances and maintains a steep concentration gradient at the surfaces where diffusion takes place.
Its made up of a fluid, a pumping organ and vessels
There are two types of circulatory system: open and closed circulatory systems
- Open circulatory system
The transport fluid is contained in the general body cavity/ coelom/ sinuses. This type of system is common in invertebrates especially arthropods.
The transporting fluid in the body cavity is known as haemocoel
Cavities are free spaces between the body wall and organs. The fluid in the cavities is in contact with body tissues.
The fluid distributes oxygen, nutrients and hormones to tissues while removing CO₂ and nitrogenous wastes from the tissues.
- Closed circulatory system
The transporting fluid (blood) is conveyed in special tubes referred to as blood vessels.
- Differences between open and closed circulatory system

– Open	– Closed
– Blood flows under low pressure	– Blood flows under high pressure
– Blood circulates over a short distance at a slower rate	– Blood circulates over a long distance at a faster rate
– Fluid is not involved in the transport of O₂and CO₂	_–Blood transports O₂ and CO₂
– Is less efficient at supplying tissues and organs with nutrients and removing nitrogenous wastes	– More efficient at supplying O₂ and nutrients to the tissues
– Organisms with open circulatory systems are generally less active	– Animals with closed circulatory systems are more active

Transport in insects

In a cockroach there is a tubular heart just above the alimentary canal. The heart has 13 chambers, 3 in the thorax and 10 in the abdominal segments.
The anterior segment is joined to the aorta that empties the blood into sinuses of the head. Each chamber contains a pair of valves at the anterior part which prevent back flow of the blood.
Each chamber has a pair of lateral openings called Ostia which are closed by valves.
The valves allow blood to flow into the heart through the Ostia but not out of it.
- Mammalian circulatory system
Mammals have a closed circulatory system where a powerful muscular heart pumps blood into the arteries.
The arteries divide into even much smaller vessels called arterioles which in turn divide into even much smaller vessels called capillaries
Capillaries spread out in a network fashion in the tissues.
The capillaries eventually reunite to form venules that in turn form larger vessels called veins. Veins take blood back to the heart.
- Single circulatory system
This is where the blood flows only once through the heart for every complete circuit hence the heart has only one atrium a ventricle e.g. fish
- Double circulatory system
Blood flows into the heart twice for every complete circulation i.e. blood from the body tissues is pumped to the lungs and then back to the heart. This is called pulmonary circulation.
From the heart, blood is then pumped to the rest of the body organs. This is called systemic circulation.
The double circulatory system is found in birds, mammals and also crocodile (reptile). The other reptiles and amphibians have double circulatory system but the ventricle is not fully divided into the left and right ventricles.
Therefore efficiency of gaseous exchange is not fully realized due to mixing of oxygenated and deoxygenated blood.
- Structure and function of the heart
- External structure of the heart
The mammalian heart is broad at the anterior and narrower at the posterior end. Its made up of two auricles (left and right) and two ventricles (left and right)
The coronary artery which branches from the aorta supplies O₂ and nutrients to the heart tissues.
The two coronary veins transport CO₂ and the metabolic wastes away from the heart.
The heart is covered by a translucent membrane known as the
Pericardium prevents the heart from being overstretched as it pumps blood. It secretes pericardial fluid which reduces friction between the heart and the adjacent tissues when the heart beats.
At the anterior end of the heart are vessels i.e. aorta and pulmonary artery which take blood away from the heart and vena cava and pulmonary vein which return blood to the heart from the rest of the body.
- Internal structure of the heart
The heart is a muscular organ about the size of the fist.
It lies inside the chest cavity between two lungs.
Internally the heart is surrounded by a tough membrane called pericardium which covers and protects it.
It’s divided into two sides i.e. the left and the right sides which are completely separated by a wall called
Septum prevents the blood on the right side mixing with that on the left side. Each side consists of a small upper chamber called atrium (plural atria) and a larger lower chamber called ventricle. This makes the mammalian heart a 4 chambered organ
The atria are also called auricles and are thin walled and receive blood into the heart which they pump into the ventricles. Ventricles are thick walled and pump blood out of the heart.
The heart is made of special muscles called cardiac muscles. This muscle is special in 2 ways:
- -It can contract continuously without fatigue- the heart can beat for a life time without taking a rest.
- -Cardiac muscle is also myogenic e. its contractions are started by the muscle itself and not by nerves as the case with other muscle tissue in the body.
Four flap like valves control the direction of blood flow inside the heart. Two of these valves are called atrio- ventricular which allows the blood to flow only from the atria to the ventricles.
The one found in the right side of the heart is called tricuspid valve because it has three flaps.
In the left side of the heart is the bicuspid valve because it has two flaps. It is also called mitral valve.
The other two valves found in the heart are the semi – lunar valves. They are found at the base of the aorta and pulmonary artery. When open they allow blood to move from the ventricles into the arteries and away from the heart.
NB: –
Valves are attached to the walls of the ventricle by valve tendons or tendinous cords (cordae tendinae). The tendons allow the valves to open but prevent inversion of the flaps of the valves when blood attempts to flow back.
The wall of the left ventricle has thicker walls muscles than that of the right ventricle because the left ventricle pumps blood a further distance to all parts of the body while the right ventricle pumps the blood to the lungs.
Circulation of blood in the heart
The right atrium receives blood coming from the body tissues through the vena cava. This blood has very little oxygen dissolved in it hence it is described as deoxygenated blood. It is rich in CO₂ and appears dull red in colour.
The right atrium then pumps the blood into the right ventricle via the tricuspid valve. When full the right ventricle pumps blood into the pulmonary artery. Semi- lunar valves at the base of pulmonary artery prevent back flow into the right ventricle. At the same time tricuspid valve prevents any backflow into right Atrium.
Tendons (heart strings) hold the valve in a closed position preventing them from turning into the atrium.
The pulmonary artery carries the blood into the lungs where it picks up O₂ and gives up CO₂. It is now said to be oxygenated and appears bright red in colour. It goes to the left atrium of the heart via the pulmonary vein. This portion of the circulatory system that sends the blood to the lungs from the heart and back is called the pulmonary circulation.
X
The left atrium pumps blood into the left ventricle via the bicuspid valve. The left ventricle pumps blood to all parts of the body except the lungs. This blood leaves the left ventricle through the aorta. Semi- lunar valves that open into the aorta prevent back flow of blood.
The left ventricle walls are much thicker than the right ventricle walls in order to prevent develop a high enough pressure to pump blood to all parts of the body. The circulation of the blood from the heart to the tissues and back is called systemic circulation.
The mammalian heart therefore acts as a double pump. The left side sends blood rich in O₂ to the rest of the body and the right side sends blood poor in oxygen to the lungs.
The heart tissue itself receives food nutrients and O₂ via a vessel known as coronary artery which branches from the aorta and spreads through the heart muscle.
The function of the heart is to receive and pump blood. The heart receives blood when its muscle relax and it pumps the blood when the muscles contract. These two processes take place in a repeated sequence or cycle known as heart or cardiac cycle.
Adaptations of mammalian heart to its functions
It has valves which open to allow blood to flow in one direction and close when blood tries to flow back.
It has muscular walls which contract to pump blood and ensure its continuous flow.
It has a septum which separates oxygenated from deoxygenated blood.
It has an inbuilt system that controls contraction and relaxation of the muscles.
It has 4 chambers which store blood briefly before it is pumped to the rest of the body.
Its muscles contract and relax continuously without fatigue.
- The heart beat
The heartbeat can be felt as a pulse in various parts of the body where an artery is close to the skin surface such as wrist.
A pulse is a series of waves of dilation that pass along the arteries caused by the pressure of the blood pumped from the heart through contractions of the left ventricle.
A complete cycle of a heart beat takes less than one second. The human heartbeats at about 70-75 times/minute when one is at rest.
The heartbeat can increase up to 200times per minute during:
Exercise
Fever
Emotional disturbances (fear)
An increased heartbeat circulates blood with oxygen and glucose needed to produce energy for the vigorous activity in the muscle tissues faster and takes away Carbon iv oxide and other wastes away.
- Control of heartbeat
Heartbeat is started by collection of cells in the wall of the right atrium called pacemaker (Sino atrio node) SAN) it is controlled by nerve messages which come from a part of the brain called medulla oblongata.
The heart will continue to beat even if the nerves (vagus nerve) from the brain are cut but it will only beat at one rate.
Nerve impulses from the brain are needed to change the rate of heartbeat.
NB: Individuals who have a heartbeat which is too slow or faster can have it regulated by the fitting of an artificial pacemaker which takes over from normal pacemaker.
One heart beat consists of a systole and diastole phase i.e.
- Diastole (relaxation)
It refers to the phase when the ventricles relax in order to allow blood to flow in. During this phase, the ventricular volume increase and the pressure decreases.
When the right atrium contracts the tricuspid valve opens to allow deoxygenated blood to flow into the right ventricle.
At the same time the left atrium contracts and the bicuspid valve opens to allow oxygenated blood to flow into the left ventricle.
The semi-lunar valves close preventing blood from flowing back into the relaxed ventricles.
- Systole (contraction)
It refers to the phase when the ventricles contract to force the blood into the arteries while atria are relaxed.
When the left ventricle muscles contract the bicuspid valves close to prevent blood from flowing back into the relaxed atria.
The volume of the ventricles decreases while the pressure increases forcing blood to flow out of the heart.
Deoxygenated blood flows through the semi lunar valve through the pulmonary artery to the lungs while oxygenated blood flows through the semi lunar of the aorta and into the tissues of the body.
The sphygmomanometer is used for measuring blood pressure. Blood pressure is obtained by placing systolic pressure of the left ventricle over the diastolic pressure of the left ventricle i.e.

Average human blood pressure = 120mm Hg (systole)
- 80 mm Hg (Diastolic)
Blood vessels

The mammalian blood vessels are arteries, veins and capillaries.
The walls of veins and arteries consist of the following three layers
(i)inner lining of a single layer of epithelial cells called endothelium
(ii)middle layer of smooth muscles and elastic fibres. Its this layer that brings about dilation and constrictions of blood vessels
(iii)outer layer made up of fibrous connective tissue
Arteries
They take blood from the heart to the body tissues and organs. Due to the pumping action of the heart, blood from the heart enters the arteries at high pressure.
- Properties of arteries
Thick muscular walls to withstand and maintain higher pressure of blood.
An outer fibrous coat for strength and protection
A thick layer of muscle and elastic fibres which contract and relax to adjust their diameter as blood flows through them. Arteries have an inner lining of cells known as an
A narrow lumen to maintain the pressure of blood inside them.
Most arteries are located deep within our bodies for protection against injury. The size of the lumen in arteries can be adjusted by nerve control of muscles in their walls e.g. the amount of blood passing through the arteries can be adjusted during exercise so that more blood flows to the legs and less blood to small intestines. This ensures that blood is properly utilized by only the parts of the body that need it most.
Pumping of the blood can be felt on an artery if pressure is put on it with a finger. This pressure makes blood to flow in only one direction.
When the ventricles contract, the muscular layer of arteries stretch to reduce resistance to blood flow. When the ventricles relax the muscular layer of arteries contract compressing the blood and forcing it flow forward in one direction.
All arteries carry oxygenated blood except the pulmonary artery which carries deoxygenated blood.
Arteries branch out to form narrower vessels called arterioles, which branch further within the tissues into finer vessels called
Some arteries are specialized to perform certain functions e.g. arteries of the lungs have thin walls due to lower pressure in pulmonary circulation. Aorta and pulmonary arteries have cardiac muscles extending to their bases.
With age arteries change in structure. In old age elastic fibres have;
Irregular thickening
- -fat is deposited between the cells
- -calcification occurs between arterial walls thus making the walls brittle.
Veins
They carry blood under low pressure from the tissues towards the heart.
They have thin walls which are composed of a thin outer fibrous coat, a thin middle layer of muscle and elastic fibres and an inner layer of cells (endothelium)
They have pocket valves at intervals in their walls which allow blood to flow in one direction towards the heart. They carry deoxygenated blood except the pulmonary vein which carries oxygenated blood.
Portal veins have capillaries at both ends. They are unique veins that carry blood from one organ to another i.e. hepatic portal vein which carries blood from the small intestine to the liver.
Most veins are found between the skeletal muscles and may be visible. The skeletal muscles contract squeezing veins and forcing blood to flow towards the heart.
When breathing in the pressure in the chest cavity reduces. The volume of the heart increases and the blood in the veins is sucked up towards the heart.
Valves are found in the heart, at the junction of major arteries and the heart and also in the veins. The veins of the lower limbs have more valves.
Open valves allow blood to flow in one direction only. Closed valves prevent the back flow of blood.
Structural differences between arteries and veins
Arteries Veins VEINS

– Have thick muscular walls	– Have thin and less muscular walls
– Have no valves except pulmonary artery	– Valve present at intervals throughout their length
– have narrow lumen	– Have wide lumen

Functional differences

– Arteries	– Veins
– Transport blood away from the heart	– Carry blood towards the heart.
– Carry oxygenated blood except pulmonary artery	– Carry deoxygenated blood except pulmonary vein.
– Blood flows rapidly at high pressure	– Blood flows slowly at low pressure
– Blood flows in pulses	– Blood flows smoothly

Blood pressure in the arteries is greater than in veins for the following reasons
Arteries
Receive blood directly from the heart pumped under high pressure
Have relatively narrower lumen, which maintains high pressure
Have thick muscular wall, which resists and generates pressure
Veins
Receives blood whose pressure has been reduced by capillary resistance
Have relatively wider lumen, which reduces pressure
Have thin less muscular wall, which reduces pressure
Capillaries
They are narrow blood vessels whose walls are one cell thick
Capillaries have certain characteristics which make them a region suitable for exchange of substances between blood and the tissues.
- Characteristics of capillaries
They are numerous in number to increase their surface area for exchange of materials
Have thin walls(one cell thick) to allow rapid exchange of substances
They form a dense network in all the tissues in the body. This creates a large surface area over which the exchange takes place.
They are narrow t for high pressure build-up within them. This ensures faster movement of substances.
Have sphincter muscles at the arterioles end, which enables regulation of blood flow
The intensity of metabolism determines the density of Capillary network in the tissues and organs e.g. there is dense network of blood capillaries in the lungs, liver, kidney, skeletal muscles etc
The walls of the capillaries are said to be permeable i.e. allow the passage of molecules through them.
A fluid is formed which is referred to as tissue fluid. The cells obtain their requirements through diffusion from the tissue fluid e.g. water, glucose, mineral salts, and hormones. The cells are bathed by the tissue fluid and they release waste products into the tissue fluid e.g. nitrogenous waste, mineral salts, CO2 and heat.
Capillaries unite to form venules which unite further to form veins.
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Diseases and defects of the circulatory system
Arteriosclerosis(atheroma)
This is the hardening of the arteries. As the arteries age the body reacts by depositing cholesterol and calcium in their walls. This causes them to thicken and harden and to become less flexible or less elastic i.e. they become sclerotic. This forces the heart to work harder in order to pump blood efficiently throughout the body.
It also causes an increase in the blood pressure. High blood pressure can lead to a stroke or a heart attack.
- Prevention
Exercises
Avoid alcohol and smoking.
Avoid fatty foods
- Treatment
Take medication that lowers blood pressure.
Coronary thrombosis
Thrombosis is the formation of blood clots in the blood vessels. Coronary thrombosis refers to the clotting of blood in a coronary artery resulting in a heart attack.
Coronary arteries supply the heart muscles with oxygen and nutrients. When a clot blocks blood from reaching the tissues of the heart, the tissues experience shortage of oxygen and nutrients supply. CO₂ and nitrogenous wastes are not efficiently removed. This result in heart attack
Symptoms
Sharp pains especially on the left side of the chest.
Difficulty in breathing
Irregular heartbeats and swelling of the feet.oedema.
Cardiac cells die leading to heart failure and death.
Prevented in the same way as arteriosclerosis
- Treatment
Take medication to prevent blood clot formation.
Cerebral thrombosis /stroke
It occurs when a blood clot is formed in the vessels of the brain.
A stroke is caused by high blood pressure in the capillaries and arteries of the brain. Arteries supplying blood to the brain have thinner walls and the high blood pressure can burst the capillaries serving the brain tissues. The brain cells in the affected area die. Some parts of the body especially the left side maybe paralyzed.
Prevented same way as arteriosclerosis
Atherosclerosis
It’s a condition similar to arteriosclerosis but it is caused when cholesterol, fat and calcium are deposited along the inner walls of the arteries. This reduces the diameter of their lumen and causes high blood pressure as the heart is forced to pump harder.
Factors that increases risk of atherosclerosis
High level of blood cholesterol
Smoking
Obesity
Diabetes
Sedentary lifestyle which does not involve much physical activity
Varicose veins
It refers to the prominently swollen veins which may appear below the knees or at the back of the legs. This condition is brought about by failure of some valves in veins to function. Blood accumulates in the veins.
Some pregnant women develop this condition albeit temporarily. Also common in men soldiers who carry out parade drills.
Varicose veins can be caused by standing or sitting for a long time. To prevent varicose veins, shift your weight from one leg to another and stretch your limbs.
- Treatment
Wear special firm stockings every morning before getting out of bed.
Congenital heart defects
At birth, the blood circulatory systems of the mother and the foetus become independent. The pulmonary artery takes very little blood to the lungs of the foetus because they are not used for gaseous exchange
Blood flows between the right and left auricle through an opening in the wall between the two auricles called foramen ovale. The passage normally seals after birth.
When it fails to seal, lungs are denied adequate blood and gaseous exchange is not efficient. Blood transports less oxygen and removes less co₂ from the tissues .The baby turns dark and may die. This condition can be surgically corrected
Also when the valves within the heart fail to close adequately, the results a backflow of blood. The condition is said to be a murmur of the heart. It’s diagnosed by the sounds of the heart as the valves close. This condition can be corrected surgically.
Also the connecting vessel between the pulmonary artery and aorta (Ducts arteriosus) may not be sealed. The vessel normally seals at birth. Blood flow to the lungs is cut off and enters the aorta hence blood flow to the lungs is inadequate.
Gaseous exchange is impaired and tissues lack enough oxygen. The baby may turn dark. This condition can be corrected surgical.
Hypertension(High blood pressure)
Normal blood pressure varies between 90/60 and 140/90mmHg
It is caused by:
Heavy drinking
Smoking
Taking large quantities of salt in the food
General body stress
The heart of a hypertensive person is overworked and the person is prone to heart failure
Hypertension may lead to bursting of arteries and capillaries. If the blood vessels in the brain burst, a stroke results and brain cells die in the affected area. Paralysis for some parts of the body usually accompanies stroke
This disorder is more common in individuals aged over 40 years
- Control
Having regular exercises
Intake of less salt
Avoiding excessive drinking of alcohol.
Avoid smoking.
Avoiding general body stress
Structures and functions of blood
Blood is liquid which transports materials in mammals.
It has 3 major functions i.e.
A medium of transport of ,materials to and from other tissues
Regulation of body temperature
Protection against disease germs
Mammalian blood forms up to 10% of the body weight. An average human adult has 5-6 litres of blood in the body
- Composition of blood
Blood is composed of:
Cellular components which form 45% i.e.
Red blood cells(erythrocytes)
White blood cells(leucocytes)
Blood platelets(thrombocytes)
- Blood plasma
Plasma makes up about 55% of the total volume of blood
It’s a pale yellow fluid
90% of blood plasma is made up of water and the other remaining 10% consists of a variety of substances that are dissolved in the water. These substances are:
Food substances e.g. glucose, amino acids and fatty acids.
Waste substances like CO₂ and urea
Hormones like adrenaline and insulin
Enzymes and antibodies
Blood plasma without fibrinogen is called serum
- Functions of blood plasma
Transport red blood cells which contain oxyhaemoglobin to the tissues hence facilitate transport of O
Transports food nutrients from the alimentary canal to the liver and other tissues.
Transports metabolic wastes such as CO₂.
Transports hormones to target organs.
Transports small amounts of CO₂ in the form of carbonic acid or bicarbonate to the lungs.
Transports mineral ions or salts such as chlorides.
Transports antigens and antibodies to the site where they are required.
Regulation of body temperature by distributing heat generated in the liver to other parts of the body.
- Cellular components
- Red blood cells (erythrocytes)
They are biconcave in shape i.e. thinner in the centre than around the edge.
- Adaptations of red blood cells to their function
They have a biconcave shape to increase the surface area over which O₂ and CO₂ diffuse
Absence of nucleus increases the space in which hemoglobin is packed.
Has haemoglobin which has a high affinity for oxygen
They are small in size hence have a large surface area to volume ratio for the diffusion of oxygen.
The small size enables them to squeeze through the narrow capillaries.
They are pliable which enables them to move through capillaries
Have enzyme carbonic anhydrase which enables them to transport carbon iv oxide
Have thin plasma membrane, which allows rapid diffusion of gases
Red blood cells are produced in the bone marrow of ribs, sternum and vertebrae. In an embryo RBC are produced in the liver and the spleen.
Since the mature RBC lack a nucleus and other cell organelles such as mitochondria, they have a short life span. They survive for about 100-120 days.
Old blood cells are destroyed in the liver and spleen. The iron component of haemoglobin is released for the formation of new red blood cells
There are about 5 million red blood cells in every cubic millimeter (mm³) of human blood. However the number of red blood cells varies depending on any of the following factors :
Altitude- the higher the altitude the more there will be
State of health of a person –people with severe anaemia or malaria have much fewer red blood cells in their blood.
- Functions of red blood cells
Transport of oxygen-this is the main function of red blood cells. They transport O₂ from the lungs to the body tissues.
The haemoglobin found in these cells readily combines with O₂ when the blood passes through the lungs to form oxyhaemoglobin.
When blood reaches a region with low oxygen levels like in the tissues, the oxyhaemoglobin readily gives up the oxygen it was carrying, it then reverts back to haemoglobin. The cells take up the oxygen while hemoglobin is free to be used again to carry more oxygen i.e.
- Haemoglobin + Oxygen
- lungs
  - Oxyhaemoglobin
- Under low oxygen concentration e.g. in high altitude areas the bone marrow produces more RBC. When one moves from a low to a high altitude area, more RBC are manufactured to increase the oxygen carrying capacity of blood. In this way one becomes acclimatized to the high altitude. e.g. –Kenyan athletes train in high altitude areas like Nyahururu and Eldoret to increase the O₂ carrying capacity by increasing the number of their RBC.
- Foetal haemoglobin – it’s a pigment found in foetus.
- It has a high affinity for O₂ than the mother’s haemoglobin. This enables the foetus haemoglobin to obtain enough O₂ from the mother’s blood even at low O₂concentration
- After birth RBC containing foetal haemoglobin are destroyed in large numbers. The large number of RBC destroyed causes a lot of pigment in the blood hence the baby maybe slightly yellow, jaundiced due to the pigment – this occurs in the first two weeks of birth.
- Myoglobin – it is a pigment found in the muscles and it has high affinity for O₂ than haemoglobin. Thus oxyhaemoglobin readily release the o₂ to myoglobin which then releases o₂ to the cells in muscles.
- Haemoglobin can combine even more readily with (carbon ii oxide) gas than with O₂ to form carboxyhaemoglobin
- However carboxyhaemoglobin does not split to release haemoglobin. This prevents adequate O₂ from being supplied. This makes carbon ii Oxide a dangerous gas because a person who has inhaled even small quantity of it especially in a room with poor ventilation can die of suffocation
- Sources of carbon ii Oxide include:-
- Burning charcoal stoves(jikos)
- Exhaust fumes from vehicles
- Transport of carbon IV oxide (CO₂)
- About 95% of CO₂is transported by RBC. Most of the CO₂ from the tissues enter the RBC where an enzyme called carbonic anhydrase speeds up the dissolving of CO₂ to form carbonic acid This acid dissociates to form hydrogen ions(H⁺) and hydrogen carbonate (HCO₃) ions.

CO ₂ + H₂O ^Carbonic H₂CO₃
- _{Anhydrase (carbonic acid)}
The hydrogen carbonate ions leave the RBC and enter the plasma where they are eventually transported to lungs
In the lungs hydrogen carbonate ions are converted back to CO₂ which is released to the air when breathing out
White blood cells (leucocytes)
They are larger than RBC colourless and are fewer than in number. There are about 6000 per cm³of blood. This number increases during infections
They have a nucleus but reduce in the case of HIV infection.
They are formed in the bone marrow of long bones and lymph nodes. Their function is to protect the body against pathogenic micro-organisms such as bacteria, protozoa, viruses etc
Types of white blood cells
Granulocyte
They are also called phagocytes or polymorphs
They have a large lobed nucleus and cytoplasm containing granules hence the name granulocytes
They can change their shapes as they actively seek and engulf diseases causing germs in a process called phagocytosis hence the name phagocytes
Some white blood cells may die in the course of phagocytosis. The dead phagocytes, together with dead micro-organisms and damaged tissues form pus
They can squeeze through capillaries walls in order to reach infected tissues. They are made in the bone marrow
Agranulocytes
They have large rounded nuclei. Their cytoplasm is also non-granular
Types of agranulocytes
Monocytes
They are formed in the bone marrow. They
destroy micro-organism such as bacteria by engulfing them
Lymphocytes
They are formed in the lymph nodes and produce antibodies that protect the body from infections in the following ways:-
Antibodies which are anti-toxic neutralize the toxins (antigens)produced by the pathogenic organisms
Some antibodies such as agglutinins cause clumping together of micro-organism.
This stops the micro-organism from multiplying and eventually they die. They are then ingested by phagocytes.
Lysins destroy micro-organisms by digesting their cell membrane or walls
Opsonins are anti bodies which adhere to the outer surfaces of micro-organisms thus making it easy for phagocytes to ingest them Opsonins are only produced during infection
Platelets (thrombocytes)
They are very small and have no nucleus. They are fragments of RBC and they are made in the bone marrow.
They are approximately 250,000 platelets per mm³ of blood. They live for about 7 days.
Platelets produce an enzyme known as thromboplastin which plays a key role in blood clotting.
Blood clotting
Blood clot is a seal that forms to close blood vessels that are cut or damaged. This has 2 functions:
Stops further bleeding at the wound and therefore prevent excessive blood loss.
Prevents entry of harmful bacteria into the body through the damaged tissue.
Process of blood clotting
When the blood vessels are damaged, the damaged tissue and platelets release an enzyme called thromboplastin (thrombokinase).
Thromboplastin initiates the process of blood clotting by neutralizing the anticoagulants factor known as heparin which occurs naturally in blood.
Thromboplastin activates the conversion of prothrombin (blood protein) to thrombin in the presence of calcium ions. Vitamin K is required in the formation of prothrombin
Thrombin activates conversion of soluble fibrinogen which is an inactive protein to insoluble fibrin which forms a meshwork of fibres on the cut surface to trap RBC to form a clot.
Blood platelets

Thromboplastin/ Thrombokinase (Enzyme)

Prothrombin Vitamin K

Ca²⁺

Thrombin

Fibrinogen

Fibrin

BLOOD GROUPS

Human blood can be grouped using the ABC system and Rhesus factor.
- THE ABO SYSTEM
The ABC of humans has special types of protein called antigens. There are two types of antigens i.e. antigen A and antigen B
Antigens determine the blood type or blood group of a person.
A person with only antigen A on their RBC is said to belong to blood group A. people with antigen B
Sometimes both antigens A & B are found on the RBC of the individual. In such a case a person is said to belong to blood group AB.
In other people the blood has no antigens on the RBC such people have blood group O i.e.

Antigen present on RBC BLOOD GROUP
AA
BB
A&AB
NONO
In addition to the antigens on the RBC, blood plasma contains other types of proteins called antibodies. These are complementary to the antigens A & B
Antibodies are named a and b respectively. Antibodies and antigens do not correspond to each other e.g.
A person with antigen A will have antibody b in the plasma.
A person with antigen B will have antibody a in the plasma.
If both antigens are present as in blood type AB, then no antibodies will be present in the plasma.
If none of the antigens is present then both antibodies are present e.g. in blood group O e.g.

– Blood group	– Antigens	– Antibody
– A	– A	– B
– B	– B	– A
– AB	– A & B	– None
– O	– None	– A & b

NB The presence of an antigen and its corresponding antibody in the blood of an individual, would lead to clumping of RBC. This is referred to agglutination
BLOOD TRANSFUSION
It’s the process of putting donated blood into a receipt. A blood donor is someone who voluntarily goes to a hospital or heath centre to give blood. The donor should be a healthy individual between 18- 65 years.
Blood is taken from the donor through a vein in the arm and passed into a bag containing anti – clotting substances. The blood is kept I bank under suitable conditions to be given to a patient who needs it (within I month) because RBC will have died after I month.
A blood transfusion may be necessary in situations such as:
When a person loses too mush blood due to an injury that may result from motor accident, war e.t.c.
When a person becomes anaemic due to diseases such as malaria
When a woman loses too much blood after child birth.
When the blood of the donor and recipient mix freely without agglutination the blood from the two individuals is said to be compatible. The blood from two individuals is said to be incompatible if agglutination occurs when the two blood are mixed.
TABLE SHOWING BLOOD TRANSFUSION IN HUMANS
- DONOR

– RECIPIENT	–	– A	– B	– AB	– O
	– A	–	– X	– X	–
	– B	– X	–	– X	–
	– AB	–	–	–	–
	– O	– X	– X	– X	–

From the table above it shows that a person with blood group O can donate blood to receipts of all the 4 blood groups. This is because the type O blood lacks antigens on the RBC that could be agglutinated by the antibodies from the receipts plasma. Therefore referred to as universal donor.
Individual with blood group AB can receive blood from all the 4 blood groups because AB has no antibodies to agglutinate the receipts blood hence referred to as universal recipient
Precautions before transfusion
The recipient must be given compatible blood i.e. blood received by recipient, should not agglutinate. Compatibility of blood is determined by A & B antigens and rhesus antigens.
After somebody has donated blood, it’s first screened before it is kept in a blood bank or transfused into a recipient.
During screening doctors test blood for:
Presence of any infective micro- organisms e.g. HIV if blood is infected, it’s normally thrown away.
After somebody has donated blood he/she receives a blood donor card bearing the name of donor and hi/her blood type.
RHESUS FACTOR
The RBC may also have another antigen on their membrane known as Rhesus factor.
Individuals with Rhesus antigens on the membrane of RBC are said to be Rhesus positive (Rh+) while individuals without the Rhesus antigens are said to be Rhesus negative (Rh-)
When a Rh- woman marries a Rh + man the woman will conceive a Rh+ foetus. The Rh+ antigens of the foetus pass across the placenta into the mother’s bloodstream during the last month of pregnancy. The mother responds by producing Rh antibodies which cross the placenta into the foetal circulation. The Rhesus antibodies destroy some of the RBC of the foetus.
The first born child has a higher chance of survival because the destruction of RBC is minimal. But in subsequent pregnancies massive destruction of RBC occurs leading to the death of foetus. This condition of is referred to as erythroblastosis foetalis or haemolytic disease of the new born.
The mother can be treated with a Rhesus globulin which prevents her from producing antibodies against the foetal antigens. This will protect the RBC of the foetus in subsequent pregnancies.
Also the baby is transfused with Rh – blood after birth due to the extensive breakdown of RBC
LYMPHATIC SYSTEM
Animals particularly vertebrates have an additional transport system besides the blood system.
This is known as lymphatic system and it supplies all the regions of the body just like the blood system.
The lymphatic system is made of narrow, thin walled tubes known as lymph vessels which branch to form lymph capillaries in which a fluid known as lymph is transported.
LYMPH
This is a fluid similar to blood plasma except that it contains less protein.
It’s formed as a result of ultra- filtration of blood from the narrow blood capillaries.
As blood circulates it reaches the body tissues through the blood capillaries that form a network throughout the tissues. The pumping force from the heart together with the narrow lumens of the capillaries exert a high pressure that forces the fluid part of the blood to filter out of the capillary walls into the surrounding tissues.
This filtrate consists of all the constituents of blood plasma except the blood cells proteins. This is because the blood cells and proteins are too large to filter out of the capillary walls. The fluid is known as tissue fluid or intercellular fluid.
Once formed the tissue fluid bathes the cells of the tissues supplying them with O₂, food and other useful substances.
The cells absorb these substances and pass out CO₂ and other waste products in exchange.
Most of the tissue fluid then return to the blood system through the venule end of the blood at the capillary.
The excess tissue fluid drains into the lymph vessels where it is known as
Lymph vessels have a swelling along their length called lymph nodes. They contain lymphocytes which defend the body against infection by producing antibodies that kill bacteria.
Also in the lymph nodes there are phagocytes that engulf bacteria
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IMMUNE RESPONSES
The micro- organisms that cause diseases are called
The production of antibodies by special cells that inactivate foreign substances is called the immune response
The ability of the body to defend against infection by producing antibodies or cells that destroy pathogens is called immunity.
The immune system includes all the parts of the body that are involved in the recognition and destruction of foreign substances. Its made up of:
Bone marrow which produces white blood cells
White blood cells especially phagocytes and lymphocytes
Various tissues of the lymphatic system such as lymph nodes, tonsils, thymus and spleen which accommodate lymphocytes.
TYPES OF IMMUNITY
They are classified into 2 major groups’ i.e.
Innate(inborn)/inherited
Acquired
Innate immune responses
Refers to natural a natural body defense like the skin, sebum and mucus and sickle cell anaemia
This type of immunity is dependent on genetic constitution of an individual e.g. blacks are generally less susceptible to malaria than whites.
Acquired immunity
Natural acquired immunity
This occurs when the body naturally overcomes an nfection e.g.
(i) Natural active immunity – this is the type of resistance which is built- up in a person after suffering and then recovering from a disease. The person develops specific antibodies against future attack of these same pathogens. E.g. when a patient recovers from chicken pox, measles he develops immunity against these diseases. A patient can not suffer from re-infection
(ii) Natural passive immunity- it’s the resistance which is inherited i.e. passed on from parents to offspring via placenta or onto a new born baby through colostrum.
(b) Artificial acquired immunity
This is the immunity acquired when the antibodies are artificially introduced into the body or weakened pathogens are introduced in the body.
Its divided into:
(i) Artificial active immunity
its developed by introducing a weakened dose of a micro- organism into a healthy person to stimulate the immune system to produce antibodies and anti- toxins.
The process of weakening the disease causing micro- organism is known
The weakened micro–organisms such as bacteria and viruses are given in the form of a vaccine.
The immunity developed lasts for a certain period of time e.g. immunity against cholera last 6 months while that for small pox lasts several years.
- (ii) Artificial passive immunity
This is the immunity that comes from using antibodies produced in one organism to protect another organism from a specific disease.ie the immunity acquired when preformed antibodies are artificially introduced into the body of a patient. This antibodies are called antisera eg anti tetanus, antirabies and antivenom antisera
In this type of immunity antibodies are administered to the body when it cannot form its own antibodies this is common during a disease outbreak.
The immunization is provided in the form of anti- serum.
An anti- serum is a serum containing antibodies. It is administered in the case of tetanus, diphtheria, rabies and cholera. Immunity acquired this way lasts for a short time.
VACCINATION
ROLE
Protects individuals from infections e.g. small pox, tuberculosis (TB) e.t.c.
- Prevents the spread of diseases
A vaccine is a weakened or dead form of a disease causing micro- organisms vaccines are administered orally or by infection.
The immunization programme is carried out nation wide by the Kenya expanded programme of immunization (KEPI)

Name of disease	– Causative agent	– Age when administered	– Method of vaccination
– Tuberculosis (TB)	– Bacterium	– At birth	– injection
– Poliomyelitis (polio)	– Virus	– At birth, 6 10, 14 weeks	– Oral inoculation
– Diphtheria	– Bacterium	– 6,10, 14 weeks	– injection
– Whooping cough	– Bacterium	– 6 & 14 weeks	– injection
– Measles	– Virus	– 9 months	– injection

ALLERGIC REACTIONS

At times the body’s natural defense system may over- react against even harmless substances such as dust, pollen, certain food, insect stings and bites such substances are referred to as allergens and they provoke the cells to produce and release chemicals such as histamine which causes inflammation itchiness and pain.
Allergic reactions may cause skin rashes itching, sneezing, vomiting, coughing and swelling of the body.
A severe condition called anaphylaxis sometimes occurs in which the blood vessels get dilated and this lower the blood pressure to the extent of causing death. This is how the bee stings can cause death.
Doctor usually prescribe an anti- histamine treatment to counteract the effect of histamine.
ORGAN TRANSPLANT
Surgeons can replace damaged tissues of organs using similar organs from other persons or animals e.g. the pig in transplant operations.
It has also been possible to transplant kidneys, liver, spleen, reproduction organs or tissues transplanted onto larger parts of recipients are called
In some cases grafts may be reject by the receipt but in most cases grafts involving identical twins or those from the same individual are not rejected.
The grafts may be rejected because the body of the host recognizes the new tissues or organ as foreign to it.
Some transplant of the heart, kidney, cornea of the eye, lungs and bone marrow have been carried out by using drugs that suppress the immune response of the host.
A substance called interferon is also used to suppress rejection of grafts. In organ transplants sophisticated mechanic used to keep the organs to be transplanted and he patient alive.

GASEOUS EXCHANGE

This is the process by which respiratory gases (oxygen and carbon IV Oxide) and water vapor are passed across the respiratory system
Importance gaseous exchange
It promotes oxygen intake for respiration in living organism
It facilitates carbon IV oxide removal from the body. Accumulation of large amounts of carbon IV oxide in the tissues is toxic to cells.
Enables green plants to obtain carbon IV Oxide for photosynthesis
Excess water is expelled from the plants through transpiration. Higher animals expel it partly in its gaseous form i.e. water vapor in the exhaled air.
Gaseous exchange in plants
It involves two main respiratory gases i.e. CO₂ and O₂
CO₂ taken in is used for photosynthesis and O₂ produced as a by-product of photosynthesis is released into the atmosphere.
However some of O₂ produced is used in the same plant for respiration. But since the rate of photosynthesis proceeds faster than respiration at daytime, excess O₂ produced is removed.
At night respiration proceeds in the absence of photosynthesis in green plants hence plants take in O₂ for respiration but give out CO₂.
In flowering plants stomata in the woody stems and roots provide surfaces for gaseous exchange
- Structure and function of stomata
Stomata are tiny pores scattered on the surface of the leaf.
Stoma comprise of a stomatal pore and two specialized guard cells -curved sausage-shaped (bean shaped).
The guard cells are structurally adapted to their function by;
(i) Having chloroplasts
(ii) The uneven thickness of their cell walls i.e. the outer walls of the guard cells is thin and stretches easily. The inner wall is thicker and does not easily stretch.
Two guard cells join at the end walls to leave apore in the middle
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Mechanism of opening and closing stomata
In the presence of light, stomatal pores open. They close during darkness.
The opening and closing of the stomatal pores is due to a change in turgidity of guard cells surrounding the pores combined with the uneven thickness of their cell walls. When guard cells are turgid the stomatal pores opens, when they are flaccid the pore closes.
- Theories supporting opening and closing of stomata
Photosynthetic theory
In the presence of light the guard cells carry out photosynthesis using the chloroplast. The accumulation of sugar in their cytoplasm raises the osmotic potential and so water enters the guard cells by osmosis.
This leads to an increase in turgidity of guard cells which then curve more due to their uneven thickness and cause the stomatal pore to open.
During darkness the guard cells cease to photosynthesize. Their osmotic potential is lowered as their sugars are transported out and so water leaves the guard cells which then become flaccid causing pore to close.
Starch –sugar interconversion theory
It has been observed that in the presence of light the guard calls take up potassium ions (K+). This causes water to enter the guard cells making them more turgid and so cause the stoma to open.
In darkness K+ ions move out of the guard cells, and also water moves out of guard cells leaving them flaccid hence their pore closes.
The starch- sugar conversion theory is under the influence of PH through enzyme action e.g. during the day photosynthesis take place in guard cells using CO The PH in the guard cells tends to rise hence becomes less acidic. This less acidic condition favours the conversion of starch into glucose. The glucose being more osmotically active brings about an osmotic effect that result in water being drawn into the guard cells. Consequently the guars cells become turgid and bulge outwards making the stomata to open.
At night CO₂ is not used up because photosynthesis doesn’t take place hence PH decreases thus favouring the conversion of glucose into starch. Starch is osmotically inactive and therefore the guard cells do not gain water. Due to the resulting flaccid state of guard cells the stomata close.
The potassium ions theory/K+theory
Guard cells have chloroplasts hence in the presence of light ATP is produced
ATP drives a K+ pump on guard cell membrane which actively transports K+ from adjacent epidermal cells into guard cells
The accumulation of potassium ions raises osmotic pressure of guard cells
Guard cells absorb water from the adjacent epidermal cells becoming turgid
The inner walls are thicker than the outer walls so the outer walls stretch more than the inner walls causing the guard cells to stretch outwards and stomata open
In absence of light (at night) ATP rapidly decreases so that there is no energy to sustain the potassium ions pump
Potassium ions migrate by diffusion from the guard cells to the adjacent epidermal cells
This lowers the osmotic pressure of guard cells which lose water to the adjacent epidermal cells becoming flaccid
The thinner outer walls of the guard cells shrink and the curvature of the thicker inner walls reduces thus closing the stomata
Mechanism of gaseous exchange in plants
Terrestrial plants
They are those growing on land under ordinary soil conditions.
Gaseous exchange occurs in the following:
Spongy mesophyll
CO₂and O₂ diffuse in and out of the leaf through
the stomata. Most of the gaseous exchange occurs through the spongy mesophyll
The rapid gaseous exchange through the leaves is due to:
Numerous stomata that increase the volume of diffusing gases.
The large air spaces within the mesophyll which increase the surface area for gaseous exchange.
Epidermis is the outer layer of the pant. The epidermal layer is one cell thick. This reduces the distance over which the gases diffuse.
Oxygen which is at higher concentration in the atmosphere or soil diffuses into the plant tissues through the epidermis of the older stems and roots peel off thus gaseous exchange only occurs through the epidermis in young stems and roots.
Gaseous exchange through lenticels
Stems of woody terrestrial trees and shrubs have areas of loosely arranged cells with large areas of loosely arranged cells with large air spaces between them. These cells together form a structure called
Lenticels are formed when the epidermis is replaced by the bark. Lenticels appear scattered on the surface of the stem as well as raised openings.
They allow gaseous exchange of O₂ and CO₂ between the atmosphere and internal tissues of the stem.
Gaseous exchange through the roots
The roots of plants such as ficus are modified to carry out gaseous exchange.
The epidermal layer of the ficus roots is thin. O₂ diffuses from the atmosphere where it is at a lower concentration.
Gaseous exchange in aquatic plants
CO₂ and O₂ gases are dissolved in water in which aquatic plants grow. Aquatic plants may either be submerged, emergent or floating.
Submerged plants
They obtain carbon dioxide and oxygen from water by diffusion through the epidermis. They don’t have stomata.
Their leaves are generally thin with large air spaces and lack cuticle e.g. Elodea spp and ceratopyllum and spirogyra
They are able to carry out photosynthesis under low carbon dioxide concentration.
Emergent plants
These are plants whose roots are firmly anchored on substratum such as a rock. The rest of the plant emerges from the water e.g. potomageton and nymphae, reeds, sedges
These plants have most of their stomata on the upper surface of their leaves. In some cases e.g. Nymphae the stomata are on the upper side only.
The plant tissues are made up of cells with thin walls of large air spaces called aerenchyma tissue which is found in stems and leaves. A lot of air is stored in the aerenchyma tissue making the stem and leaves of the plant buoyant.
Gaseous exchange occurs through the stomata. The aerenchyma tissue provides a large surface area over which gaseous exchange takes place.
Floating plants
They are those plants whose roots hang freely in water while the leaves float.
They have most of their stomata on the upper surface of leaves, gaseous exchange occurs through the stomata e.g. water hyacinth, salvia molesta, water fern, water lily, water lettuce, duck weed.
- NB The roots of these plants have aerenchyma tissue that enables plants to float.
Gaseous exchange in plants found in marine water and estuaries
Some plants growing in waterlogged soil develop breathing roots. The roots emerge from the soil. The roots are called pneumatophores
Gaseous exchange occurs through the epidermis e.g. white mangrove (Avicania spp)
PRACTICAL ACTIVITIES
- Activity 1: To investigate presence of stomata on leaves
- Materials
Water in a beaker
Leaves of various plants
Bunsen burner
Procedure
Heat the water to boiling point. Turn off the burner and wait for the water to stop boiling.
Immerse a leaf into the hot water and notice the air bubbles emerging from the leaf.
Repeat the procedure using other leaves
Compare the average numbers of air bubbles from the upper and lower epidermis of different leaves.
- Leaf of bubbles
- Activity 2: To investigate the shape of guard cells and distribution of stomata on leaves
- Materials
Clear nail varnish
Microscope
Cover slip
Forceps
Microscope slide
Leaves of various plants (maize)
Beans, hibiscus, zebrine, water lily
Procedure
Apply a thin coat of clear nail varnish on the upper and lower epidermis of a leaf, let it dry
Peel off the varnish off the leaf using a pair of forceps.
Place the varnish with the imprint of stomata on a microscope slide.
Add a drop of water and gently lower a cover slip on the specimen
Observe the specimen under the microscope starting with the low power objective lens and then shift to the medium power objective lens.
Note the arrangement of the guard cells
Count the number of stomata in the field of view under the medium power objective lens.

– Plant	– Number of stomata		– Likely habitat
–	– Upper epidermis	– Lower epidermis	–
– Water lily	–	–	–
– Maize	–	–	–
– Zebrina	–	–	–
– Tradescantia	–	–	–
– Hibiscus	–	–	–

Activity 3: To investigate internal structures of stems or leaf stalks in aerial and aquatic plants
Materials
Water lily leaf stalks
Bougainvillea twig
Beaker containing water
Scalpel
- Procedure
Cut off the apex of the bougainvillea twig and pluck the leaves
Insert one end of the stem into the water and try to blow into or suck water from the beaker
Repeat the above procedure using water lily leaf stalk
Account for the difference
GASEOUS EXCHANGE IN ANIMAlS
Types of respiratory surfaces in animals

– Type Of Respiratory Surface	– Environment	– Example
– Cell membrane	– Water	– Amoeba
– Gill filaments	– Water	– Fish
– Tracheoles	– Air	– insects
– Alveoli/ lungs	– Air	– Mammals, birds, frogs, reptiles
– Skin	– Water, – air	– Frog – earthworm
– Buccal cavity	– air	– frog

Respiratory surface is the basic unit of any breathing system upon which gaseous exchange takes place by diffusion.
- Characteristics of respiratory surfaces
Covered with a thin epithelium for faster diffusion of gases across it.
It’s moist to dissolve gases as they diffuse across it.
It has a large surface area for maximum gaseous exchange
It should posses a rich capillary network to quickly transport gases to and from cells
Mechanism of gaseous exchange
Protozoa
These are single- celled organisms e.g. Amoeba, plasmodium
Trypanasoma, these are microscopic organisms
They are mainly found in water or in the body fluids of other organisms. The respiratory surface of protozoa is the cell membrane. Gaseous exchange occurs across the cell membrane directly by diffusion.
Due to the respiration the concentration of carbon dioxide inside the unicellular organisms is higher than that in the surrounding water therefore carbon dioxide diffuses out of organisms into the surrounding. The concentration of oxygen is higher in the surrounding water than inside the organism. Oxygen therefore diffuses from the surrounding water into the organism.
- X
- Gaseous exchange in insects
The respiratory system in insects is called the tracheal system; it consists of spiracles, trachea and Tracheoles.
- X
The spiracles are found only on the sides of the thorax and abdomen. There are no spiracles on the head.

The spiracles have a muscular valve which can be opened or closed to regulate the flow of air.
There are also hairs in the spiracle which prevent excessive loss of water by evaporation from the tissues.
The spiracles open into large tracheal tubes called tracheae (singular trachea). These tubes are strengthened with spiral bands of chitin to keep them open at all times.
There are several large air sacs which are connected to tracheal tubes, which act as reservoirs.
The tracheae are subdivided into microscopic tubes called Tracheoles. Tracheoles penetrate the body tissues and are in direct contact with all the living cells. They lack the spiral bands of chitin and their ends are filled with a fluid. Those ends act as respiratory surfaces between the cells and the Tracheoles.
Inspiration (breathing in) in a grasshopper
During inspiration (breathing in) air enters the body of the insect. Inspiration takes place when the internal muscles in the abdomen of the grasshopper relax. This makes the abdomen and the tracheal system to expand and increase in volume.
The pressure of the tracheal system decreases compared to that of the atmosphere. This causes air to be sucked into the tracheal tubes via the spiracles in the thorax which are open at the time. This air travels through to the Tracheoles.
Oxygen from the air dissolves in the fluid in the Tracheoles and diffuses directly into the cells. Carbon dioxide which is at a higher concentration in the cells than in the Tracheoles.
Expiration (breathing out) in a grasshopper
In order to expel the used air, internal muscles in the abdomen of the grasshopper contract and compress the abdomen. This causes a compression of the tracheal system. The reduce volume and increased pressure in the tracheal system forces air (CO₂) out of the system through the spiracles.
In the grasshopper the 4 anterior (front) spiracles close while 6 pairs of posterior spiracles open so that air flows from the front to the rear end and then out of the insect
Insects which live in water also carry out gaseous exchange in water. Insects such as the dragonfly or may fly larvae (nymphs) use tracheal gills that are seen as paired plates on either side of the abdomen e.g.
X
However most of the aquatic insects have an elaborate tracheal system and are not truly aquatic because they need to come to the surface to breathe e.g. mosquito larvae have the spiracles near the rectum carried on a tube called respiratory siphon.
The siphon is opened when the larva comes to the surface of the water to take in air and closed by valves when the larva submerges. Larvae come to the surface of water periodically to breathe and position themselves.
X
In the pupa stage, a pair of siphons open just behind the head, pierce through the water surface to allow for gaseous exchange.
X
Some adult insect like water beetles and water bugs use bubbles of air trapped by hairs. The air bubbles give these insects a silvery appearance.
Some insects use the respiratory device, plastron for gaseous exchange. A plastron is a pile of very fine non- washable hairs which cover the cuticle for some distance around the spiracle to hold off water and also maintain a film of air over the body surface.
Gaseous exchange in a fish (bony fish)
In a bony fish the respiratory structures are the gills.
The bony fish ahs 4 gills on each side of the body
The gills are located inside a cavity in the head region known as operculum cavity
Each side of the fish has an operculum cavity which has an opening to the outside of the fish called operculum opening.
The gills are protected by a thick gill cover or operculum on both sides of the body near the head.
X
The gills of a fish consist of a long curved bony structure called gill bar/ gill rakers. It provides attachment to the gill filaments and gill rakers.
Gill rakers- they are teeth like structures. They prevent food and other solid materials in water from reaching the delicate filaments.
Gill filaments – they are membranous protection on gill bar. The filaments are richly supplied with blood due to the presence of many capillaries.
Each gill filament sub divides into gill lamellae
The two rows of gill filaments provide a large surface area for gaseous exchange.
Mechanism of gaseous exchange in the gills of a bony fish.
Inspiration: flow of water into the mouth cavity
The process below brings water into the mouth cavity
The mouth opens
Muscular contractions in the mouth lower the floor of the mouth. This increases volume in the mouth cavity and decreases the pressure inside it.
The water outside is at a higher pressure and it rushes in through the open mouth.
Each operculum on the side of the fish bulges outwards by muscular action. This increases the volume in the operculum cavity and lowers the pressure there. Water from the mouth is sucked into the opercular cavity. Meanalbile body wall of he fish. This prevents water outside the fish from entering through the operculum cavity.
X
Expiration 🙁 flow of water over the gills during expiration)
The mouth closes
The floor of the mouth is raised. This (space) in the mouth cavity and increases the pressure.
The operculum presses inwards by muscular action decreasing.
The volume in the operculum cavity but increasing its pressure.
The free edge of the operculum moves away from the body wall of the fish to open the operculum cavity.
Water rushes from the operculum cavity and flows out of the fish via the operculum opening.
X
Exchange of gases between the water and gill filaments
Gaseous exchange in fish takes place on the gill filament as water passes over the gills.
Blood in the capillaries in the gill filaments has a lower concentration of o₂than the water entering the mouth. Therefore O₂diffuses from the water flowing over the gill filaments into the blood through the thin walls of the capillaries.
On the other hand blood in the capillary has no higher concentration of CO₂than the water entering the mouth cavity. Therefore co₂diffuses from the blood through the walls of the capillaries into the water flowing over the gill filaments.
In order to have maximum gaseous exchange between the blood in the gill filaments and the flowing water a steep concentration gradient must be maintained across the respiratory surfaces. This is achieved by the flow of water and blood in opposite directions. This is called counter current flow system.
As the movement of blood and water continues in opposite directions within the respiratory surface, o₂diffuses out of the water into the blood and co₂ from the blood leaves the respiratory surface into the water. By the time the blood leaves the respiratory surface, it has as much o₂as the water. This is so because water moves along, less and less o₂diffuses out of it as blood becomes more and more concentrated with o
X
NB reasons why fish cannot live in air
When out of water, the gill filaments stick together. This reduces the surface area compared to when its in the water hence gaseous exchange is more efficient in water….
In air, the moisture evaporates fast from the gill filaments. Since gaseous exchange requires moist surfaces, the diffusion of o2 and co2 cannot take place.
- Gaseous exchange in amphibians
Amphibians by nature they both live in land and in water. This double habitation calls for special adaptation in gaseous exchange. The respiratory structures are:
Buccal cavity
Lungs
Skin
Buccal (mouth) cavity
Air is taken or expelled from the mouth cavity by raising and lowering the floor of the mouth.
The lining of the mouth cavity is moist and O₂ from the air dissolves in it.
Under the lining of the mouth, there is a rich supply of blood capillaries and O₂ diffuses into the blood and is carried by haemoglobin to all parts of the body.
CO₂ from the tissues is brought by the blood to the mouth cavity where it diffuses out.
- Lungs
When the nostrils are closed the air can be forced into the lungs by the pumping action of the floor of the mouth.
The air reaches the alveoli sacs of the lungs that are well supplied with blood through a large network of blood capillaries.
The o₂in the air dissolve into the moist inner lining of the alveoli. It then diffuses into the blood across the wall of the capillaries, combines with haemoglobin in the RBC and is transported to all parts of the body.
The co₂from the tissues is carried by the blood and diffuses into the alveoli then pumped out by the pumping action of the mouth cavity.
- Skin
Frogs have thinner and moist skin than the toads. Beneath the skin there is a large network of blood capillaries. O₂ from the air and from the water diffuses through the skin into the bloodstream.
CO₂in the blood diffuses out of the blood capillaries through the moist skin into the surrounding water and air.
Toads do not use the skin surface for gaseous exchange normally except when they are hibernating.
Mechanism of gaseous exchange in mammals
Nose
The nose has two openings called nostrils which let in air into the air passages (nasal cavity).
Function of nasal cavity
Nasal cavity is lined with mucus secreting cells and hairs. The mucus and hair filter and trap dust and micro-organisms from the air. So particles are prevented from entering the lungs.
Air is warmed and moistened in the nasal cavity.
The lining of the nasal cavity also houses sense organs for smell which can detect and distinguish different types of smell.
Pharynx /Throat
It’s that part where the mouth cavity and the nasal cavity meet.
Larynx /Voice box
It’s a hollow box- like structure. It’s noticeable externally by the sight projection at the front of the throat (Adams apple).
The pharynx connects with the larynx through a slit-like opening called the glottis.
The glottis has a gap known as epiglottis which closes when a person is swallowing to prevent food from entering the trachea.
Choking and coughing are reflex actions which remove any foreign particles which accidentally enter the trachea.
X
Just below the glottis there are two membranous cords called vocal cords. The vibrations of these cords caused by the movement of air out of the lungs during exhalation, results in the production of sound.
Trachea /wind pipe
It’s made up of rings of cartilage to ensure it does not collapse during breathing. Also they enable the tubes to be stretched e.g. during coughing.
The incomplete rings (c-shaped) have gaps on the side facing the oesophagus which allow smooth swallowing.
The inner lining of the trachea has mucus to trap and filter micro-organisms and dust particles preventing them from entering the lungs.
X
The trachea is lined with cilia which move mucus upwards into the pharynx. From the pharynx the foreign matter is expelled from the air passages by spitting or swallowing.
X
NB: cigarette smoke is known to inhibit the action of cilia in the respiratory tract. The result is accumulation of dust particles, bacteria and mucus.
The bacteria may invade the cells of the mucous membrane causing diseases. As a result smokers get frequent respiratory tract infection.
Also smokers cough frequently as the body tries to get rid of the accumulated mucus and other material
Bronchi
The trachea branches in to the tubes called bronchus. They are similar to trachea except that they are narrower and have other materials.
Bronchioles
Each bronchus enters a lung and extensively branches into narrow tubes called bronchioles.
Bronchioles have no rings of cartilage and each bronchiole end up into a tiny sac called alveolus (plural alveoli) hence the spongy nature of lungs.
Alveoli
The walls of epithelium are composed of thin and flat epithelium
Structural adaptation of alveoli
They provide a very large surface area. There are approximately 300million alveoli in the lungs of a human adult
NB: total area is 90m²(nearly as large as a basketball pitch)
The internal surface is moist being lined up with mucus to help in the rapid diffusion of gases
Have a rich supply of blood capillaries which allows rapid gaseous exchange between the air in the alveoli and the blood in the adjacent capillaries.
The walls of the alveoli are made up of a layer of thin epithelial cells. This thin barrier permits rapid diffusion of gases.
Lungs
Each lung is enclosed by two membranes (double membrane) known as plural membrane.
One part of the membrane adheres tightly to the lungs and the other covers the inside of the thoracic cavity. The space between these membranes is known as pleural cavity.
It’s filled with pleural fluid which reduces friction and therefore makes the lungs move freely in the chest cavity during breathing.
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Thoracic cavity
The lungs and pleural membranes are contained in the thoracic cavity.
Thoracic cavity is surrounded by ribs, sternum and vertebrae which are all held together by muscles at the lower end known as the diaphragm.
Ribs are curved bones which project from the vertebral column dorsally and ventrally with sternum
However the lower most ribs are not attached to the sternum .the ribs protect the lungs and heart.
Between the ribs are internal and external tissue referred to as interrcostal muscles. These muscles work antagonistic to each other i.e. when one set of muscles contract each other set relaxes.
Thorax
It’s an airtight cavity enclosed by the ribs and the diaphragm.
Mechanism of breathing in humans
It involves two processes: inspiration / inhalation/ breathing in and expiration/exhalation/ breathing out. These two processes are brought about by movement of the ribs and diaphragm.
Inspiration/ inhalation/ breathing in
This process occurs when the thoracic cavity increases in volume and thereafter decreases in pressure.
During inspiration the external intercostal muscles contract while the internal intercostal muscles relax. This movement pulls the ribs upwards and outwards.
The diaphragm which is dome shaped flattens by the contraction of its muscles. The flattening of the diaphragm together with the outward movement of the ribs increases the volume of the thoracic cavity and decreases the pressure inside it.
Atmospheric pressure being higher than pressure inside the thoracic cavity forces air to rush into the lungs through the nose and trachea hence inflating the lungs.
Expiration/ exhalation
This process occurs when the volume of the thoracic cavity decreases and the pressure inside it increases.
This is brought about by the following:
The external intercostal muscles relax while the internal intercostal muscles contract bringing the ribs down to their original position. At the same time the muscles of the diaphragm relax and regain its original dome shape.
These movements decrease the volume of the thoracic cavity and increase the pressure inside it. Thus air is forced out of the lungs through the air passages into the atmosphere thus deflating the lungs.
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The alveoli and blood capillaries are made of very thin walls.
The walls of the alveolus are covered by a film of moisture which dissolves O₂ in the inhaled air.
Since O₂ concentration in the blood is lower than in the alveolus it diffuses through the epithelium, the capillary wall, the plasma and into the RBC where it combines with haemoglobin.
CO₂ in the capillaries surrounding the alveoli is at a higher concentration than inside the alveoli hence it diffuses into the alveoli.
Changes during inhalation and exhalation

X
Percentage composition of gases inhaled and exhaled air

– Gas	– % in inhaled air	– % in exhaled air
– Oxygen	– 20	– 16.9
– Co₂	– 0.03	– 4.0
– Nitrogen & other gases	– 79.97	– 79.97

Regulation of breathing
The average breathing rate in human beings is 16 to 18 times per minute.
Breathing movements normally take place unconsciously
In the brain there is a region called medulla oblongata which controls the breathing movements.
As CO₂ in the blood reaches this region it triggers this part of the brain to send impulses to the rib muscles and the diaphragm which in turn respond appropriately. This makes breathing to continue on and on.
During vigorous activity the concentration of CO₂ increases into the body tissues hence more CO₂ diffuses into the blood and reaches the medulla oblongata.
The high concentration of CO₂ in blood triggers the medulla oblongata to increase the rate of breathing.
Increased rate of breathing helps to increase the amount of O₂ in the blood thereby meeting the demands of the increased tissue respiration.

– inhalation	– Exhalation
– External intercostal muscles relax	– External intercostal muscles relax
–	–
– Internal intercostal muscles relax	– Internal intercostal muscles contract
– Ribcage is lifted up outwards	– Ribcage moves downwards and inwards
– Diaphragm muscles contract	– Diaphragm muscles relax
– Diaphragm flattens	– Diaphragm archs upwards and becomes dome – shaped
– Volume of thoracic cavity increases	– Volume of thoracic cavity decreases
– Air pressure decreases	– Air pressure increase
– Air moves into the lung through the nostrils, pharynx, glottis, the trachea and into the alveoli	– Air is forced out of the alveoli into the trachea, glottis, pharynx, nostrils and into the atmosphere
– Lungs inflate	– Lungs deflate

Factors affecting the rate of breathing
Exercise
During vigorous physical activity the rate of breathing increases so as to meet the increased demand of O₂
Faster breathing also eliminates the extra CO₂ produced by the increased respiration.
Age
Young people have a higher demand of O₂. They therefore have faster breathing rate. This is because young people are actively growing hence the faster rate of breathing is to supply tissues with O₂
Emotions
Generally the body emotions affect the production of hormone adrenaline which increases the general metabolism and hence increased rate of breathing e.g. fear anxiety and fright
Temperature
When the temperature is high there is a tendency in the rate of gashouse exchange to increase. However if temperature is too high the breathing rate will reduce
Health
During sickness the rate of breathing increases. The faster rate of breathing enables the liver to remove toxins in drugs those released by diseases causing micro-organism
The faster rate of breathing also enables the kidneys to excrete waste products of body metabolism through urine
Altitude
At high altitude the rate of breathing is faster than at low altitude. At high altitude O₂ concentration is low thus faster rate of breathing helps supply tissues with sufficient oxygen.
LUNG VOLUMES
Lungs of an adult can hold approximately 5500cm³ of air when completely filled. The volume is known as lung capacity
During normal breathing a small volume of air 500cm³is taken in and out of the lungs. This volume of air is referred as to as the tidal volume.
In addition to tidal volume a person can have a forced inhalation. this additional volume is called inspiratory reserve volume (200cm³)
Tidal volume +inspiratory reserve volume =inspiratory capacity
After normal exhalation it is possible to force out extra volume of air. This volume is called expiratory reserve volume (1300cm³)
It is possible to have deepest possible exhalation. Such volume of air which can only be forcibly pushed out of lungs is called vital capacity.
After the deepest possible exhalation some air normally remains in the lungs. His volume of air is called residual volume (1500cm³)
- Respiratory diseases
These are the diseases that affect the breathing structures and make gaseous exchange in animals difficult.
Asthma
This is a disease which mainly affects the air passages
- Causes
Allergy which can be due to pollen grains dust, spores, flowers, fur of animal e.t.c.
Constant lung infection caused by viruses and bacteria.
Emotional and mental stress such as anxiety, anger and fear
Mild or extreme cold weather
Certain hereditary diseases especially those affecting respiratory organs increase the chances of infection.
- Treatment and control
The spraying of a muscle- relaxant directly into the bronchial tubes.
Injection of drugs or oral application pills prescribed by a health physician.
Avoiding the causative agents.
Bronchitis
This is the inflammation of the bronchial tubes. There are two types of bronchitis .i.e.
Acute bronchitis
This is widespread illness in children and frail adults. It is caused by.
A complication of the common cold. It results into the chilling of the body giving way to bacterial infection.
A complication resulting from a previous disease attack e.g. measles, whooping cough, influenza and dengue fever.
Coughing
Head aches
Fever
Pain beneath the sternum
Breathing fast
- Control
Keep warm
Seek prompt treatment for infections
Chronic bronchitis
Result from heavy cigarette smoking and constant attacks by acute bronchitis.
- Symptoms
Production of thick sputum (phlegm) that is green or yellow in colour due to pus from respiratory surface.
Difficulties in breathing
- Control
Avoid smoking or smoking places and continuous exposure to dusty places.
Keep warm and live in well ventilated places.
Seek medical attention immediately the symptoms are observed.
- (iv) Whooping cough
Results from an acute infection of the respiratory tract by a bacterium called Bordetella pertussis.
The disease is endemic in Kenya .i.e. it’s regularly found in a specific group of people.
- Symptoms
Prolonged coughing and vomiting
Bleeding of the eyes (conjunctival haemorrhage)
Convulsions and coma
Ulcers and cardiac failure
Malnutrition especially protein and calory deficiency due to repeated vomiting and difficulty in eating.
Treatment
Patients with complications should be admitted to the hospital for special care and treatment.
Patients should be fed well during the time of sickness.
Immunization should be given soon after birth.
Pneumonia
It’s a disease caused by the bacterium Streptocococcus The chances of attack are increased by other chest infections e.g. bronchitis, whooping cough e.t.c.
- Symptoms
Shallow and difficult breathing.
Coughing with production of sputum
Fevers and chest pains
Lungs become inflamed and alveoli are filled with fluid.
- Control and treatment
Avoid overcrowded and poorly ventilated places
Use antibiotics eg penicillin and sulphonamides prescribed by a doctor.
- (v) Pulmonary Tuberculosis
It is caused by the bacterium Mycobacterium tuberculosis
This disease affects any part of the body.
Its an air borne disease and its spread through saliva droplets, sputum and infected milk.
Symptoms
Fevers and fatigue
Deep coughing sometimes with sputum containing blood
Loss of body weight.
Slight afternoon fever
The bacterium destroys lung tissues making it hard for the patient to breathe. It may eventually result into death.
Control
Suspected sufferers should have a medical check up and can be detected in its early stages by radiographical method
Avoid overcrowded and dirty places.
Vaccination of the population using BCG(Bacille calmette Guerin)
- (vi) Common cold
It’s a mild disease of the upper respiratory tract caused by a large variety of viruses.
Each year about ¾ of human population suffer one or more colds hence the name common cold.
Symptoms
Stuffy nose
Watery eyes
Sneezing, coughing and fever. In severe cases there may be a headache, backache and muscle ache.
Transmission
It’s through close contact with infected people especially through coughing and sneezing. It can also be transmitted through contaminated eating utensils.
Treatment
So far there is no cure for common cold.
The disease normally cares itself within a few days.
The patient should however lie in bed and have plenty of fluids.
Painkillers like aspirin may be taken to relieve various aches and to relieve fever.
If the disease does not cure within few days or if there are persistent aches, then it is advisable tosee the doctor.

RESPIRATION
It is the process by which food substances are chemically broken down in all living cells to release energy, CO₂, water or alcohol.
It takes place in all living cells and involves a series of complex enzyme catalyzed reactions.
NB– Respiration is a chemical process taking place inside tissue cells while gaseous exchange is a purely physical process which takes place at respiratory surfaces
Respiration is also called tissue respiration or internal respiration
SIGNIFICANCE OF RESPIRATION
It provides energy which is obtained due to the break down of food. Foods which can provide lots of energy are carbohydrates (starch and glucose) and fats.
The energy derived from these food substances is used for activities such as muscular contraction, conduction of nerve impulses, secretion of enzymes, growth etc.
Tissue respiration takes place mainly in cell organelles called
Structure of Mitochondria
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Mitochondria are small round or rod shaped cell organelles found in cells and provide sites for respiratory activity.
Living cells such as the kidney cells, the flight muscles of insects and birds, the sperm cells and muscle cells have high energy requirements and consequently posses’ large number of mitochondria.
Mitochondrion has two membranes, the outer and inner membrane that are separated by fluid filled spaces.
The inner membrane folds into projections inside the area for respiratory activities. Enzymes are bound to the cristae.
TYPES OF RESPIRATION
There are two types:
- -Aerobic respiration
- -Anaerobic respiration
Aerobic respiration
It is the process in which food substances such as glucose are broken down in the presence of oxygen in tissue cells to release energy water and carbon iv oxide.
The total energy released at the end of respiration (oxidation) is very high.
If all the energy were released at once in the form of heat it would burn the body cells. To protect the cells from burning, the heat energy is released in small quantities in stages.
This energy is used to bring about a chemical reaction in which a compound in the cell called adenosine diphosphate (ADP) combines with an inorganic phosphate molecule to form another compound called Adenosine Triphosphate (ATP)
ADP +(PO₄)^3- +Energy ATP (High Energymolecule)
- _(Adenosine _Triphosphate
_{(Adenosine Diphosphate)}
Aerobic respiration can be summarized by the following equation
C₆H₁₂O₆+6O₂6CO₂+6H₂O +Energy (ATP)
Molecules of ATP store the energy released in respiration in their bonds and avails it to cells readily when required.
Activity 1: To investigate what gas is given off when food is burnt
Materials
Food sample (starch powder)
Source of heat
Boiling tubes
One holed rubber stopper
Delivery tube
Calcium hydroxide
Solution (lime water)
Procedure
Place some food sample in a dry boiling tube and insert a one-holed rubber stopper into the mouth of the tube.
Hold the boiling tube containing the food sample horizontally.
Pour a little calcium hydroxide solution into another boiling tube and support it. Using a delivery tube connect the two boiling tube into the lime water is as illustrated in the figure below:-
Heat the boiling tube containing the food substance strongly.
Observe and record what happens to the food sample, lime water and the upper sides of the test tube with the food sample
Disconnect the apparatus and rub anhydrous cobalt II Chloride paper on the inner upper side of the boiling tube containing the food sample.
Record the colour change observed on the cobalt II Chloride
Discussion
When the starch was heated at the beginning, some drops of water were deposited on the walls of the test tube. This water comes from starch
When the food sample was heated strongly, it turned into a black substance. This substance is
When carbon was heated with the delivery tube dipped in the boiling tube containing lime water, the lime water turned to a white precipitate/ became cloudy/turbid. This is due to the presence of carbon iv oxide in the gas that was produced.
The results indicate that the food sample contains Carbon, Hydrogen and Oxygen (CHO)
Respiration takes place in two major phases i.e.
- First phase (Glycolysis)
The earliest stages of respiration takes place without using oxygen. These stages involve a series of chemical reactions which occur in the cytoplasm of the cell.
A compound with a 3-carbon molecule called pyruvic acid is formed from glucose.
After pyruvic acid has been formed and oxygen is not supplied to the cell, pyruvic acid is partially broken down to lactic acid in animals or ethyl alcohol (ethanol) and CO₂ in plants e.g.

NB- In Glycolysis one molecule of glucose yields 8 molecules of ATP
C₆ H₁₂O₆ ^enzyme CH₃COCOOH + 6O₂ 6CO₃ +6H₂O + ATP _{(Pyruvic acid) controlled reactions in cytoplasm}

Second phase (Kreb’s cycle) citric acid cycle
This phase takes place in the matrix of the mitochondria.
It involves a series of enzyme controlled reactions that require oxygen
The pyruvic acid formed in first phase is further oxidized by oxygen in a series of enzymatic reactions into CO2, Energy and water as end products i.e.
CH₃COCOOH + 6O₂ ^enzyme 6CO₂ + 6H₂O + ATP
_{(Pyruvic acid) Oxygen controlled reactions carbon iv oxide water30 molecules}

At the end of this cycle, 30 molecules of ATP are produces thus at the end of aerobic respiration, 38 ATP molecules are produced i.e. 8 ATPS – Glycolysis

For the above process to be maintained in the living cells, the following conditions are necessary: –
Cells must be provided with glucose/food
Oxygen must be taken in and react with glucose
There must be respiratory enzymes to catalyze the reaction
Favorable temperature should be maintained for efficient enzyme functioning
The end products of the reaction i.e. CO₂, water and energy must constantly be removed from the mitochondrion.
Expt: To show that heat is produced during respiration
Soak seeds for 24 hours and then divide them into 2 equal portions
Boil one portion of the seeds for 10 minutes, let them cool and wash them in 10% formalin
Place a thermometer in each flask such that the bulb is surrounded by seed
Hold each thermometer with cotton wool and record the initial temperature
Record the temperature every morning and evening for a week.
Germinating seeds break down stored carbohydrates in the process of respiration in order to get energy which they require for growth.
Some of the energy is released as heat; hence there will be temperature rise in the flask containing germinating seeds.
The boiled seeds did not produce any heat because they didn’t carry out any respiration.
Before the experiment the seeds were disinfected with 10% formalin to kill bacteria that would cause decay of seeds
The flasks were inverted in order to prevent loss of heat. Warm air rises up and if the flasks are not upside down, warm air in the flask would rise and lead to heat loss from the germinating seeds
Anaerobic respiration
It occurs in the absence of oxygen. In the plants glucose is oxidized in the absence of O₂ to give ethanol, CO₂and energy e.g.
C₆H₁₂O₆2C₂H₅OH +2CO+ energy(Glucose) (Ethanol) (Carbon IV oxide)
Anaerobic respiration in plants is also referred to as Fermentation Fermentation occurs when bacteria or fungi breakdown glucose to form alcohol, CO₂and energy.
In animals anaerobic respiration leads to the formation of lactic acids and energy e.g.

C₆H₁₂O₆2C₃H₆O₃+2O₂ + energy
(Glucose) (lactic acid)
NB the incomplete breakdown of glucose in anaerobic respiration result in the production of less energy than in case of aerobic respiration.
In the absence of O₂, most plant and animal tissues can respire anaerobically for a limited period.
It is essential that they get rid of the end products (lactic acid in animals and ethanol in plants) immediately. This is because these end products become toxic to the organism if left to accumulate within the cells.
Oxygen debt
This is the O₂ required to get rid of the lactic acid that accumulates in the body tissues when the supply of O₂ is less than the demand.
Under these conditions the animal’s tissues respire through anaerobic respiration and this causes lactic acid to accumulate in the tissues.
The lactic acid might cause fatigue and result in muscular cramps e.g. When a short distance runner or driver holds his/her breathe while running or diving. The O₂debt incurred here is “paid” back by the person breathing more quickly and more deeply in order to increase the supply of O₂ during the recovery period after the race.
During the process of paying back the O₂ debt, lactic acid is oxidized to CO₂, water and energy when O₂is available.
Anaerobic organisms
The organisms that carry out anaerobic respiration are called anaerobes. There are 2 types of anaerobes.
Obligate anaerobes.
Respire in the absence of O₂ and die in the presence of O₂. They lack the enzyme catalysis which breaks down hydrogen peroxide. (H₂O₂) e.g.
Escherichia coli
Bacillus subtilis
Clostridium botulinum
Clostridium tetan.
Facultative anaerobes
Respire in the presence or absence of O₂g. yeast, most bacteria, parasites or fungi.
- Comparison between aerobic and anaerobic

– Aerobic respiration	– Anaerobic respiration
– O₂ is necessary for the process to take place hence a complete oxidation of the substrate	– O₂ is not required hence substrate is not broken down completely
– More energy released ( 38 ATP molecules) from one glucose molecule	– Less energy released (2 ATP molecules) from one glucose molecule.
– Substrate is completely broken down to CO₂ and water	– Substrate is not completely broken down producing lactic acid and alcohol
– The end products are water and CO₂	– End products are alcohol in plants. Lactic acid in animals.
– Occurs in cytoplasm. Mitochondria	– Occurs in cytoplasm

Application of anaerobic respiration
Commercial production of alcohol

In the brewing industry, barley is fermented with yeast to produce beer.
In the wine industry, sugar from grapes is the source of germination. Different strains of yeast are used during the anaerobic process to produce wines of different flavours.
Distillation of some of the products of fermentation gives rise to stronger alcoholic drinks called spirits e.g. Distilling wine makes brandy.
- In the dairy industry
Milk contains lactic acid bacteria which anaerobiccaly breaks down milk sugar called lactose to form lactic acid which makes the milk sour. The dairy products include: cheese, butter, yoghurt and cream.
- Sewage treatment plants
Certain bacteria are introduced into the sewage to break it down by anaerobic respiration. These reduce the bulk of the sewage which on further treatment is purified and is safe for release into rivers or water bodies.
- In agriculture
The making of silage is an anaerobic fermentation process which is carried out on farms.
Silage is prepared by allowing bacteria to ferment vegetation giving it a good flavour and scent. The silage is used as animals feed.
- Production of biogas and gasohol
Manures from cows or other waste plant materials can be used as a substrate for fermentation, producing biogas which contains 70% methane. The gas can be used for cooking and lighting.
Cane sugar is used to produce gasohol in the presence of yeast. Gasohol can also be pressed from ethanol. Gasohol can be used on its own to run engines.
- In the home.
In bread production for domestic and commercial use. During fermentation using yeast, CO₂ produced in the dough mixtures causes the dough to rise as bubbles of the gas are produced. Thus the bread becomes porous.
- Commercial production of oxalic acid, citric acid and vinegar.
These are produced through anaerobic respiration. Those products are used in food processing.
- Fossil fuel formation.
As the organic remains take many years to decay. Fossil fuel such as natural oil, gas, coal and peat are formed.
- Expt: to investigate the gas produced during fermentation
Boil about 20 cm³ of glucose in a tube, cool to 40 O^c and add some yeast.
Pour onto the glucose and yeast suspension some kerosene oil. Leave for about one hour. ( several minutes)
Put some lime water (calcium hydroxide) in a test tube and connect this test tube to the boiling tube using the delivery tube. Rubber
stopper as shown in the diagram below
- Discussion
The water was first boiled to expel any dissolved oxygen to prevent any aerobic respiration form taking place.
Yeast being a living organism would be killed or its enzymes denatured with hot water. This is the reason for first cooling the water before the yeast is added to it.
As the yeast respires in the absence of O₂, it uses up some of the sugar and produces a gas and ethanol.
The gas causes a lot of frothing in the conical flask and some of it goes up the delivery tube and makes the lime water appears turbid (white precipitate is formed). This is carbon iv oxide
- Respiratory substrates
They are substances that are oxidized to release energy. They are:
- Carbohydrates
They are the common oxidized substrates. Excess carbohydrates are stored in plants in the form of starch and in animals in the form of glycogen.
Carbohydrates are broken down into simple forms of glucose and fructose before being oxidized
19% carbohydrates release 17kj when oxidized.
- Fats
They are oxidized when carbohydrates resources are depleted. Fats are broken down by enzymes called lipases into glycerol and fatty acids before being oxidized.
Ig of fat yields 38kj when oxidized. Most food stored in plants and animals is in the form of fats and lipids.
Fats are not the main substrates of respiration because:
They are not very soluble and therefore not easily transported to the sites of respiration.
It will also require more O₂to oxidize one gram of fat than one gram of glucose
- Proteins
They are oxidized when both carbohydrates and fat reserves are exhausted especially during prolonged starvation. Proteins are hydrolyzed by enzyme protease into amino acids. The amino acids are denominated to urea and a carboxyl group.
One gram of protein produces 22kj when oxidized.
- Respiratory quotient (R.Q)
It’s a ratio showing the relationship between the amount of carbon IV oxide produced against the amount of oxygen used in respiration i.e.
Q= volume of CO₂ produced
Volume of O₂consumed
During aerobic respiration of carbohydrates the RQ=I i.e.

Q of fat= 0.7
Q of proteins= 0.9
Q of Carbohydratest= 1.0
- Factors affecting Respiratory quotient (R.Q)
Type of respiration– Aerobic respiration gives an R.Q of 1.0 or less while anaerobic respiration gives an R.Q greater than 1.0.
Type of substrate– Oxidation of carbohydrates gives an R, Q of 1.0, proteins 0.9, lipids 0.7.
Metabolic processes
Synthesis of fats, carbohydrates and organic acids use a lot of oxygen to produce low volumes of CO2.
During seed dormancy the R.Q is more than 1.0.During germination the value of R.Q reduces to 0.7.
Hibernation-It’s the state of inactivity during winter when animals burrow underground to escape the low temperatures.
During this time the animals are less active. The major substrate respired are fats.
Aestivation- It’s the state of inactivity during which some animals burrow to escape hot weather.
(g) Age–R.Q increases when one becomes
Temperature of the surrounding-R.Q will be lower in low temperatures and higher during higher temperatures.
Health status of organism-During sickness R.Q increases due to the effect of the infection such as the presence of toxins.
Factors affecting the rate of respiration
Age
Young people are more active than old people. The rate of respiration or metabolism is faster in young people than in old people.
Young plants have a faster rate of respiration than old plants.
State of health
The rate of metabolism increases during illness so as to remove toxic substances released by pathogens.
Size
Small animals have a large surface area to volume ratio compared to bigger animals. Small animals lose heat at a faster rate thus they respire at a faster rate to replace the lost heat.
Temperature
Respiration is an enzyme controlled reaction. At low temperatures, the rate of respiration is low. An increase in temperature increases the rate of respiration.
Above the optimal temperature, enzymes become denatured and the reaction stops.
Activity
An organism at rest mainly requires energy for sustaining of life processes e.g. breathing and circulation of blood. This energy is referred to as Basal Metabolic Rate (BMR)
BMR increases in active organisms. In humans, males have a higher BMR than females.
Hormones
Certain hormones in the body such as adrenaline and Thyroxine increase respiratory activities.
Substrate concentration
The primary respiratory substrate in the tissues is sugar. When sugar concentration increases, the rate of respiration also increases. The reverse is also true.
Oxygen concentration
Respiration is affected by the amount of oxygen available in the tissues. When the amount of oxygen is low the rate of respiration slows down. When the amount of oxygen is high the rate of respiration increases.
In diving animals the oxygen concentration in their environment is low hence as soon as they dive, cardiac frequency drastically decreases (bradycardia) and the arterioles of all the vital body organs constrict so that oxygen can be delivered to the vital organs that cannot endure oxygen deprivation e.g. the brain and the heart.
As a result of this, less oxygen reaches other body tissues and organs hence their respiration rate reduces.
Expt; To demonstrate that respiration takes place in plants
Procedure
Set up the apparatus as shown below

The delivery tubes should be arranged so that one arm forms the inlet and the other outlet.
Use petroleum jelly or wax to seal off any gaps in the tubes to stop air entering into the apparatus.
The potted plant is placed under the bell jar and the bell jar is covered with the impermeable materials to exclude any carbon iv oxide from the soil.
When the filter pump is switched on, air flows through the whole set up.
NB Soda lime is used to remove any carbon iv oxide in the air entering conical flask A hence lime water in A remains clear.
After 30 minutes, a white precipitate forms in the lime water in conical flask B. this shows that the potted plant is respiring producing carbon iv oxide which reacts with lime water to produce the white precipitate.
The black cotton prevents the green plant from carrying out photosynthesis.
- Expt; To show aerobic respiration in animals
Put some 2cm³ of bicarbonate indicator solution in 2 conical flasks and label them A and B.
Put 2 grasshoppers on a muslin cloth or wire net and place them in conical flask A. Cover the conical flask immediately with the rubber bung.

In conical flask B place only a muslin cloth or wire net. Cover the conical flask immediately with the rubber bung and leave the set-up for 30 minutes.

After 30 minutes the bicarbonate indicator solution in A turns from red to yellow. This shows that grasshoppers are releasing carbon iv oxide through respiration.
The bicarbonate indicator solution in B remains red. Set-up B acts as a control experimen

EXCRETION AND HOMEOSTASIS
Excretion- It’s the process by which living organisms get rid of metabolic waste products. In plants some waste products are removed while others are reused or stored as harmless substances.
In animals, waste products resulting from the metabolic processes are generally removed from the body.
- Homeostasis– It’s the control and maintenance of a constant internal environment around the cells in body despite the fluctuations in the external environment.
Egestion- It’s the removal of indigestible and undigested food substances from the body.
Secretion- It’s the release of substances from the cells into the body fluids such as blood and the tissue fluid or to the outside of the body. Examples of secretions; hormones, enzymes, mucus, sebum etc
Ecretion in Plants
Metabolic processes in plants occur at a slower rate than in animals. Some of the waste products prodused in one procces are used in another process eg CO₂ released during respiration is utilised in photosynthesis.
Most of the substances that are broken down in plants are carbohydtrates in nature. Waste products from carbohydrates are not harmful to the plants.
Some of the waste products eg resins, gums are stored in dead tissues of plants such as xylem.
Methods of Excretion
Diffusion– Eliminate waste products that are in gaseous form eg CO₂, Oxygen and water vapour.
(ii) Transpiration– water vapour.
Guttation– water and dissolved mineral salts.
Exudation– It’s the release of a fluid from a plant at a slow rate eg gums, latex, mucilage, rubber, resins and Caicium pectate and oxalates.
Deposition– Resins, tannins, caffeine, nicotine, quinine etc are deposited in the Xylem, bark, seeds, fruits, flowers and leaves of plants.
Storage of excretory substances in plant parts
Some plant waste substances that may be toxic to the plant are converted to less harmful substances which are then stored in different parts of the plant such as petals, leaves, fruits and seeds. Some of these plant parts are eventually shed by the plant.
Some plant waste substances are stored in the vacuoles of plant cells. Some are stored in in dead permanent tissues such as the wood or barks or leaves which are shed seasonally. In this state they have no harmful effects on the activities of living tissues.
Most perennial plants store excretory materials in dead tissues.
Aquatic plants lose most of their waste substances by diffusion directly into the surrounding water.
Useful Excretory Products
Anthocyanin
Gives colour to petals and leaves in plants. The dominant colours are red, purple and blue.
These colours are of great aesthetic value and are extracted to make dyes.
Tannins
They are deposited in the dead tissues of trees such as wood and bark. They are common in conifers and mangroves.
Tannins are used in the treatment of leather and manufacture of ink.
They are also used in cosmetics eg henna which is a plant extract used to colour the nails, feet and hair.
Latex
It’s a milky substance that is produced by some plants. Latex from the rubber tree is used to make rubber.
Gums
They are produced by different plants such as arabic, ghath and carob. These gums are edible and are used to thicken food and creams.
Sapodilla gum is used in the manufacture of chewing gum.
Alkaloids
They are produced in many forms and are stored in different organs of plants eg
- Quinine
It’s obtained from the bark of cinchona tree
It’s used in the treatment of malaria.
Also its added in drinks as a stimulant.
- Cannabis
It’s stored in flowers, fruits and also leaves of Cannabis sativa.
It’s normally extracted and used in the manufacture of drugs such as painkillers.
Cannabis sativa induces hallucinations ie seeing or hearing unreal things.
- Cocaine
It’s obtained from the leaves of a south American plant called coca plant.
It’s used as a local anaesthetic . when taken in large quatities it causes great physical or mental effects such as convulsions or hallucinations.
It’s addictive when taken in large amounts and can lead to ailments of the heart.
- Nicotine
Occurs in the leaves of the tobacco plant. Its used to manufacture insecticides and narcotic drugs.
- NB Narcotic drugs are substances that cause one to sleep or become very relaxed and feel no pain.
The tar from the tobacco is poisonous and cause lung cancer in human beings.
- Caffeine
It’s stored in coffee beans and tea leaves.
It’s a mild stimulant which is refreshing. It increases mental activity and reduces fatigue.
Excessive intake of caffeine can cause sleeplessness and so may cause mental illness.
It can cause changes in cells of the foetus.
It increases the activity of adrenaline.
- Morphine
It’s extracted from the poppy plant and is used to make narcotic drugs.
It’s also a painkiller and muscle relaxant.
- Papain
Its extracted from pawpaw trees and used as a meat tenderiser.
- Colchicine
Its obtained from the roots of crocus plant. Its used to bring about mutation in genetic materials thus useful in plant breeding.
Its carcinogenic ie it can cause cancer.
- Khat
Also reffered to as miraa (Khat edulis). Its extracted by chewing the leaves and the twigs of the tree.
Its used as a stimulant.
- Pyrithrin
Its extracted from pyrethrum flowers. Its used to make insecticides.
- Alkaloids
Produced in irish potatoes when exposed to sunlight turn the tubers green. They are bitter and can be poisonous if ingested in large quantities. Naturally, the alkaloids protect tubers exposed on the groundfrom being fed on.
Excretion in animals
Unlike plants, animals have more problems of getting rid of waste substances for several reasons;
-Animals are more active than the plants threfore their metabolic processes takesplace at a higher rate producing large quantities of waste products.
-Animals do not put most of their waste products to other uses the way the plants do.
-Animals take in certain substances in their food in excess of their needs. These extra substances eg proteins are broken down with the formation of toxic substances such as ammonia.
Excretion in unicellular organisms
Most simple organisms such as protozoa live in aquatic environments. Their waste products include CO₂and nitrogenous wastes.
Protozoa such as amoeba and paramecium depend on diffusion as a means of excretion.
Their bodies have high surface Area to volume ratio that provide a large surface area for gaseous exchange and excretion to take place by simple diffusion. These waste products diffuse from the cytoplasm where they are at a higher concentration across the cell membrane into the surrounding water where their concentration is low.
Another method of excretion is by use of contractile vacuole.
Amoeba and paramecium live in an aquatic environment that is hypotonic to their body fluid hence there is excess inflow of water by osmosis. Excess water and dissolved chemicals accumlate in the contractile vacuole.
On reaching the maximum size, contractile vacuole moves to the cell surface and bursts releasing the contents to the surrounding.
Soon afterwards other contractile vacuoles form in the cytoplasm, accumlate more waste contents and the process continues eg

Excretion in animals
Excretion in animals is carried out by elaborate systems made up of specialized tissues and organs. This is because their bodies are complex and have greater number of cells.
The excretory tissues and organs include;
-Flame cells-Platyhelminthes
-Nephridia-Annelida
-Malphigian tubules-Insects
- -Gills, lungs, liver and kidney- Vertebrates
These organs are specialized to function in different environments such as aquatic (marine and fresh) and terrestrial.
Excretion in mammals
The main excretory organs in mammals are;
(a) Skin
This is the largest body organ as it covers the whole body surface and even continues into many body openings like nostrils, mouth and ears.
Functions
- Protection of the uderlying tissues from entry of micro-organisms, physical damage and ultra-vilet rays from the sun.
-Since the outermost layer is waterproof, the skin preventsthe body from drying up.
- Regulation of body temperature.
- Excretion of salts, excess water and traces of urea.
- Reception of stimuli of heat, cold, pain, touch and pressure.
- Synthesis of vitamin D.
- Storage of fat.
The skin consists of two main layers; outer epidermis and inner dermis.
The Epidermis
It’s the upper layer of the skin and its made up of 3 layers of cells i.e.
The cornified layer
It’s the outermost layer and it’s made up of flattened dead cells that become filled with a tough flexible substance called keratin. This layer provides protection against mechanical damage and invasion of bacteria.
It also reduces the loss of water by evaporation. Cells of this layer are continuously lost through friction and replaced from beneath by granular layer.
Its thickness varies in the body e.g. its thickest in areas of high friction like palms of hands and soles of feet, but thinnest on lips and eyeballs.
Granular layer
It’s the middle layer of epidermis and consists of living cells that have granules. It gives rise to the cornified layer.
Malphigian layer
It’s the innermost layer of cells and is made up of actively dividing cells that give rise to new epidermis.
The cells have pigment granules called melanin that gives colour to the skin. The more it is, the darker the skin colour. It also gives protection against harmful ultra-violet rays from the sun.
Dermis
This is thicker than the epidermis and is located below it. It contains the following;
Sweat glands
These are tiny coiled tubes which secrete and release sweat through the pores on the surface of the skin.
Sweat consists of water and mineral salts such as sodium chloride and traces of urea and lactic acid. The liquid that forms sweat is absorbed by the sweat glands from the blood capillaries supplied to each gland.
It reaches the surface of the skin through the pore and water in it evaporates into the air. This cools the body.
Sweat glands function when the body temperature rises above the normal by between 0.2 ºC-0.5 °C.
Blood vessels and Lymphatic vessels
Blood vessels contain blood that supplies nutrients and O₂ to the skin tissues and remove waste products and CO
Blood also helps in temperature regulation.
Lymphatic vessels drain excess tissue fluid.
Nerve endings
The nerve cells that detect changes from the external environment thus creating awareness within the body of the changes in temperature (cold and heat), pressure and touch.
Hair
Originates from a deep infolding of the epidermis that forms the hair follicle. The hair follicle is lined with granular and malphigian layers of epidermis.
At the base of the hair is a dermal or hair papilla from which the hair root develops.
The hair follicle is supplied with sensory nerve to increase sensitivity of the skin and blood vessels, for the supply of nutrients and removal of waste products.
Each hair is made up of a base called hair root and hair shaft which protrudes outwards.
‘Growth of hair’ is due to continuous addition of new dead cells at the base of the hair.
Erector pili muscles are attached to the follicle at one end and on the other end to the epidermis. These muscles undergo contraction and relaxation to alter the angle between the hair shaft and the skin and therefore vary the amount of air trapped between the hair and the skin.
NB Certain hairs have become specially specialized adapted e.g.
-Eye lashes and the hairs inside the human nose which help to keep out dust particles.
-Cats, dogs, cats etc have long whiskers which help with the sense of touch.
-The long stiff spines of porcupines, the horns of rhinoceros and the pangolin’s scales are examples of modified hairs.
Sebaceous glands
They are attached to the follicle and the gland opens into the follicle. They secrete sebum which keeps the hair and epidermis flexible and waterproof (water repelling property).
Also sebum contains antiseptic substances for protection against bacteria.
Also keeps epidermis supple and reduces the tendency for it to become dry due to evaporation.
Subcutaneous layer
This is a layer of fat beneath the dermis and binds the skin to the muscles and other organs deep in the body.
It acts as a storage region for fats and an insulation layer against heat loss.
NB Skin lightening creams contain among other chemicals, mercury. They destroy;
Malphigian layer- this leads to the destruction of melanin producing cells making skin appear lighter, but this exposes the skin to harmful U.V rays which cause cancer.
Cornified layer- its destruction gives the impression of a softer skin but this exposes the skin to mechanical injury and microbial attack.

Lungs
In mammals , birds, reptiles and amphibians, CO2 formed during tissue respiration is removed from the body by the lungs.
The Kidney
The functions of kidney are;
-Excretion
-Osmoregulation
-Ionic balance
-Regulation of PH
The kidney is an organ found in vertebrates and each organism has two kidneys.
Kidneys are bean-shaped and are red in colour. They lie near the back of the abdominal cavity about the level of the waistline.
Each kidney weighs approximately 142.5g, ie about the size of a clenched fist. The right kidney is generally slightly lower than the left. The kidney is surrounded by a layer of fat which helps to cushion it from mechanical or physical injury.
The kidney is supplied with blood from the general circulatory system via the renal artery which branches off the aorta.
Blood from the kidneys goes back to the general circulation through the renal vein which joins the vena cava.
A tube called the ureter connects each kidney to the bladder located in the lower abdomen. From the bladder another tube called the urethra opens to the exterior of the organism.
In males, the urethra is long and is joined to the reproductive system unlike in females hence refered to as urinogenital system.
Two rings of sphincter muscles encircle the urethra and they control the emtying of the bladder. The two kidneys, two ureters, the bladder and the urethra make up the urinary system.
Structure ot the kidney
The kidney has two main functions;
Excretion-They remove excess salts, water and nitrogenous wastes from the blood.
Osmoregulation-They regulate the concentration of water and salts found in the body fluids.
A longitudinal section of mammalian kidney shows 3 distinct regions i.e.
-Cortex– Its dark red in colour and found to the outside.
-Medulla– Its red in colour and lies to the center of the kidney and extends to form conical structures called pyramids. These pyramids open into swollen cavity called pelvis.
-Pelvis– Its white in colour and narrows to form ureter.

Nephron
It’s the basic functional unit of the kidney. Each kidney has about 1.25 million nephrons.
Each nephron is made up of two main parts namely;
- Renal tubule
-Glomerulus
Renal tubule
It has 5 main parts i.e.
Bowmans capsule-It’s a thin double-walled and cup-shaped structure.

Proximal convoluted tubule-Its coiled and extends into a U-shaped part.
Loop of henle-It’s the U-shaped part.
Distal convoluted tubule– Its coiled and extends into a collecting tubule.
Collecting tubule– Drains into a collecting duct into which Collecting tubules from several nephrons drain thus forming an outlet of urine through a pyramid into the pelvis.
Glomerulus
It’s a fine network of blood capillaries enclosed by the Bowman’s capsule. Glomerulus is formed from the;
-Afferent arteriole– It’s a branch from renal artery.
-Efferent arteriole– It collects blood from the glomerulus and extends to the renal tubule where it divides into capillaries that ramify the tubule.
It channels blood away from the glomerulus.
Functions of the glomerulus
Excretion in the nephron is carried out in two stages i.e.
-Ultra-filtration
lumen-Reabsorption
Ultra-filtration
This is the process by which the useful substances enter the nephron.
Reabsorption
This is the process by which the useful substances are taken back into the blood so that they are not lost.
Kidneys receive blood from the renal artery and branch off the dorsal aorta. This blood is rich in nitrogenous wastes e.g urea. It also contains dissolved food substances, plasma, proteins, mineral ions, hormones and oxygen.
The Afferent arteriole entering the Glomerulus has a wider lumen than the Efferent arteriole leaving it.
The narrowness of the Efferent arteriole produces both resistance to blood flow and back pressure which create extremely high pressure in the glomerulus.
Also the renal artery branches directly from the dorsal aorta whose blood flow is at a high pressure.
This pressure forces water, mineral ions and small molecules like glucose, amino acids and urea out of the the glomerulus. These pass through the tiny pores in the walls of the glomerular capillaries into the Bowman’s capsule. This process is known as ultra-filtration and the liquid collected in Bowman’s capsule is called glomerular filtrate.
The larger molecules in the blood eg blood proteins, white blood cells, red blood cells and platelets cannot pass through the capillary walls of the glomerulus hence the blood which remains is rich in plasma proteins and little water.
The glomerular filtrate then flows from the capsular space into the Proximal convoluted tubule of the nephron. As the glomerular filtrate flows along, most of the filtered substances which are useful to the body are selectively reabsorbed back into the blood.
In the Proximal convoluted tubule, all glucose , amino acids, some water (80%) and mineral salts are actively reabsorbed against the concentration gradient, a process that requires energy (active transport).
NB The substances reabsorbed are those which are useful to the body hence refered to as selective Reabsorption
Adaptations of Proximal convoluted tubule for efficient Reabsorption
-Cells lining the tubules have numerous mitochondria which provides the necessary energy in the form of ATP.
- -Cells of the tubules have micro-villiwhich increases the surface area.
-The tube is long and highly coiled to provide a large surface area for Reabsorption.
- -The coiling of the tubule reduces the speed of flow of the filtrate thereby giving more time for efficient Reabsorption.
-The tubule is well supplied with blood capillaries.
The glomerular filtrate flows into the loop of henle, which has a unique U –shape feature with a descending and an ascending limb.Salts especially sodium chloride are reabsorbed into the blood.
The U-shape loop is generally longer and has a counter-current flow established between the flow of the filtrate and the blood supply in vessels.
Active transport is involved in the reabsorption of sodium salts.To regulate the intake of sodium salt, a hormone called aldosterone is secreted by the adrenal glands.
Low content of salt in the blood stimulates adrenal glands to secrete more aldosterone hormone and therefore more salt is reabsorbed from the filtrate and vice versa.
The glomerular filtrate flows into the distal convoluted tubule where controlled amount of water is reasorbed into the blood by osmosis .This process is enhanced in 2 ways:
(i) Due to the active intake of sodium salt into the blood at the loop of henle which increases the osmotic potential of the blood.
(ii) A hormone known as antidiuretic hormone (ADH)/vasopressin. This hormone is secreted by the pituitary gland.
ADH increases the permeability of the tubule and blood capillaries to water. When there is excess water in the body eg as a result of excessive intake of fluids, osmotic potential of the blood falls causing the pituitary gland to reduce its secretion of ADH into the blood. Water reabsorption in the tubule is thereby reduced and results in the production of large amounts of dilute urine.
If the body loses a lot of water through sweating, the blood pressure is raised hence the pituitary gland release more ADH which results in increased water reabsorption from the tubule into the blood. This results in the production of little amounts of concentrated urine.
NB Adaptations of distal convoluted tubule are similar to those of proximal convoluted tubule.
The glomerular filtrate flows into the collecting tubule from where more water is reabsorbed. The glomerular filtrate now becomes urine and trickles down into the collecting duct where it joins urine from the collecting tubules of other nephrons.
The urine then flows into the pelvis via the pyramid and is finally emptied into the urinary bladder through the ureter.
About 1-2 litres of urine trickles into the urinary bladder in a day. In the urinary bladder, about 250ml of urine will initiate the urge to urinate. The sphincter muscles relax and the urine is passed out.
The resultant urine composition of a healthy person maybe as follows;
Water——————-95%
Urea———————2%
Uric acid—————-0.03%
Creatinine————–0.1%
Salts (Na+, K+, cl-)—1.4%
Ammonia—————0.04%
Proteins—————–0.0%
Glucose—————–0.0%
The quatinty and concentration of urine in animals is affected by terrestrial, aquatic, desert conditions, the physiological and structural adaptations of the animals eg in a desert rat, water reabsorption is maximised by the development of a long loop of henle.
Kidney Diseases and Disorders
- Nephritis
This is a condition which affects the glomerulus. It is due to the poisons released during infection by certain bacteria called streptococci in various parts of the body.
It can also be caused by small pox, measles, typhoid and sore throat.
The glomeruli become so swollen that they are unable to carry out fitration of the blood.
Symptoms
Headaches, fever, vomiting and weakness.
Swelling of the body called oedema.
Urine is highly coloured and cloudy due to the presence of albumen.
Treatment and Control
Dietary restrictions especially salts and proteins.
Administration of drugs.
- Kidney Stones
There are various causes;
Improper balance of diet, lacking certain vitamin and inadequate intake of water.
Chemical salts in urine eg oxalates, phosphates, urates and uric acid. These may undergo precipitation and form hard deposits or stones in pelvis, ureter hence causing blockage of urine.
Symptoms
Increased frequency in passing out urine.
Pain and soreness in the upper back side.
Pain, chills and fever.
Difficulty in passing out urine.
Treatment and Control
Consult a physician.
Take balanced diet with plenty of water.
Take hot baths and massage the back with hot soft material.
Dialysis or artificial washing out of wastes.
Use of laser beams to disintegrate the stones.
In severe cases, surgical treatment which may involve kidney transplant.
- Albuminuria (Protein in urine)
This disorder is also called proteinuria. It’s a condition in which protein, mainly albumen, is found in urine.
This is due to increased permeability of glomerular capillaries which may be caused by bacterial infections.
Symptoms
Fluid accumulation in tissues (oedema). Its fatal if not treated.
- Kidney failure/Renal failure
The failure of the kidneys to function may occur as a result of a drop in blood pressure due to heart failure, haemorrhage or shock. Haemorrhage means excessive bleeding.
Due to the drop in blood pressure, the filtration rate in each glomerulus is reduced. In some cases the blood pressure is so low that no urine is formed and the kidneys stop working.
If one kidney fails, a person can still lead a normal life using the other kidney. However, if both kidneys malfunction, the individual will still survive if treated promptly. Such treatment can be administered in two forms i.e.
-Kidney dialysis
-Kidney transplant
- Pyelonephritis
This is a bacterial infection of the renal pelvis. The infection may spread to the urethra and bladder.
The kidney becomes swollen and filled with pus. It can be treated with antibiotics.
- Uremia (Uraemia)
It’s a condition in which there is excess urea in the blood.
It occurs when the kidneys are not working properly and the poisonous nitrogen-containing waste products accumulate in blood.
Symptoms
-Convulsions
-Coma
-Vomiting
-Diarrhoea
-Lethargy
-Mental disorientation and confusion.
-Difficulty in breathing
- Gout
This is a disorder caused by the absorption of uric acid salts into the blood.
In high concentrations, uric acid salts form crystals in joints in the toes, fingers and even the kidney itself. Its very painful for the patient to make any movements including walking.
Gout is caused by a diet that has too much organ meat eg kidneys or red meat.
Treatment and Control
Patients are put on medications that break up uric acid into harmless compounds.
They are advised to have a diet low in protein.
Avoid red meat.
Drink plenty of water.

The liver
It’s the 2^nd largest organ after the skin (Adult 2-3% of body weight-1.5kg) and it’s a special organ of excretion because many excretory products are produced by it.
It lies immediately beneath the diaphragm and is made up of several lobes.
It receives blood from the blood vessels i.e hepatic portal vein and hepatic artery. Blood flows out of the liver through the hepatic vein.
The liver consists of a large number of lobules. Each lobule is made up of many liver cells. The blood supply to each lobule is from two sources e. hepatic portal vein and hepatic artery. These vessels branch between the liver lobules.
Between the plates of liver cells are channels called canaliculi which receive blood. The bile moves outwards to the periphery of the lobules where it collects into bile salts.
Functions of the liver
- Deamination
It’s the removal of the amino group from an amino acid. Proteins which are taken in by the body are digested producing amino acids. Excess amino acids are are not stored in the body but are deaminated.
The amino group deaminated enters the ornithine cycle where it combines with CO₂ to form urea, which is excreted from the body through the kidney e.g

2NH₃ +CO_{2 ornithine cycle} → CO(NH₂)_{2 (Urea)} +H₂O

NB Reptiles and birds need to conserve their water. Their ammonia is converted to uric acid that does not need water to eliminate. They are refered to as uricotellic organisns and they produce white droppings instead of urine.
Animals that excrete mainly ammonia live in aquatic environments. CO₂and the toxic ammonia can be diluted to harmless concentrations with plenty of water hence refered to as ammonotelic eg fresh water fish.

_{Enzyme orginase}

Terrestrial animals produce more urea since it does not need to much water for dilution hence refered to as ureotelic eg mammals.
- Detoxification
It’s the process by which harmful compounds such as drugs or poisons are converted to less toxic compounds in the liver.
The toxic substances are subjected to biochemical reactions. The toxins are rendered harmless through oxidation and reduction.
Detoxification can also involve combining the toxin with another compound. The toxic substances are then excreted in the urine.
Toxic compounds in the body may arise from medication, drugs and micro-organisms.
(c ) Heat production
Many metabolic activities take place in the liver. These metabolic activities release heat energy which is distributed by the blood to the other parts of the body.
Haemoglobin elimination
Haemoglobin from the worn-out red blood cells is broken down in the liver and the residual pigments, urochrome which gives urine a yellow tinge, is eliminated by the kidney.
- (e) Regulation of plasma proteins
Plasma proteins are synthesised from amino acids in the liver eg prothrombin and fibrinogen which are involved in blood clotting.
Other plasma proteins eg serum, albumen contribute to the maintenance of osmotic pressure in the body. Also non-essential amino acids are synthesised in the liver.
Haemoglobin is broken down into haem and globin. Globin is digested into amino acids and enters the amino acid pool while the haem group is changed into biliverdin and bilirubin and taken to the gall bladder. These are later released into the gut as bile and then passed out through the faeces. These two substances give faeces its characteristic brown colour.
- (f) Storage of vitamins and mineral
The liver stores vitamins A, B, D, E and K. The liver of cod fish is a rich source of vitamin A and D. When the RBC are broken down, iron is released and stored in the liver in the form of a compound called ferritin.
Regulation of blood sugar level
Excess glucose is converted into glycogen and fat under the influence of insulin. If the blood reaching the liver has less glucose, the stored glycogen is converted to glucose.
- (h) Storage of blood
The liver is highly vascularised and therefore able to hold a large volume of blood. This is achieved through the dilation of blood vessels to accommodate more blood.
Formation of erythrocytes
Erythrocytes are formed in the liver of the foetus. As the foetus develops, the role of the liver in the formation of erythrocytes declines. The liver breaks down old erythrocytes.
Diseases of the liver
- Liver cirrhosis
This disease is also called liver rot.
Its caused by alcoholism i.e. taking too much alcohol over a long period causes the liver cells to die and they are replaced by fibrous scar tissue. The normal functions of the liver are greatly reduced.
- Signs and Symptoms
Loss of appetite and indigestion.
Abdominal pain around the location of the liver.
Haemorrhage evident in the blood stained vomit.
Treatment and Control
There are no drugs for curing cirrhosis. Most peopple with severe cirrhosis die from it.
If the feet are swollen, the patient should stop taking salt in the food.
Strict diet containing easily digestible foods.
- Hepatitis
It’s caused by viruses. There are 3 types i.e.
-Hepatitis A
-Hepatitis B
-Hepatitis C
Hepatitis A is common among children and young adults.
It’s infectious and transmitted through contact, food and water contaminated with faeces of infected peopple.
Hepatitis B
It’s common among adults and transmitted through body fluids eg saliva, blood and semen. Also transmitted through dry blood.
Hepatitis C
Transmitted in blood causing chronic liver disease.
Symptoms of Hepatitis B
Inflammation of the liver.
Loss of appetite, nausea and fatigue.
Abdominal discomfort.
Jaundice of mucous membranes especially in the eyes.
Treatment and Control
Hygienic processing of food.
Proper disposal of sewage.
Treatment of water.
Vaccination against the disease
Proper handling of the blood products.
Screening of all blood and blood products to be transfused .
Use properly sterilised needles and syringes.
- Jaundice
Its caused by an increase in bile pigment called bilirubin in the blood. This may be due to;
-Damage of the liver cells by toxic or infectious materials. This blocks the bile canals in the liver and can not be transported to the gall bladder. As a result, bile pigments are reabsorbed into the blood.
-Excessive destruction of red blood cells.
-Obstruction of bile flow between the liver and duodenum. This occurs when gall stones block the bile duct. Gall stones are formed as a result of accumulation of excess insoluble cholesterol in the gall bladder.
Symptoms
Patient may have itching caused by retention of bile salt in the blood.
The presence of bile pigment in the blood makes the eyes look yellow.
Activity; To investigate effect of catalase on Hydrogen peroxide
Requirements
Test tubes
Labels
Measuring cylinder
Hydrogen peroxide
Liver
Muscle tissue
Potato
Water bath
Source of heat
Procedure
- Label 4 test tubes A, B, C and D.
Measure 2cm³of Hydrogen peroxide and put in test tube A. Repeat the same procedure for test tube B and C.
Cut a small piece of liver and place in test tube A. Immediately introduce a glowing splint into the mouth of the test tube.
Repeat step III using muscle tissue (in test tube B) and a potato (in test tube C).
Repeat step III using boiled liver (in test tube D) and make sure that the liver is thoroughly boiled for about 5 mins. Tabulate your results e.g.

– Test tube	– Observation	– Conclusion
– A-Hydrogen peroxide+ raw liver	– -Relights glowing splint – -Vigorous production of bubbles	– A lot of catalase enzyme present
– B-Hydrogen peroxide+ muscle tissue	– -Relights glowing splint – -A lot of bubbles produced	– Medium amount of catalase enzyme present
– C-Hydrogen peroxide+ potato	– -Relights glowing splint – – Production of bubbles	– Little amount of catalase enzyme present
– D- Hydrogen peroxide+ boiled liver	– -No bubbles	– Enzymes denatured

Discussion
Living things contain an enzyme called catalase which breaks down hydrogen peroxide to water and oxygen. The oxygen produced relights a glowing splint i.e.
2H₂O₂→ 2H₂O + O₂
_{Hydrogen peroxide catalase water oxygen}

Homeostasis
It’s a process that adjusts changes in the body of an organism to optimum standards or levels and threfore brings about a steady state.
- External environment-It’s the immediate surrounding of the organism. It may be aquatic or terrestrial.
- Internal environment– It’s the immediate surrounding of the body cells.
Neuro-endocrine system and homeostasis
Neuro-endocrine system comprises of the nervous and endocrine system.
Nervous system comprises of the receptors and nerve fibres that make up the nervous tissue.
Receptors detect the changes in the internal or external environment. An impulse passes through fibres to the Central Nervous System (CNS). The CNS in turn initiates the correct response. The CNS sends an impulse to the organ which responds appropriately.
Receptors also send nerve impulses up the endocrine glands which comprises of the glands that secrete hormones. Endocrine system is also known as hormonal system. The hormones secreted are transported in the bloodstream to the target organs.
Principles of homeostasis
Inorder to maintain a state of balance in the internal environment, the various systems in the body work on a feedback mechanism eg
- Negative feedback
When a factor in the body such as temperature drops below or shoots above the normal, it is detected and corrective action is taken. Such an action is either;
-An increase in the level if it was dropping or
-A decrease in the level if it was increasing. This feedback restores the condition to normal.
- Positive feedback
In Positive feedback, a change below or above the normal is not corrected, instead, what is meant to be corrective action leads to further undesirable change from the normal e.g

Role of hypothalamus in thermoregulation

Hypothalamus is a small region between the cerebrum and cerebellum part of the brain. It acts as a thermoregulatory centre.
It has numerous temperature receptor cells which detect the slightest changes in the body temperature. The external temperature affecting the body is determined by the thermoreceptors in the skin.
Thermoreceptors relay the impulse to the hypothalamus through the sensory nerves.
The internal temperatures are detected by the hypothalamus as the blood flows in the brain.

Role of the liver in homeostasis

Regulation of blood glucose

The normal amount of glucose in blood is about 90-100mg /100cm³of
The liver carries out the control of the blood sugar level through two hormones produced by the pancreas i.e insulin and glucagon which are produced by the interstitial cells of the pancreas in the islets of langerhans and released into the bloodstream. The functions of insulin are antagonistic to those of glucagon eg
After a meal, carbohydrates are digested forming glucose, thereby increasing glucose level in the liver. The high glucose level in the liver is detected by the brain which sends impulses to the pancreas to secrete insulin, which carries out corrective measures as follows;
-Converts glucose into glycogen which is then stored in the liver and muscles.
-Changes glucose into fats which is then stored under the skin.
- -Breaks down glucose into CO₂and water in a process of tissue respiration.
When there is decreased glucose concentration in the blood eg during fasting, the pancreas is stimulated to release a hormone called glucagon which affects the liver ie
-Converts glycogen to glucose.
-Converts fats to glucose.
- -Reduces respiration i.e. reduces rate at which glucose is being broken down to form water and CO
Also another hormone called adrenaline produced by the adrenal gland causes increased hydrolysis of glycogen and this results in increase in blood sugar. This hormone is produced during emergencies to increase available glucose for respiration and release of energy for the emergencies.
Diabetes mellitus (sugar disease)
_{From Greek –meaning sweet urine.}
This is a condition in which the pancreas fails to produce insulin or produces inadequate amounts. This may be due to hereditary reasons or disease affecting the islets of langerhans.
A person with Diabetes mellitus has an abnormally high level of glucose in the blood (hyperglycaemia). The kidney eliminates some glucose in the urine, a condition called glycosuria (sweet urine).
Symptoms
Passing large amounts of urine.
Excessive excretion of glucose in the urine.
Loss of body weight due to the breakdown of proteins and fats.
Chronic starvation.
Feeling of thirst.
Treatment and Control
Eating foods with less carbohydrate.
Taking tablets that activate islets of langerhans in the pancreas to produce sufficient insulin.
Administering injections of insulin everyday.
NB insulin cannot be taken by mouth because it is a protein and hence will be digested in the alimentary canal before reaching the liver.
Avoid excessive intake of alcohol.
NB when a higher than normal amount of insulin is introduced in the blood, the patient;
-Feels hungry
-Sweats
-Becomes irritable
-Has double vision

Deamination;

The liver breaks down excess amino acids; The amino group is removed as ammonia; and the remaining carbon skeleton oxidized to carbon IV oxide and water; This process leads to release of energy. The carbon skeleton may be converted to glucose to be used during respiration;

Detoxification;

Ammonia from the process of deamination is converted in the liver into urea; which is less toxic. Bacterial toxins are converted to less toxic substances by liver cells;

Regulation of plasma proteins;

The liver produces most of the proteins found in blood; fibrinogen and prothrombin which play a role in blood clotting. Albumin and globulins are also produced by the liver. Globulins act as antibodies;. Albumin contributes to the maintenance of osmotic pressure in the body; Non essential amino acids are synthesized by the liver;

Heat production;

The various metabolic activities of the liver lead to release of heat energy; This energy is distributed by the blood to other parts of the body hence contributing to maintenance of constant body temperature;

Regulation of fat metabolism;

When carbohydrates are in short supply in the body, fats in different parts of the body are mobilized and taken to the liver; The fats are oxidized to carbon (IV) oxide and water with the production of energy or modified and sent to tissues for oxidation;

Role of kidney in homeostasis
- Osmoregulation
It’s the mechanism of regulating water in the body. It attempts to maintain an optimum osmotic pressure in the body tissues and fluids that is favourable to normal functioning of cells.
When the osmotic pressure of the body rises as a result of dehydration, the hypothalamus is stimulated and sends impulses to the pituitary gland which releases a hormone called Antidiuretic hormone (ADH)/vasopressin into the blood.on reaching the kidney, the diastal convoluted tubule and the collecting tubules become more permeable to water which is then reabsorbed into the bloodstream thus lowering the osmotic pressure of the blood. This leads to the production of concentrated urine.
When the osmotic pressure of the blood falls due to large intake of water, pituitary gland is less stimulated. This leads to reduced release of ADH into the bloodstream. The kidney tubules become less permeable to water and less reabsorption of water into the bloodstream takes place. The osmotic pressure of the blood rises and dilute urine is produced.
Diabetes Insipidus
When pituitary glandreleases very little ADH or fails to release it completely, the kidney nephrons are unable to reabsorb the required amounts of water. This leads to the production of excessively large volumes of dilute urine. This is known as diuresis. Patients may excrete upto 20 litres of urine per day.
The urine can also be described as ‘tasteless’ or insipid thus the name Diabetes Insipidus.
Symptoms
Frequent urination .
Secretion of a lot of urine.
Production of dilute urine.
Frequent thirst sensation.
Treatment
Administration of synthetic or natural ADH.
Regulation of ionic content
A hormone called aldosterone which is produced by the adrenal glands regulates the level of sodium ions.
When the level of the sodium ions is low in the blood, adrenal glands are stimulated to release aldosterone into the blood which then stimulates loop of henle of the kidney and the gut to reabsorb Na+ into the blood.
If the sodium concentration in the blood rises above the optimum level, adrenal glands produce less aldosterone into the blood and less amount of Na+ are reabsorbed.
Role of the skin in homeostasis
- Salt and water balance
Skin has sweat glands which secrete waste products of metabolism such as water, mineral salts especially sodium chloride. These waste products are lost in the form of sweat through the pores in the skin.
About 99% of the sweat is water while the remaining 1% is mainly mineral salts. The water and mineral salts lost in the sweat contribute to osmotic changes of the body cells and fluids.
On a hot day, the body loses a lot of water and mineral salts resulting in a sensation of thirst being felt due to tissue dehydration. The osmotic balance is however restored by drinking large volumes of water and intake of mineral salts in the diet.
- Temperature regulation
- Homeotherms/Endotherms– They are organisms whose body temperature is maintained at a constant despite the wide fluctuations in the temperature of the external environment.
Poikilotherms/Ectotherms- Their body temperatures fluctuates with that of the external environment.
Thermoregulation in humans
Heat loss
The body loses heat to the environment when it’s in a cold environment. The heat is lost by;
-Radiation
-Conduction
-Convection
-Evaporation
Radiation– It’s the transfer of heat by diffusion through the air between a warmer body and a colder one.
Conduction– It’s the transfer of heat from a hot body to a colder one when the two are in contact.
Evaporation– It’s the change of liquid to vapour.
Convection– It’s the movement of air in which warm air in one place rises and cooler air replaces it.
Heat loss occurs through;
-Sweating and breathing
-Passing out of urine and faeces.
- -Mammals such as cats lose heat by licking fur on their limbs and bellies.
Heat gain
The body gains heat from metabolic activities such as respiration and by muscle contraction.
The body uses physiological and behavioural means to regulate the temperature.
When cold
- Physiological mechanisms
Decrease in sweat production-This leads to less heat lost through the latent heat of vapourisation.
Shivering- It involves the rapid contraction of skeletal muscles to generate heat.
Increased metabolism yields heat to raise the body temperature. Increase in secretion of the hormone Thyroxine raises metabolism and heat production.
Arterioles beneath the skin constricts which decreases the blood flow to the skin hence less heat is brought close to the skin surface and this reduces heat loss. This is called vasoconstriction. White people appear pale/white
The liver and spleen store some of the blood which should be in the general body circulation. Thus heat is retained in the body.
Erector pili muscle contract and pull the hair follicles. This way, the hair is raised to trap a layer of air which is a good insulator against heat loss.
- Behavioural mechanisms
Dressing in warm heavy clothing enables the body to conserve heat.
Basking in the sun or warming of the body using a source of heat.
Increased muscular activity such as rubbing hands and stamping feet
- NB Some animals hibernate i.e. go into deep sleep due to cold conditions.
- When hot
- Physiological mechanisms
Increase in sweat production– It leads to heat loss through latent heat of vapourisation.
Arterioles beneath the skin dilate and this increases the blood flow to the skin hence more heat is brought close to the skin surface. This increases heat loss to the atmosphere. This is called White people appear pink.
Erector pili muscles relax and this makes the hair to lie flat on the skin. This way, air is not trapped beneath the hair and a lot of heat is lost to the environment
- Behavioural mechanisms
Dressing in light clothes which do not retain much heat.
Moving to a shade to avoid exposure to direct sunshine.
Some homeotherms such as elephants have large ears which are flapped vigorously to create air currents which take heat away from the body of the animal.
Some animals aestivate i.e. a state of inactivity by some animals that occur during prolonged period of heat e.g. Bats and lungfish. Some animals are only active around sunrise, sunset and at night.
Decreased muscular activity.
Parts of the skin concerned with thermoregulation
Sweat glands
They are coiled tubular glands in the dermis. When the body temperature increases, the sweat glands increase the rate of sweat production. Water in the sweat evaporates by absorbing heat (latent heat of vapourisation) from the body and a cooling effect results.
- NB (i) Birds do not have sweat glands.
Dogs only have them on the pads of the feet.
Hair and Erector pili muscles
When the body temperature lowers, Erector pili muscle contract and pull the hair follicles. This way, the hair is raised to trap a layer of air which is a good insulator against heat loss.
When its hot, the Erector pili muscles relax thus trapping little air hence heat can be lost from the body surface.
Blood vessels
When the body temperature lowers, the blood vessels in the skin constrict (vasoconstriction) and blood is diverted to a shunt system. This reduces the blood flow to the skin and more blood is stored in the spleen as an adaptation to lose less heat.
Dilation of blood vessels (vasodilation) increases blood flow to the skin encouraging heat loss when the body temperature is too high.
Subcutaneous fat
It’s a good insulator against heat loss. Animals in cold areas have thick cutaneous fatty layer for this purpose.
Organisms in warm areas have thin fatty layer to encourage more heat loss to the environment.
Once the temperature changes have been detected by the hypothalamus, the hypothalamus sends impulses to the appropriate responding tissues of the skin.
When the hypothalamus fails to register an increase in the body temperature above normal level, a further rise in body temperature occurs. This causes fever in humans.
If this condition is not corrected, abnormally high body temperature occurs (Hyperthermia). This leads to death if body temperature goes above 43ºC.
If a decrease in body temperature below normal continues, without correction due to the failure of homeostatic mechanisms, abnormally low body temperature occurs (Hypothermia). Death occurs if body temperature falls below 26°C.
Temperature regulation in other animals
- Camels
The camel is able to withstand high environmental temperatures without sweating and will only start to sweat when its body temperature goes beyond 40ºC.
Its hump stores fat which can be metabolized to provide water in times of shortage.
The camel goes for a long time without drinking water and survives as much as 30% reduction in body weight due to dehydration. Under such conditions, a man would die in 2 days.
When a dehydrated camel finds water, it drinks very fast and can drink water equivalent of 30% of its body weight in about 10 minutes.
A camel has a long loop of henle and collecting ducts. These enable it to secrete scanty but highly concentrated urine.
- Kangaroo rat
It has fewer and smaller glomeruli and Long loops of henle. This reduces ultra filtration while increasing the reabsorption of water
It releases insoluble uric acid thus conserving water in the body.
It metabolizes fats and retains the water resulting from the oxidation of fats.
- Birds
They are homeotherms and use physiological and behavioural mechanisms to regulate body temperature
- Reptiles
They are ectotherms and its body is cooled when water evaporates from its skin surface.
.when the temperature is high; the reptiles open their mouths and pant. Panting leads to heat loss through evaporation of water from its mouth.
- Amphibians
They have moist skin and lose heat through evaporation of water. They lose heat rapidly to the dry atmosphere.
- Fish
They are aquatic ectotherms. The body temperature is in equilibrium with the temperature of the water.
Size of animal and body size
Small animals such as rats have a large surface area to volume ratio hence they tend to lose heat at faster rate than the large animals.
Large animals e.g. elephants have a small surface area to volume ratio hence they tend to retain most of their body heat. Hence small animals eat a lot of food to increase their metabolism. This produces heat which replaces the lost heat.

Teachers' Resources

Download Business Studies lesson plans for all topics

October 5, 2025 Maverick John Leave a comment

Here are all the Business Studies secondary school lesson plans for all topics. You can also download the editable and pdf lesson plans below.

FREE BUSINESS STUDIES LESSON PLANS (FORM 1-4)

TEACHER’S NAME……………………………………………………..TSC NO…………….

SCHOOL/INSTITUTION……………………………………………………………………….

CLASS: 1

SUBJECT: BUSINESS TOPIC INTRODUCTION TO BUSINESS STUDIES

SUB TOPIC: MEANING AND IMPORTANCE OF BUSINESS STUDIES

WEEK: 4 LESSON NO: 1, 2

DATE……………………………………………….TIME……………………………………….

OBJECTIVES: By the end of the lesson, the learner should be able to explain the meaning of business studies, Explain the importance of business studies in the society

LESSON PRESENTATION

TIME

CONTENT

LEARNING ACTIVITIES

RESOURCE MATERIALS

5 Minutes

30 Minutes

5 Minutes

Introduction

Introduction to new topic on business studies, trying to define business studies

Content

Defining business studies, identifying key word in business studies such as goods, services, production, distribution, consumption, economic, commerce etc. explaining the meaning of key word in business studies.

Discussing the importance of business studies in society

Conclusion

Summary of the lesson giving a brief summary on meaning of business studies and major key words.

Discussion

Explaining

Teacher/learner discussions

Visit to relevant business ventures in the immediate environment

Answering relevant questions

Group work on relevant tasks

Discussion

Explaining

Diagrams

Charts

Resource persons

Chalkboard and pieces of chalk

Inventor KLB book 1 pages 1-3

Charts

Resource persons

Chalkboard and pieces of chalk

Diagrams

Charts

Resource persons

Chalkboard and pieces of chalk

SELF EVALUATION:

TEACHER’S NAME……………………………………………………..TSC NO…………….

SCHOOL/INSTITUTION……………………………………………………………………….

CLASS: 1

SUBJECT: BUSINESS TOPIC BUSINESS AND its ENVIRONMENT

SUB TOPIC: MEANING AND PURPOSE OF A BUSINESS

WEEK: 4 LESSON NO: 3

DATE……………………………………………….TIME……………………………………….

OBJECTIVES: By the end of the lesson, the learner should be able to explain the meaning and purpose of a business

LESSON PRESENTATION

TIME

CONTENT

LEARNING ACTIVITIES

RESOURCE MATERIALS

5 Minutes

30 Minutes

5 Minutes

Introduction

Recap of previous lesson content on meaning of business studies and major key words i.e. goods, services, production, distribution, consumption , economic, commerce

Content

Defining business and its environment, discussing the purpose of business. Discussing the types of business activities i.e. extraction, processing of raw materials, manufacturing, construction, distribution, trade, provision.

Discussing the business environment and their effects on the business

Conclusion

Summary of the lesson by giving a brief summary on business and its environment, the purpose of business. The types of business activities i.e. extraction, processing of raw materials, manufacturing, construction, distribution, trade, provision.

Discussion

Explaining

Teacher/learner discussions

Visit to relevant business ventures in the immediate environment

Answering relevant questions

Group work on relevant tasks

Discussion

Explaining

Diagrams

Charts

Resource persons

Chalkboard and pieces of chalk

Inventor KLB book 1 pages 5-7

Charts

Resource persons

Chalkboard and pieces of chalk

Diagrams

Charts

Resource persons

Chalkboard and pieces of chalk

SELF EVALUATION:

TEACHER’S NAME……………………………………………………..TSC NO…………….

SCHOOL/INSTITUTION……………………………………………………………………….

CLASS: 1

SUBJECT: BUSINESS TOPIC BUSINESS AND its ENVIRONMENT

SUB TOPIC: MEANING AND PURPOSE OF A BUSINESS

WEEK: 4 LESSON NO: 3

DATE……………………………………………….TIME……………………………………….

OBJECTIVES: By the end of the lesson, the learner should be able to explain the meaning and purpose of a business

LESSON PRESENTATION

TIME

CONTENT

LEARNING ACTIVITIES

RESOURCE MATERIALS

5 Minutes

30 Minutes

5 Minutes

Introduction

Recap of previous lesson content on meaning of business studies and major key words i.e. goods, services, production, distribution, consumption , economic and commerce

Content

Discussing the business environment and their effects on the business

Conclusion

Discussion

Explaining

Teacher/learner discussions

Visit to relevant business ventures in the immediate environment

Answering relevant questions

Group work on relevant tasks

Discussion

Explaining

Diagrams

Charts

Resource persons

Chalkboard and pieces of chalk

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Biology Paper 1 Exams and Marking Schemes Kassu Jet Exams

October 5, 2025 Maverick John Leave a comment

Name………………………………………………… adm no. ……………class…….

School …………………………………………………

231/1

BIOLOGY

PAPER 1

Time: 2 HOURS

KASSU JET EXAMINATION

231/1

BIOLOGY PAPER 1

Time: 2 HOURS

Sept 2021

INSTRUCTIONS TO CANDIDATES

Answer ALL the questions.
Answers must be written in the spaces provided in the question paper.
Additional pages must not be inserted.
The paper consists of 14 printed pages.

FOR EXAMINERS USE ONLY

Question	Maximum score	Candidate’s score
1-29	80

This paper consists of 13 printed pages. Candidates should check the question paper to ascertain that all the pages are printed as indicated and no questions are missing

How does growth as a characteristic of living organisms differ in plants and animals?

(2marks

In plants growth occurs at meristematic tissues only ;while in animals growth occurs all over the body ;

a) State the role of active transport in animal nutrition (1mark)

Reabsorbtion of sugars and some salts in the kidney

Absorption of digested food from the alimentary canal into the blood stream.

Excretion of waste products from body cells.

b) Cyanide lowers the rate of active transport. Explain? (2marks)

cyanide is an enzyme inhibitor, it affects the rate of active transport

The figure below is a diagram of a vertical section of a mammalian tooth.

(i) Name the part labelled A and B. (2 marks)

A Enamel

B Gum

(ii) State two ways in which structure D is adapted to its functions. (2 marks)

– Contains blood vessels which supplies food nutrients and oxygen and remove carbon IV oxide and nitrogenous waste products;

– Contains nerve endings for sensitivity;

(iii) List two ways of preventing gingivitis. (2 marks)

– Regular brushing of teeth;

-Proper exercise of the teeth by eating tough fibrous food;

-Eating foods that are rich in vitamin C

The figure below shows % saturation of oxygen in blood in fish as water passes along the gill plate.

(a) (i) Name the type of blood flow shown in the gill plate. (1mark)

Counter current flow

(ii) Explain the advantage of the type of flow named in a (i) above (2marks)

Creates a steep diffusion gradient; that enhances the rate of gaseous exchange across the gill filaments/ increases rate of gaseous exchange; through diffusion (Rej increases rate of gaseous exchange/enhances gaseous exchange alone

(b) State two organs in humans which display the type of flow named in a (i) above Kidney; Placenta; Ileum (2marks)

– Stomata found only on upper epidermis to allow efficient gaseous exchange;

– Presence of large air-spaces/aerechyma tissue to enable floating and storage of air;

– Lack cuticle to enhance gaseous exchange;

The equation below shows an oxidation reaction of food substances.

C₅₁H₉₈O₆ + 145O2 ——- X CO2 + 98 H2O + energy

a) What do you understand by the term respiratory quotient? (1mark)

Volumetric relationship between Carbon (IV) oxide produced and oxygen consumed

b) Determine respiratory quotient of the oxidation of food substance. (2marks)

R.Q = CO₂ produced

O₂ used up

R.Q = ¹⁰²/₁₄₅

R.Q = 0.7

c) Identify the food substances. (1mark)

Fat/ Lipid

When any one of the growth parameters such as growth in size or weight, increase in number of cells are plotted in a graph against time like below, a clear curve is obtained

State its name sigmoid curve (1mark)

The graph below represents the growth in a certain phylum.

How does this differ from growth in humans? (1mark)

In humans it’s continuous from time of birth to maturity when it slightly tails off while for

this phylum it’s discontinuous /in intervals with shedding of exoskeleton

The embryo of a dry, fully developed seed usually passes through a period of rest after ripening period and it cannot germinate even when provided with all favorable conditions. State the significance of this. (2marks)

Provides the seeds with enough time for dispersal so that they can germinate in suitable environment

Enables seeds to survive during adverse seeds to survive during adverse environmental conditions

The embryo has time to develop until favourable conditions are available

a) Cowpeas seeds were place in a vacuum flask and left for five days. What is the expected change in composition of gases in the flask on the sixth day? (1mark)

Decrease in oxygen and increase in carbon(1V)oxide

b) Give a reason for your answer in (a) above (1mark)

Germinating seeds respire using oxygen and release carbon(IV)oxide

Biotechnologist works day a night to curb food insecurity using the knowledge of polyploidy in genetics. Explain the economic importance of such practice? (2marks)

Increases yields in plants

Cause early maturity in plants

Enhance resistance to pest, disease and drought

b) Define a backcross? (1 mark)

A cross between an offspring with one of its parents

The structure below was obtained from an animal cell
What is the name of the hair like processes and state its function? (2marks)

Name

Cilia

Function

Propel mucus or a fertilised ovum

From which parts of the mammalian body are these structures found? (1mark)

Trachea and oviduct

State the effect of cigarette smoking to the structure? (1mark)

Cigarettes contain tar which impairs/inhibit the action of cilia in the respiratory tract leading to cancer.

A student was found to have blood group B+
a) What type of antibody is present in his plasma? (1mark)

b) Which antigens are present in this blood group? (1mark)

B and Rhesus antigens

Plants relatively have less waste to excrete than animals. Give two reasons to explain this observation (2marks)

Plants reuse some of their waste products;

Plants produce their waste products slowly compared to animals that produce slowly ;

State two methods by which plants get rid of their waste products (2marks)

Diffusion

Transpiration

Guttation

Exudation

Deposition ( any two)

To estimate the population size of mosquitoes in Banji village that covers an area of 25km², visiting researchers caught 400 mosquitoes which they marked and released. After 24 hours, 200 mosquitoes were caught out of which 120 had not been marked.

(a) Suggest the sampling method described above. (1 mark)

Capture recapture method

(b) What are the disadvantages of this method? (2 marks)

–Some organism may die during the study period;

-The mark may come the out during the study period;owtt

The table below shows stomatal distribution on leaves A and B and their surface area. Use the information to answer the questions.

	Leaf surface	A	B
Number of stomata	Upper leaf surface	20	5
Number of stomata	Lower leaf surface	0	15
Surface area		25 cm²	18cm²

Identify with reasons the habitats of the plant from which the leaves were obtained.

Leaf A: Habitat Fresh water; (1 mark)

Reason; Maximum number of stomata on the upper leaf surface for quick loss of excess water by transpiration; (1 mark)

Leaf B: Habitat: Savannah/ Rain forest; (1 mark)

Reason : More stomata on loer surface than on the upper to reduce the surface area exposed to excessive loss of water by transpiration

(1 mark)

Name the causative agent of the following diseases (2 marks)

(i) Trichomoniasis.

Trichomonas vaginalis

(ii) Gonorrhea

Neisseria gonorrhoea

The diagram below shows a pollen tube as it develops down the style. Use it to answer the questions that follows;

(i) Name the part labelled G. (1 mark)

Pollen tube;

(ii) State two functions of structure labelled E. (2 marks)

-Fuse with the egg cell nucleus to form zygote

–Fuse with polar nuclei to form a triploid endosperm nucleus;

(a) Define parthenogenesis? (1 mark)

A type of asexual reproduction in insects where eggs produced without being fertilized are able to hatch into adult insects;

(b) Name the plant hormone that induces fruit ripening. (1 mark) Ethylene;

A group of Form Three students collected a certain specimen for study as shown below. Study it carefully and use it to answer the questions that follow.

(i) Name the type of metamorphosis in the above specimen. (1 mark)

Complete metarmophosis;

(ii) Give any two advantages of the above metamorphosis. (2 marks)

Each stage occupies a different ecological niche; hence there’s no competition for resources e.g food;

(i) Give two structural features in a leaf that adapts it to absorb Carbon (IV) Oxide. Broad lamina;

– Many stomata

(ii) Name the cell organelle in which Carbon (IV) oxide combines with water to form a complex organic compound takes place (1 mark)

Chloroplast

In an experiment to investigate a factor affecting photosynthesis; leaf of a potted plant, which had been kept in the dark overnight was covered with an aluminum foil as shown in the diagram below. The set up was kept in the sunlight for three hours after which a food test was carried out on the leaf.

(a) Which factor was being investigated in the experiment? (1 mark)

Light;

(b) Which food test was carried out? (1 mark)

Starch test;

Starch absent/Iodine retains its brown colour/starch test negative;

Explain how the following plant adaptations minimizes rate of transpiration (2marks)
a) Sunken stomata

Water vapour accumulates in the pits reducing water vapour diffusion gradient hence reduced transpiration rate;

b) Thick cuticle

Reduces permeability of the leaf to water thus reducing water loss;

Explain how drooping of leaves on a hot sunny day is advantageous to a plant (2marks)

The leaves expose a smaller surface area to the sun thus reducing excessive water;

Name two tissues in plants which are thickened with lignin (2marks)

Parenchyma cells;

Xylem vessels;

The diagram below shows the front view of a male reproductive system.

Give the functions of the structures labelled X and V (2marks)
Provide an alkaline fluid which contains nutrients for the spermatozoa;
Seminiferous tubules which provides a large surface area for production of sperms;
What is the role of Follicle Stimulating Hormone in male reproduction? (1mark)

Stimulates the synthesis and maturation of sperms;

27. Explain why the concentration of insecticides in fish eating birds may be hundreds of times greater than its concentration in the water where the fish live (3marks)

Aquatic plants in water absorb the pesticides that drain into water bodies and so accumulate ,fish consume small quantities every time the ingest the water this also accumulates in the fish with time, as the birds continue feeding the contaminated fish they pesticides increase in levels

The diagram below shows a stage in meiosis

State the biological significance of the stage represented on the diagram above (1mark)

The exchange of DNA hence increases genetic variation;

How do the following factors hinder self-pollination in flowering plants? (3marks)
a) Self-sterility

It is a condition where pollen grains from the anthers cannot grow on the stigma of the same flower of plant/ are incompatible to stigma of the same plant/flower;

b) Heterostyly

Is a condition of having different arrangements of style and stigma i.e. shorter stamen than pistil;

c)Protogyny

it’s a condition where the female matures an its ready to receive the pollen grains before the male parts mature;

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KCSE CRE Paper 2 Best Revision Booklet (Mostly Set 1000 Questions and Answers)

October 5, 2025 Maverick John Leave a comment

PIVOT POINT REVISION MAGIC BULLETS

REVISION

C.R.E.

PAPER 2 (313/2)

SAMPLE 1

313/2

CHRISTIAN RELIGIOUS EDUCATION PAPER 2

(a) With reference to St. Luke’s Gospel state the mission of John the Baptist as prophesied by his

father in the Benedictus 4mks)

(b) Give the teachings of John the Baptist about the Messiah (8mks)

(a) Describe the reasons why Jesus taught in parables (6mks)

(b) Explain the qualities of God from the parable of the prodigal son (8mks)

(a) Explain the role of the Holy Spirit as taught by Jesus Christ (8mks)

(b) Relate the message of Peter on the day of Pentecost (7mks)

(a) State the factors that lead to sexual immorality among the youth today (8mks)

(b) Give guidelines that can help boys and girls to lead responsible sexual lives (6mks)

(a) Give the qualities a Christian would look for in a prospective marriage partner (7mks)

(b) Give the characteristics of responsible parenthood (8mks)

(a) Give reasons why people work (6mks)

(b) Explain some of the factors that bring about unemployment in Kenya today (8mks)

(6mks)

MARKING SCHEME

SAMPLE 1

313/2

a) With reference to St. Luke’s Gospel state the Mission of John the Baptist as prophesied by his

father in the Benedictus (4mks)

i) He would be the prophet of the most high
ii) He would go before the Lord to prepare the road for him/ he would be the forerunner of the Messiah

His duty would involve imparting the knowledge of salvation to the people

He would call people to repentance and forgiveness in order to restore the right relationship with God
He would give ‘light’ to those living in darkness and guide them into path of peace

Any 4×1=4 mks

b) Give the teachings of John the Baptist about the Messiah (8mks)
i) He was going to be famous and greater than John the Baptist
ii) He would baptise with the Holy Spirit and fire

He would bring God’s salvation/ saviour/ save humanity

He will proclaim judgement
He will punish the sinners and reward the righteous/ give eternal life
Jesus is the Messiah/ Christ is the anointed one

Jesus as the lamb who takes away the sins of the world/ Christ would die for the sins of humanity
Jesus will be a Lord and a judge

Jesus will lead the new community of believers
Christ was sent by God/ He is from God Any 8×1=8 mks
c) How do Christians today play the role of John the Baptist? (8mks)
i) John baptised the believers; Christians today baptise those who believe
ii) John preached and called people to repentance; Christians today preach and call people to repentance

Like John Christians should announce the consequences of not believing in Christ’s salvation

Like John Christians should be bold enough to condemn evil in society
Like John advised the tax collectors to be honest, Christians should advice people to be honest and not practice corruption
John proclaimed Christ as the lamb of God who takes away the sins of the world, Christians should do the same today

John taught against accusing people falsely, Christians should preach and practice justice
John taught on sharing, therefore Christians should not only preach but also practice it

Christians should realise that Christianity alone without sincerity cannot justify their entry into the kingdom of God Any 8×1=8 mks
a) Describe the reasons why Jesus taught in parables (6mks)
i) To distinguish the serious seekers of the kingdom from those who were just eager to

listen to the story

To explain unknown issues in a language people could easily understand

To make people think critically about issues concerning the Kingdom

They were used by Jesus to teach about God’s Kingdom
To teach about God’s mercy to sinners i.e. parable of the debtors/ parable of lost coin
To conceal his true identity as the Messiah

To teach about the requirement of God’s Kingdom
Parables were avenues of passing important teachings to the people, how to relate to one another, how to use wealth and even how to pray

It was a method of teaching which was common those days
To teach his disciples that they should be persistent and never be discouraged (Lk18:1-8)
He wanted to make an indirect attack on his opponents like the Pharisees, the scribes and the Sadducees Any 6×1=6 mks
b) Explain the qualities of God from the parable of the prodigal son (8mks)
i) God is a God of freedom who has given man freedom to choose as the father granted the

wishes of the younger son

God responds to people’s prayer and request as the father listened to the wishes of his younger son

God is always concerned about sinners as the father was looking forward to the return of his son

God is merciful and compassionate and ready to receive those who repent as the father welcomed his son
God is a loving father as he loves us unlimitedly the way the father loved his younger son
God is happy and rejoices when the repentant sinner go back to him i.e. when the lost one is found as the father rejoiced when the son came back

God is forgiving as the father forgave his son
God does not discriminate against some people. All people are his children just like the father did not discriminate against his son

God is generous as shown by the father who gave his younger son part of his property and told the older one that all my wealth is yours
God is mysterious i.e. the older son could not understand what the father was doing
God provides as he is the provider and giver; this is seen in the father giving his younger son a ring and clothing

God knows and understands each individual needs as the father knew the needs of his sons Any 8×1=8 mks

c) Identify ways in which the disciples of Jesus reacted to his parables (6mks)
i) They were surprised at the meaning and reasons for the use of parables
ii) They failed to understand the message of some of the parables

They asked Jesus to explain the parables

They felt privileged
They understood some parables clearly
They marvelled at Jesus ability to teach using parables

They enjoyed listening to Jesus parables
They felt challenged by Jesus parables

a) Explain the role of the Holy Spirit as taught by Jesus Christ (8mks)
i) He would be a counsellor/ advocate/ a comforter / aiding believers / helpers
ii) He would convict people about sins, righteousness and the coming judgement

He would remind the believers everything that Jesus had taught them

He would reveal the truth concerning God the father and Jesus the son
He would pass judgement to the sinners
He would declare things to come through the believers

He will glorify Jesus by declaring what belongs to Jesus Christ from the father
He would teach the believers of all things and reveal the sins of the world

He would give authority to the believers to forgive sin
He would give the believers power to become witnesses of Jesus Christ
He would enable the believers to discern and expose the secret heart of sinful people

He would affirm the right of Jesus as the son of God Any 8×1=8 mks

b) Relate the message of peter on the day of Pentecost (7mks)
i) What was happening was the fulfillment of Joel’s prophecy about the outpouring of

God’s spirit

The outpouring of the Holy Spirit was a positive proof that the Messianic age had arrived through Jesus Christ

Jesus was from Nazareth through whom God worked miracles

Jesus suffered and was crucified by sinful people in accordance with God’s plan
God raised Jesus up to fulfill the prophecy of King David
The apostles are living witnesses to the resurrection of Jesus

God has exalted Jesus and made him both God and saviour
Jesus is a descendant of David

Peter told the people to repent so that they could be forgiven and receive the gift of the Holy Spirit Any 8×1=8 mks
c) How are the gifts of the Holy Spirit manifested in the Church today? (5mks)
i) People speaking in tongues
ii) People healing the sick

Performing miracles e.g. casting out demons

Proper understanding and interpretation of God’s message
Proper/ wise decision making and right believing
Giving of prophetic messages to the church

Ability to withstand challenges of life/ persecution
Helps believers to repent and seek forgiveness for their sins

The believers loving one another
Charity work done by believers Any 5×1=5 mks
a) State the factors that lead to sexual immorality among the youth today
i) Human weakness
ii) Sexual curiosity

Testing fertility

Fear of being jilted by boy/ girl friend
Poverty
Availability of contraceptives

Negative peer pressure and bad company
Influence from print and electronic media

Frustrations
Unemployment
Drug and alcohol abuse

Proving manhood Any 8×1=8 mks

b) Give guidelines that can help boys and girl to lead responsible sexual lives (6mks)
i) Avoid physical contact like kissing and hugging
ii) Avoid conversations on topics that are sexually suggestive

Dwell on positive thoughts about your special friend as a way of controlling your desires

Avoid meeting in lonely and isolated places
Avoid social functions such as discos and parties that may create room for sexual abuse
Avoid visiting each other frequently

Avoid pornographic literature, films and music
Seek guidance from Christian leaders and counselors on your relationship

Study the Bible teachings on sex
Pray for God’s guidance Any 6×1=6 mks
c) Describe the Social and Moral effects of HIV/AIDS on the family
i) Economic contribution of the AIDS patient is drastically reduced
ii) It increases the medical expenses as the person is often sick

The AIDS patient eventually dies and this brings sorrow to the family

Emergence of children- headed households
Dropping out of school by the children
Discrimination against people living with AIDS in work place, and may not be hired for jobs

Stigmatisation of AIDS sufferers
Feeling of loneliness, isolation and helplessness

Grandparents are burdened with taking care of orphans
Work load for women in the family set-up increases because they have to provide for the family Any 6×1=6 mks
a) Give the qualities a Christian would look for in a prospective marriage partner (7mks)
i) Commitment to ones faith
ii) Moral uprightness

Common interest

Ability to provide for the family
Socio- economic status
Level of education

Common faith
Good management of family finances

Health status Any 7×1=7 mks
b) Give the characteristics of responsible parenthood (8mks)
i) Understanding parental roles and responsibilities
ii) Educating children in all aspects of life

Teaching and training children in religious matters

Helping their children to grow physically, socially, psychologically and emotionally
Helping children to develop intellectually by providing opportunities for them to learn
Teaching morals and right behaviour to their children

Teaching children on how to relate with another as brothers and sisters, among other relatives and kinsmen and among their peers
Teaching them social and civic responsibilities

Being hardworking so that they are able to provide for the needs of their families
Parents should remember not to discriminate against any of their children
Showing understanding and tolerance to their children Any 8×1=8 mks
c) Family life today is faced with innumerable problems. List some of them (5mks)
i) Child abuse and domestic violence
ii) Diseases and sicknesses

Unfaithfulness

Separation and divorce
Childlessness
Single- parent families

Jealousy
Misuse of family resources

Leadership style Any 5×1=5 mks
a) Give reasons why people work ( 6mks)
i) For self satisfaction and fulfillment
ii) For personal development

To acquire basic essentials of life e.g. food, clothing and shelter

To give life direction, meaning and dignity
To provide services for the community and fellow human beings and to develop and care for the environment
To acquire wealth and status in society

To socialize and grow as a member of a community e.g. people participate in communal work and projects such as building schools, bridges, churches e.t.c.
To attain independence. Young people look forward to the time when they will not depend on their parents and guardians for the necessity of life Any 6×1=6 mks

b) Explain some of the factors that bring about unemployment in Kenya today
i) The available job opportunities are too limited to absorb the great number of school

leavers and college graduates

Most young school leavers despise “blue collar” jobs (The manual jobs) and opt for “

White collar jobs”

Bribery and corruption. This denies job those who are more qualified in preference for the unqualified who might have given bribes

Lack of skills and capital to start self-employment
Bad governance and poor economic policies by the government
Increase in poverty since 1990’s due to Structural Adjustment Programmes (SAPs) imposed by the world bank and IMF on developing countries like Kenya

International policies of globalisation that have affected the agricultural sector which has been the greatest employer in Kenya.
Retrenchment of workers in the civil service and private sectors

Increase in crime that has aggravated insecurity in the country. This has discouraged local and international investors Any 8×1=8 mks
c) What are the negative results brought about by a strike action in a school situation?
i) May lead to the inciters and culprits being expelled
ii) It may be very expensive to parents as they pay for damages

May lead to injuries among students in case of riots

It leads to loss of study time due to suspension
May lead to poor academic results
May lead to mistrust between the teachers and students

Injuries and death may occur when police confronts the strikers
Loss of employment if teachers are involved

Victimisation of some individuals especially the ring leader may occur

Any 6×1=6 mks

SAMPLE 2

313/2

CHRISTIAN RELIGIOUS EDUCATION PAPER 2

a) What is the concept of the Messiah in the New Testament. 9mks
b) State the expectations of the Jews about the Messiah. 6mks
c) Explain the Christian understanding of the concept of the Messiah to day. 5mks
a) Identify the three Temptations of Jesus in the wilderness and how Jesus responded to each one of them. 9mks
b) Explain the relevance of Jesus Temptations to Christians today. 6mks
c) State some of the Temptations that you face today as young people. 5mks
a) Why did Jesus find it necessary to pray during his public ministry. 8mks
b) Explain the ways in which God’s divine power is demonstrated in the healing miracles. 6mks
c) Give ways in which the church in the New Testament demonstrated unity. 6mks
a) Briefly describe ways in which an employer can motivate his employees. 7mks
b) State ways in which alcoholism can deter one’s career. 8mks
c) Which criteria can Christians use to evaluate the use of leisure? 5mks
a) Describe the social and moral effects of HIV/AIDS on the family. 10mks
b) How do you identify an individual infected with HIV/AIDS? 5mks
c) Explain the way the church in Kenya is handling the challenge posed by HIV/AIDS. 5mks
a) Explain the factors that cause disaster in society. 6mks
b) Why are Christians opposed to capital punishment? 6mks
c) How do Christian contribute to the maintenance of law and order in Kenya 8mks

MARKING SCHEME

SAMPLE 2

313/2

a) – He was born of a virgin Mary

– He was a descendant of David (Joseph his foster father was a descendant of David)

– He was born in Bethlehem the city of David

– The Tittles given by prophet Isaiah were reported by the angel Gabriel to Mary e.g. “the son of the most high”

– Jesus himself at the beginning of his ministry affirms that, he is the Messiah, when he repeats that Isaiah said, “The spirit of the lord is upon me” Isaiah 61:1-3

– His life, death and resurrection fulfills what the Old Testament prophets said about him

– His shameful death on the cross fulfills what Deutro –Isaiah 53 said about the suffering servant of God

– His resurrection fulfills what the Old Testament said that God will restore his life after his suffering 1×8=8 mks

– The Messiah was to come out of the house of David

– The messiah was to liberate the Jews from the rule of the Romans

– The messiah was to restore the kingdom of God

– The Messiah was to do great things for Israel

– The Messiah was to be a perfect King

– The Messiah was to establish ultimate peace and joy

– The Messiah was to lead them into a time of great National power and prosperity

1×7=7 mks

– Christians accept Jesus as the Messiah that was foretold by Old Testament prophets

– Christians believe that Jesus is the final revelation of God’s saving works

– Jesus the Messiah established a spiritual kingdom which is received by faith by those who believe in him

– The Messianic Kingdom is universal offered to all mankind and not just restricted to the Jews

– Through his death and resurrection, he established God’s kingdom on earth

1×5=5 mks

a) – Jesus was led to the wilderness by the holy spirit after his baptism to be tempted by satan. While

in the wilderness, Jesus entered into prayer and fasting for 40 days. At the end of 40 days, satan came to him and tempted him in three ways

i) Turn stones into bread

satan wanted Jesus to use his divine power to turn stones into bread. Jesus told satan that man does not live on bread but by the word that procedeth from his mouth

ii) satan took Him to high ground and showed him all the earthly wealth and told Jesus that all will be given to him, if he paid homage to satan. Jesus told him that he should not tempt God. Homage is only given to God

satan took Jesus to the highest part of the temple (pinnacle) and told Jesus that if he jumped down, God will provide angels to protect him from being hurt. Jesus rebuked satan. He departed from Jesus at opportune

NB Introduction before temptation 3 mks

First temptation 1 mk

Response 1 mk

2^nd temptation 1 mk

Response 1 mk

3^rd temptation 1 mk

Response 1 mk

Total 9 mks

– Being tempted is not a sin unless you give in to the temptations

– Temptations will always come to those who follow Jesus

– Since Jesus was tempted; he understands our problems/ our weakness

– When Christians turn to Christ in prayer, he answers and assists us

– Temptations can be overcome through the word of God as Jesus did 2×3=6 mks

– Drug abuse (cocaine, heroin)

– Lack of finance leads to commercial sex/ prostitution

– Pre-marital sex/ fornication

– Early pregnancies

– Abortion

– Drunkard-ness/ smoking

– Gossiping

– Stealing/ theft

– Robbery Any other relevant points1×5=5 mks

a) – To get assurance from his father/ God

– To ensure that he was within the will of God

– To get in touch with the father

– To renew his strength in order to face temptations

– To give thanks to God

– To set examples to his disciples

– To teach his disciples how to pray

– To intercede for others

– To honor God

– To seek for guidance from God. 1×8=8 mks

b) – The spoken word of Jesus during the healing of the paralytic, “your sins are forgiven” shows that

Jesus was interested in curing people of their spiritual illnesses

– Restoring the sick people back to health by liberating them from physical and psychological suffering

– The casting out of demons shows that evils and the powers of satan are overcome

– Those who witnessed the healing miracles of Jesus realised the presence of God’s power in Jesus

– The healing of non-Jews states that God’s saving power/ grace is for all communities/ universal

– The healing miracles liberated the sick from despair and restored them to holiness

– The healing was a demonstration of God’s love for the oppressed

– The healing miracles shows that Jesus had come to liberate human kind from suffering and death

1×6=6 mks

c)- They met for prayers/ prayed together in fellowship

– They prayed for one another.

– They shared meals

– They shared their belongings / property with one another.

– They sold their property and distributed the money among themselves according to each individuals needs

– They celebrated the Holy Communion together/ broke the bread together

– They showed concern/ to help the less privileged

– They welcomed each other to their homes/ offered hospitality to others

– They solved problems affecting the church

– Winning converts/ accepting members of the church without discrimination on gender, race or social class

– Preached the same Gospel of Jesus Christ

– Churches in different places kept in touch with each other through visits/ missionary work

– The unity of the church was demonstrated through the work of the Holy Spirit among Christians

1×6=6 mks

a) – By incorporating them as partners in business by allowing them to buy shares

– Sharing profit with them

– Appreciating their work

– By paying terminal benefits

– By allowing them time for entertainment/ leisure/ worship

– By allowing employees leave/ rest

– By practicing charity and good will

– By treating them humanely

– By setting good example i.e. being respectful, hard working and punctual

– By taking care of workers welfare 1×7=7 mks

– Alcoholism affects one’s efficiency at work. It reduces the productivity

– It can lead to one to embezzle public funds

– It can cause accidents especially drivers if they are under the influence of alcohol

– A lot of money meant for basic things is spent on drinking sprees hence causing poverty

– It causes misunderstanding among the family members/ workmates and the employer

– It affects one’s health rendering him in effective at work

– It may lead to break up of one’s family which will affect his performance

– It may make one to be absent from work 1×8=8 mks

c) – Christians should engage in leisure activities which promote their respect and dignity

– Christians should avoid leisure activities that may be harmful to others

– Christians should choose leisure activities that enrich their knowledge of God

– Christians should follow Jesus’ example by engaging in those activities that promote service to others

– Christians should not engage in leisure activities that are harmful to the individuals and put ones life at risk

– Christians should avoid leisure activities that lead to addiction

– Christians should avoid leisure activities that lead to sin

– Christians should engage in leisure activities that please God 1x 5=5 mks

a) – Many adults die leaving behind many orphans, by 2005 we had 1.6 million orphans

– Caring for the orphans is a burden

– Grand parents or old children get the burden of running the homes where orphans are left

– The sick person uses a lot of money medically

– The sick person becomes unproductive economically

– A lot of sorrow in families due to death

– Women are easily infested and affected

– The women are burdened with care of the home

– Poverty increases in the homes

– Children are forced to look for manual work to provide for the homes

– Children are forced to drop out of school due to lack of money or look for employment

– People with AIDS feel lonely, isolated and hopeless

– Those living with AIDS feel guilty and as punishment

– People with AIDS are discriminated at places of work and not hired for jobs

– People living with AIDS may be denied their basic human rights e.g. to travel, to have children, employment and education 1×10=10 mks

b)- Night sweats

– Fever lasting several weeks

– Diarrhoea which lasts two or more weeks

– Loss of appetite and weight e.g. loosing 5 kilos in two months

– Swelling of the glands in the neck, armpit and groin

– A feeling of tiredness lasting for weeks without apparent cause

– Skin diseases – a type of cancer known as “Kaposis Sarcoma”

– The growth of fungus in the mouth

– Prolonged cough and shortness of breath 1×5=5 mks

– Preach to those with AIDS and show love to them

– Clergy should be trained in how to handle those with AIDS

– Educational programmes should be held to give people dangers of HIV/ AIDS&STIs

– Parents should be helped by the church to educate youths and adolescents on the importance of responsible sexual behavior

– Counsel those with AIDS not to spread it

– Give hope to full blown AIDS people to live with dignity and not blame themselves

– Cultivate self worth, self image and self assertion to possible candidates of AIDS (promiscuous persons)

I cor 13 Gal 5:19-21 Eph 5:3

– Involve affected and infected in employment or income generating projects

– Advocate for protection of human rights of people with AIDS 1×5=5 mks

a) – Unjust laws and unjust punishment

– Inadequate distribution of wealth and resources

– Racism/ tribalism/ Nepotism/ Secterialism/ Apartheid/ Prejudices

– Poverty

– Lack of tolerance by leaders

– Arrogance and hypocrisy by leaders

– Oppression/ suppression and repression of the citizens

– Unemployment/ underemployment

– Selfishness and greed for money and power

– Corruption/ bribery/ rigging of elections 1x 6=6 mks

– Life is sacred/ Holy, only God should claim it

– Capital punishment does not make the person reform but clears him

– Death cannot be reversed incase a judge makes a mistake in his prosecution

– It has no economic value/ benefit to a country

– The methods used are inhuman and lowers a persons dignity

– It is contrary to God’s statutes/ laws for example “Though shall not kill”

– It undermines the Biblical teaching on forgiveness

– It distabilises the family/ deprives a family of a resourceful/ loved member

– It destroys the image of God in man 1x 6=6 mks

– Preaching/ teaching values pertaining to peace and justice

– Appealing to people to obey lawful authority (the rule of law)

– By practicing what they teach/ preach, setting good examples/ living exemplary lives

– Condemning vices which lead to disorder in the society e.g. injustice, corruption

– By playing reconciliatory roles/ mediators between warring parties or individuals

– Educating people on their right and duties as citizens of the country

– Caring/ helping the poor, needy, oppressed people in the community

– Create employment and equip people with skills so that they can be employed elsewhere

– They offer guidance and counseling services where people who have problems can get help

– They provide education whose objective is good citizenship

– Support individual/ groups who fight for justice and peace

– They pray for peace and justice and good government 1×8=8 mks

SAMPLE 3

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CHRISTIAN RELIGIOUS EDUCATION PAPER 2

Answer Any Fice Questions Below

(a) Show how prophet Isaiah presented the servant of Yahweh 8mks

(b) Explain using examples from St. Luke’s Gospel how Jesus fulfilled Jeremiah’s teachings

about the new covenant. 7mks

( c) Why do Christians commemorate the lord’s supper 5mks

(a) explain the message about God and John the Baptist as outlined in the Benedictus.

6mks

(b) How do Christians mark the birth of Christ in their lives today. 8mks

( c) What lessons can a Christian learn from the shepherds response to Angel’s message.

6mks

(a) Narrate the parable of the wicked tenants Lk 19:9 -18. 9mks

(b) Explain the meaning of the parable of the tenants. 6mks

(c ) Give reasons why some people find it difficult to accept the Gospel of Jesus Christ.

5mks

(a) Explain the New Testament teaching on love, 1 Corinthians 13: 4-8. 8mks

( b) State the criteria for discerning the spiritual gifts. 6mks

( c) What religious duties do Christians perform as a Royal Priesthood to God. 6mks

(a) Explain the social problems facing the society today. 8mks

(b) How does the state punish those who break the law. 7mks

( c) How can a young Christian use the life skills to cope up with challenges in life5mks

(a) What is a justified wage. 8mks

(b) Explain the Christian view on industrial action 7mks

( c) Suggest the government solution to child labour 5mks

MARKING SCHEME

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(a) How prophet Isaiah presented the servant of Yahweh.

(i) The servant will succeed in his work and will be highly honoured.

(ii) His success and Honour will surprise many who have witnessed his death.

(iii) The servant is despised, rejected and ignored by those who are with him.

(iv) He has nothing attractive, he is very ordinary and simple.

(v) He is harshly treated, arrested and sentenced to death and killed.

(vi) He accepts the suffering, which should have been received by other for their sins.

(vii) His body is buried with the bodies of rich men.

(viii) He endures all that is done to him in humble silence.

(ix) Through his sufferings, human beings are reconciled to God.

(x) It is the will of God that the servants suffer.

(xi) His death is a sacrifice to bring forgiveness of sins. (8 x 1 = 8mks)

b) How Jesus fulfilled Jeremiah’s teachings on the new covenant.
i) Jesus inaugurated the new covenant during the Last Supper. He said, “This cup is

poured out for you in the new covenant in my blood”. (LK 22:20)

ii) The death of Christ on the cross shows that God will not remember their sins anymore.

iii) Jesus summarized the Mosaic Law into the double commandment of love of God and love of one’s neighbour LK10: 27. The law of love is eternal.

(iv) Jesus established the kingdom of God, a new community of God’s

people based on faith and obedience is his teachings.

(v) Jesus established an everlasting covenant through his death and resurrection.

(vi) In Jesus, people have direct relationship with God

(vii) Jeremiah gave hope about restoration which found fulfillment in Jesus who came from

the Davidic lineage to set up God’s rule on earth.

(vii) Jesus established the new beginning in the New Testament, which is

the basis of his doctrine of the new covenant.

( Well explained point 7 x 1 = 7 mks)

Why Christians commemorate the Lord’s supper

(i) To renew their faith in God and bind themselves to it.

(ii) It marks the day of deliverance for Christians from the bondage of sin.

(iii) It is a way of continuing to proclaim Christ death until his Second Coming.

(iv) It provides Christians an opportunity to confess and repent their sins.

(v) Jesus presence becomes a reality through the sharing of the bread and the wine.

(vi) To thank God for his Love.

(vii) It is a time for re-dedication and self-renewal to Christ.

(viii) It is an act of obedience in Christ command. (5×1 = 5 marks)

2 (a) Message about God and John the Baptist as outlined in the Benedictus.

i) God is a redeemer

(ii) God has bought salvation through Jesus

(in) God fulfills his promises

(iv) John the Baptist would be the prophet of the most high God

(v) He would prepare the way for messiah

(vi) John the Baptist will give the knowledge of salvation to his people

(vii) John the Baptist would call people to repentance and forgiveness

(viii) John would give light to those in darkness & shadow of death

(ix) He would guide people into the part of peace

(6×1 = 6 marks) 3 points for each (God & John)

(b) How Christians mark the birth of Christ.

(i) There is the decoration of the Christmas tree

(ii) Decorations of church and homes

(iii) The Santa Claus (Father Christmas)

(iv) Special Christmas dinner/food/ cake

(v) Exchange of gifts/ cards

(vi) Christmas Carols

(vii) Attending of church services/ church meeting/ Christmas vigil

(viii) Helping the poor/ needy

(ix) Visits/Visitations (8×1=8 marks)

Lessons Christians can learn from the shepherd’s response to the Angel’s message.

(i) God demand obedience from us,

(ii) We should respond positively to God’s message and be ready to serve him.

(iii) Christians should share the good news of the kingdom with others Just like the Angel

shared the news with the shepherds.

(iv) Christians should believe in God’s message and respond to it at once,

(v) It is important to praise and glorify God,

(vi) We should serve God with our wealth,

(vii) God speaks to us despite our status in life,

(viii) God reveals his message to man.

(ix) All are called to serve God. (8 x 1 =(8 marks)

3(a) Narrate the parable of the wicked tenants,

– When Jesus was teaching and preaching in Temple during his Jerusalem ministry, the

Jewish religious leaders questioned his authority.

– Jesus told them the parable of the wicked tenants to help interpret his role and that of

his opponents.

– Jesus told them the parable of the man who planted vineyards and let it out to tenants &

went into another country for along time.

– When the time of harvest came, he sent his servants to the tenants, that they

may give him some of the fruits of the vineyard. The tenants instead beat him up and sent him empty-handed.

– He sent another servant whom they beat & treated shamefully and sent him

empty handed.

– He sent yet a third one whom they wounded and caste out.

– The owner of the vineyard then sent his own son but he was beaten up and killed.

– Jesus posed a question “what then will the owner do to the vineyard?.

He will come destroy those tenants and give the vineyard to others.

– Jesus concluded the parable that “the very stone the builders rejected has become the corner stone “. (9×1 = 9 marks)

Meaning of the parable of the wicked tenants.

(i) The parable is about the rejection of Jesus by the Jewish religious leaders, providing a preview of what will happen in Jerusalem,

(ii) The owner of the vineyard is God, the vineyard is Israel while the tenants are

the Jewish.

(iii) The servants that are beaten represent God’s prophet, messengers.

(iv) Jesus is the son referred in the parable whom the people rejected and killed,

The death of the son anticipates Jesus death at the hands of Jewish leaders,
The quotation in Psalm shows that the stone the builders rejected is Jesus himself.

(6×1 = 6 marks)

Reasons why people find it difficult is accept the Gospel of Jesus Christ.

(i) Alternative religions.

(ii) Too much wealth/riches.

(iii) It is too demanding.

(iv) Lack of faith.

(v) Science technology seems to provide solutions to man’s problems.

(vi) Lack of proper role models.

(vii) Discouragement from church leaders.

(viii) Permissiveness in the society.

(ix) Drug abuse makes people not to think about God. / pleasures of the world

(5×1 = 5 marks)

4 (a). New Testament teachings on love & 1Cori. 13:4 — 8.

(i) Love is patient and kind,

(ii) Love is not jealous or boastful.

(iii) Love is not arrogant or rude.

(iv) Love does not insist on its own way.

(v) Love is not irritable or resentful.

(vi) Love does not rejoice at wrong but rejoice at the right.

(vii) Love bears all things, believes all things, hopes all things, and endures

all things.

(viii) Love never ends.

(b) Criteria for discerning spiritual gifts’.

(i) Test of loyalty to Jesus or conformity to the faith.

(ii) Test of love. The exercise of spiritual gift should result to love.

Test of spiritual One who is lead by the spirit of God should bear the fruits

Of the Holy Spirit,

Doctrinal test. One who is inspired by the Holy Spirit can not contradict the scripture

/ doctrine

(3 x 2= 6 marks, 1 mk for mention, 1 mk for explanation Total (6 marks)

C . Religious duties that Christian performs as a Royal priesthood to God.

(i) Praising God and praying to Him.

(ii) Devoting their lives in obedience to God as a spiritual sacrifice.

(iii) Calling others to repentance.

(iv) Forgiving others.

(v) Meeting the needs of other people by sharing/ helping needy.

(vi) Creating unity/ peace/harmony among people.

(vii) Preaching the Gospel.

(viii) Leading by examples/ role models.

(6×1 = 6 marks)

5 (a). Social problems facing the society

(i) Poverty

(ii) Drug abuse

(iii) Unemployment

(iv) Child abuse / child labour

(v) Sexual Immorality / rape / adultery / prostitution

(vi) Discrimination/ Tribalism/ Racism

(vii) Wars

(viii) Epidemics/ (diseases e.g. HIV/AIDS)

(ix) Calamities such as earthquake, floods Tsunami

(x) Accidents

(ix) Corruption / bribery

(iix) Broken marriage / divorce / separation

( 1 x 8 = 8 mark) ( the points must be explained)

(b). How the state punish those who break the law.

(i) Imprisonment

(ii) Paying fines

(iii) Corporal punishment

(iv) Probation

(v) House arrest

(vi) Assigning community work

(vii) Hard labour

(viii) Capital punishment (soon it might be scrapped)

(ix) Approved schools and juvenile remands homes

(7×1= 7 marks)

How a young Christian can use life skills to cope up with various challenges

(i) Using decision-making, by identifying the best alternative to overcome

a challenge encountered.

(ii) Critical thinking, by examining and assessing a given situation impartially.

(iii) Creative thinking by using ideas imaginatively to solve a problem.

Having high self-esteem, being confident, consistent, outgoing, social and

having positive altitude, about ourselves

(v) Being Assertive in making decisions/principles and hold on to item.

( 5 x 1 = 5mks)

6 (a). What is a justified wage.

(i) Wage that is paid on time.

(ii) Equivalent to the work done.

(iii) Paid according to labour contract.

(iv) That enables the employee to meet his/ her basic needs.

(v) Extra-work extra-pay.

(vi) A wage that does not push the employer out of business.

Equal work equal pay/no discrimination.
Should not be withheld.

(ix) Should take into consideration the period of framing/ practical experiences/ skills

responsibilities assigned/energy use / importance of the work to the community.

(8×1=8 marks )

(b). Christian view on industrial action (strike)

(i) A strike must have a serious and just cause

(ii) The gain front the strike must out-weigh loss/damages.

(iii) It should he peaceful.

(iv) Must have a hope of success.

(v) There should no strike among essential service workers.

(vi) There should not be no picketing / intimidation / victimization.

(vii) The strikers should not harass the non-union workers.

There should be peaceful negotiation between the employers and workers representatives or union.

(ix) Should be the lust resort when other means have failed.

The employers and employee should fulfill their obligations towards one another

to avert strike.

(7×1 = 7 marks)

Suggest the government solutions to child labour.

(i) The government discourages child labour.

(ii) The government has introduced laws that protect children.

(iii) Introduction of free education.

(iv) There are laws governing labour contract/ no employment of under age

The government sponsorship/ giving funds to children from poor families to

Continue with schooling.

(vi) The government together with the N. G. Os sensitize the public about the child rights.

(5×1= 5 marks)

NB mark Any other relevant point raised by the student

SAMPLE 4

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CHRISTIAN RELIGIOUS EDUCATION PAPER 2

a) What were the Jewish expectations of the Messiah? (6 marks)
b) Out-line what Simeon and Anna revealed about the life of Jesus when his

parents presented him to the Temple for dedication. (8 marks)

c) Give ways in which church leaders are preparing people for the second

coming of Christ. (6 marks)

a) Give the account of the sinful woman in Luke 7:36-50 (7 marks)
b) Highlight the main teachings of Jesus on the sermon on the plain (8 marks)
c) State ways in which the church in Kenya is continuing with the healing

ministry of Jesus (5 marks)

a) Outline the testimony of the holy women regarding the resurrected Jesus

Christ. (Lk. 24:1-10) (7 marks)

b) Why did Jesus disciples respond to the news of his resurrection with fear

and disbelief? (6 marks)

c) Why is violence against women rampant in Kenya today? (7 marks)

a) Identify the symbols used to describe the unity of believer in the New

Testament. (5 marks)

b) In what ways was unity of believers demonstrated in the early church? (8 marks)
c) State the factors which threaten unity in the church today. (7 marks)

a) Identify the main sources of Christian Ethnics. (5 marks)
b) What is the significance of leisure in traditional African communities? (7 marks)
c) Give reasons why employees should be given rest. (8 marks)

6 a) Explain Christian teachings on human sexuality. (8 marks)

b) How has science and technology helped to improve human life? (7 marks)
c) Identify ways in which Christians can help to control desertification. (5 marks)

MARKING SCHEME

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1a) The Jewish expectation of the messiah. (6marks)

They expected apolitical /military leader who would overthrow their colonial rules the Roman.
They expected a messiah who would rule the whole world from Jerusalem and receive tribute/ homage from all the nations of the world.

Messiah would be a descendant of David.

The messiah comes after the return of Elijah.
He would not associate with the poor sinners, and Gentiles.
He would perform wonders and miracles.

The establishment of his kingdom would be preached with cosmic powers.
The messianic kingdom would be established through God’s judgment on his enemies.

The messiah would come from a rich /noble family. (1×6=6)

b) What Simeon and Ann revealed about the life of Jesus when his parents presented him to the Temple for dedication.
Jesus was the messiah.
Jesus would bring salvation to Israel.

Jesus mission is universal.

Jesus would be light to the revelation of the Gentile.
Jesus was going to suffer for the sake of humankind.
He would deliver the Israelites from oppression.

His mission was to reveal God to humankind.
The coming of Jesus would cause division among people in Israel.

He would be source of joy to many people.
He would restore Jerusalem back to its glory. (Any 4×2=8)

c) Ways in which church leaders prepare people for the second coming of Christ. (6marks)
By preaching the good news.
Living exemplary lives/ being role models.

Helping the needy/ doing charitable work.

Condemning evils in society.
Providing family life education.
Sharing in the Lords supper.

Encouraging people to repent / getting saved
Providing guidance and canceling people eg prisoners.

Organizing seminars / workshops/ crusades to encourage people to holy lives. (1×6=6)

2a) How John the Baptist prepared the way for the messiah.

John preached repentance of sin and warned of the coming judgment.
He invited people to be baptized and be forgiven of their sins.

Baptized the people who repented.

Encourage the rich to share with the poor.
Advised people not to bear false witness against fellow human beings.
He told tax collectors not to take more than what was required.

He introduced Jesus to the crowd as the messiah (lamb of God)
He rebuked Herod for marrying Herodians.

He baptized Jesus in River Jordan. (Any 7×1=7)

b) Why Jesus submitted to John’s Baptism and yet he was not a sinner.
He wanted to confirm and encourage listeners of John to accept his baptism.
He wanted to identify himself with those with sin and he come to save them from sins.

He wanted to symbolically take upon himself man’s sins.

He saw it as away of fulfilling the Old Testament prophecies about the messiah.
He saw it as a last act of preparing those who were ready to receive the messiah in person.
He considered it as God’s plan of saving mankind.

Baptism provided an opportunity for the manifestation of the Holy Trinity.
Through Baptism Jesus was to get an assurance and confirmation that he was God’s son.

Through baptism he was to receive the anointing and power of the Holy Spirit to guide him for the messianic mission.
To prepare him for his earthly ministry. Jesus was about to start his public ministry.

(Any 8×1=8)

c) Ways in which the church in Kenya is continuing with the healing ministry of Jesus.
Establishing hospitals.
Establishing medical college.

Visiting and praying for the sick.

Encouraging and supporting medical research both morally and financially.
Teaching the public on preventive measures against STIs and AIDs/HV. (Any 5×1=5)

a) Testimony of the holy women regarding the resurrected Jesus Christ. (LK24:1-10)
The holy women notably Mary Magdalene, Joanna and Mary the mother of James went to the tomb in the morning of Sunday carrying spices to prepare Jesus’ body.
They found the stone rolled away from the entrance to the tomb.

They went in but they did not find the body of the Lord Jesus.

They stood there puzzled about this.
Suddenly two angles appeared to them.
Full of fear, the woman bowed down to the ground.

The angles asked them why they were looking for the living among he dead.
They were told Jesus was there but had risen as he had told them he would do on the third day upon his crucifixion death and burial while in Galilee.

The women went and told all these things to the eleven disciples of Jesus. (Any 7×1=7)

b) Why did Jesus disciples respond to the news of his resurrection with fear and disbelief?
The news was first brought by women who were despised / regarded lowly.
The disciples had witnessed the death// burial of Jesus.

They lacked faith / had little faith.

They had expected a glorious / implants political messiah/ not one that would die/ did not expect a spiritual messiah.
There was conflicting message about resurrection.
The tomb was heavily guarded by the Roman soldiers/ abide rock was placed at the entrance.

Influence from their historical background/ resurrect ions was impossible/ had not happened before/ was a strange new things. (Any 6×1= 6)

c) Why violence against women rampant in Kenya today.
Male chauvinism /superiority.

Women are vulnerable

Ignorance of the law/women don’t report cases of violence.
Attitude towards women.
Cultural beliefs/ norms.

Lack of laws /legislation guarding women against violence.
Lenient punishment by the law court /laws against offenders/ corrupt legal procedures.

Drug abuse/ alcohol.
Male dominated society in leadership. (Any 7×1=7)

a) Symbols used to describe the unity of believers in the New Testament.
i) The body of Christ 1cor. 12:12 – 27 Eph 4:1-12.
ii) The vine and the branches Jn 15:1-10.

The church Eph 5:21-32.

The bride Rev. 21:1-12, 2cor 11:2.
The people of God 1 Peter 2:9-10 (5×1=5)

b) Ways in which unity of believers demonstrated in the church.
They prayed together for one another.
They shared their meals together in their homes.

They shared their property.

They helped the less privileged.
They celebrate the Holy Communion together.
They met together for apostolic teaching and instructions.

They preached the same Gospel of Jesus Christ. (7×1=7)

c) Factors which threaten unity in the church today.
Minister interpretation of the scriptures.
Struggle for leadership.

Traditional and cultural difference.
The church’s stand on certain issues eg abortion and gay marriages.
Insecurity in the country.

Political interference.
Doctrinal differences. (8×1=8mark)

a) Main sources of Christian ethics.
Holy scriptures (Bible)
Teaching of Christian community eg church

Authoritative leaders eg church ministers.

Natural law
State law.
Human reason and experience/ conscience.

African cultural law. (5×1=5)

b) The significance of leisure in traditional African communities.
Leisure is activity used to thank God for the achievement in life.
It provides an opportunity to educate the youth in the community.

It in for the celebration of the good men of life.

It is an opportunity to socialize among friends and relatives.
It is used to develop talents.
It is used to establish and renew relationship with ancestors.

Community elders used leisure to solve community matters.
It is used to identify future community leaders.

It is through leisure activities that marriage partners are identified.
It is an opportunity for one to regain lost energy during work. (7×1=7)

c) Why employees should be given rest.
To regain lost energy.
To be able to attend to their families.

To be motivated to work better.

It is a requirement from labour organization world wide.
So that they do not cause destruction due to accumulated stress.
It helps them to develop good relationship with their with their employer.

It enables them to worship God. (8×1=8)

a) Christian teaching on human sexuality.
God created both male and female/ should marry.
Male and female complement each other ie share duties/ companionship.

Male and female have distinctive role in creation.

Both shares in the image of God.
The union of male and female is consummated in marriage / virginity is highly valued.
Male and female should live in harmony.

Procreate/ fulfill God’s command to multiply.
Sexual deviation s condemned (homosexuality, lesbianism, bestiality etc.

Male and female should respect each other, should regard each other as equal.
Sex is sacred/ a gift from God. (7×1=7)

b) How science and technology has helped to improve human life.
Means of transport and communication have improved social interaction and faster movement.
It has improved efficiency at work where machines are used.

Has led to improved agricultural development hence increasing food production.

It has enhanced human beings understanding of the environment leading to its better use such as through irrigation and weather forecasting.
It has brought better health care through modern medical technology.
Human beings are now better placed in security matters by use of radar, alarms and electrical fencing.

It has led to creation of job opportunities through industrial development.
Trade has been promoted through the use of computers and the internet. (4×1=4)

c) Ways in which Christians can help to control desertification.
Practicing afforest ration and reforestation programmes.
Practicing agro-forestry.

Using alternative energy sources as opposed to the public on how to preserve the environment.

Participating in environmental conservation programmes.
Providing education to the public on how to preserve the environment.
Giving financial assistance to bodies that control desertification.

Protecting all water catchments areas
Carrying out betters methods of farming

Reporting cases of forest destruction to relevant authority. (5×1=5)

SAMPLE 5

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CHRISTIAN RELIGIOUS EDUCATION PAPER 2

(a) Narrate the parable of the unfruitful fig tree (Luke 13: 6 – 9) (8mks)

(b) Explain the reasons why Christians are baptized. (5mks)

(a) Identify the relevance of Jesus’ Baptism to Christians today. (8mks)

(b) Explain ways in which the disciples of Jesus showed their support to his ministry. (7mks)

(a) Discuss how Jesus triumphant entry into Jerusalem failed to portray him as an earthly

Messiah. (6mks)

(b) Why was it difficult for Jesus disciples to believe that Jesus had resurrected? (8mks)

(6mks)

(a) Identify the lessons that a Christian can learn from the out pouring of Holy Spirit on the

day of Pentecost. (5mks)

(b) Describe how the unity of believers is expressed in the symbol of Body of Christ. (7mks)

(a) Outline the ways in which an individual acquires life skills. (5mks)

(b) State ways in which the youth can be encouraged to practice chastity. (5mks)

(a) Explain how the birth stories of Jesus fulfil the prophecies of Isaiah. (6mks)

(b) Identify the activities that took place following the birth of John the Baptist. (7mks)

(8mks)

MARKING SCHEME

SAMPLE 5

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(a) Outline the parable of the unfruitful fig tree as outlined in Luke 13: 6-9.

A man had a fig tree growing in his vineyard
He went looking for figs on it.
But found none.
He said to his gardener, “Look, for three years I have been coming here looking for fig on this tree.
I haven’t found any.
Cut it down.
Why? Answered the gardener, leave it alone sir.
Just this year, I will dig a trench around it and fill it with fertilizer.
If the tree does bear fruit next year so much the better.
If it does not bear you have to cut it down.

8 x 1 = 8mks.

(b) Explain the reasons for Christian Baptism.

To follow Jesus example and identify with him.
To receive the Holy Spirit who teaches, guides in their lives.
To prepare one to enter Kingdom of God.
Unites Christians under one body of Christ.
It effects complete forgiveness of sins.
It is an act of being born again.
It is a symbol of death and resurrection.
Through Baptism one is considered a child of God. 5 x 1 = 5mks.

(c) State lessons that Christians Learn about Jesus in his temptations in the wilderness.

Jesus was a human i.e. he felt hungry.
Jesus was Divine / Son of God/ messiah.
Jesus had power over Satan/ temptations
He was obedient to his father
Jesus was ready for his mission.
Jesus was well versed with scriptures.
He knew temptation comes from the devil as he told the devil to leave him.

1 x 7 = 7mks

(a) Identify the areas of conflict between Jesus and Jewish religious leaders. (8mks)

Sabbath observance Jesus healed on Sabbath while Jewish leaders taught no work should be done on Sabbath day.
Jesus mixed freely with sinners e.g. tax collectors gentile and unclean people.
He exposed hypocrisy of Jewish religious leaders openly.
Jesus disciples were not fasting like those of John the Baptist.
His divine authority, Pharisee felt it blasphemous of Jesus claim to forgive sins.
His claim to be messiah.
His claim that he can destroy the temple and build it in three days.
His teachings that the first will be the last and last will be the first in the Kingdom of God. 7 x 1 = 7mks.

(b) Ways disciples of Jesus showed their support to his ministry.

They left their families and followed him.
They accepted his teaching / obeyed him.
They invited him to their homes e.g. Levi and Peter.
Accompanied him in his work.
Addressed him as rabbi / teacher
Took care of his mother.
Peter cried when he remembered he had denied Jesus.
They were present when he was crucified on the cross.
They were happy when he resurrected.
Helped him to spread good news e.g. mission of 12 and 72. 7 x 1 = 7mks

(c) The obstacles faced by Christian leaders as they do their work.

Political interference with church work.
Corruption in the society.
Immorality among the church members.
Factionalism within church members.
Death / sickness/ family problems among the members.
Struggle for leadership among the members.
Hypocrisy of some of the members.
Increasing numbers of atheist.
Modern styles of living.

Any 6 x1 = 6mrks

(a) How Jesus’ triumphant entry into Jerusalem failed to portray him as an earthly messiah

Triumphant means successful or victorious.
He rode on a colt which is a sign of humility / poverty. He should have rode on a horse.
He wept on seeing Jerusalem since he already knew destruction awaiting the magnificent city.
He was angered by what he saw in the temple such that he chased away the business men violently.
He experienced loneliness at Mt Olives on the night he was betrayed because his disciples slept while he prayed.
The entry was not victorious because Judas Iscariot betrayed him.
He suffered physical and spiritual torture in the hands of the solders.
A criminal Barnabas was preferred for release while Jesus who innocent was chosen for crucifixion.
His closest disciple Peter denied him three times.

6 x 1 = 6mks.

(b) Reasons why it was difficult for the disciples to believe that Jesus had resurrected.

Disciples of Jesus had witnessed his death and burial hence they knew he had gone forever.
The report about the resurrection of Jesus was first reported by women who were highly despised in Israel.
The disciples expected a glorious triumphant political messiah and not one that would die.
The disciples lacked faith in the teachings and sharing that Jesus had with them
There was conflicting message and information about the resurrection of Jesus. He appeared to the disciples differently hence different reports.
The tomb was heavily guarded by the Roman soldiers and a big rock was placed in the tomb. Jesus could not pass all these without being noticed.
The influence from their historical back ground about resurrection where Sadducees did not believe in resurrection of the dead.
The disciples were ignorant of the divine nature of Jesus.

8 x 1 = 8mks

(c) Ways Christians live according to Jesus will as they await the second coming.

Avoiding Sins
Living under the guidance of the Holy spirit
Repenting their Sins / confessing their Sins.
Leading prayerful lives
Doing charitable works
Encouraging false prophets who keep on cheating them.
Patiently looking for signs of end of time.
Giving hope to others by preaching the good news to them.

6 x 1 = 6mks

(a) Lessons Christians learn from the out pouring of the Holy Spirit on the day of Pentecost

It is important for Christians to meet together in fellowship as the Holy Spirit was poured on the disciples as they worshipped together.
Jesus keeps his promises, he had promised before he ascended to heaven the Holy Spirit and it came.
The Holy Spirit empowers believers to be in the forefront in preaching God’s word.
Baptism of the Holy Spirit is important in salvation.
Christianity is a universal religion this is shown by the disciples proclaiming good news in different languages.
Shows the importance of vernacular languages as the best means of spreading the good news/ Shows the importance of bible translations into vernacular languages so that everyone reads and understands.
As Peter did, Christian leaders should boldly explain to the world God’s continued work of salvation to mankind. 1 x 5 = 5mks.

(b) How the unity of believers is expressed in the symbol of the Body of Christ.

Church like a human body has many parts that functions for the good of the whole body.
As the body of Christ, the church has many members from different back grounds but all have been baptized in one spirit.
Every member of the church has a role to play for her development.
Different members have different spiritual gifts that are used for the development of the church.
All members of the body of Christ are equal because they serve the same God.
All spiritual gifts have been given by God hence none is inferior to the other.
Christians strive to remain in Christ and also invite others to become members of the body of Christ.
Unity is achieved through the practice of virtues such as humility gentleness and patience.
The spirit is the unifying power that brings Christians together as one body of Christ.
Jesus is the head of the Church.
Christians share one faith in Christ.
Christ has given Christians gifts for the purpose of building the body of Christ.

7 x 1 = 7mks

(c) How Churches discipline those who cause disunity in the church today.

They are denied leadership positions
They are denied some services of the church such as wedding rites, burial rites etc
They are given a warning
They are charged a fine
Those who cause disunity are reprimanded and asked to apologise.
Workers who cause disunity can lose their jobs.
Leaders who cause disunity are asked to resign.
Some church excommunicate members who blaspheme.
Some times they may be suspended from church duties and church service. 8 x1 = 8mks

(a) Outline the ways in which an individual acquires life skills.

Through education- basic education gives facts concerning skills such as decision making, creative thinking and interpersonal skills.
Religious instruction- this inculcates skills that contribute to spiritual and moral well being.
Observation- by observing behaviors or practical life experience of others an individual acquires desirable skills.
By practice- by deliberately putting into practice initiating desirable skills/ doing what is learnt and imernalisiny the skills.
Mass media- by reading news papers, watching educative Tv programs listening to radio, one can acquire life skills
Personal experience- the economic social and political experiences a person goes through will consciously or unconsciously influence them to develop desirable skills.

5 x 1 = 5mks

(b) Ways in which the youth can be encouraged to practice chastity.

Openly discussing issues associated with unchastely.
Seminars and work shops to give one another skills.
Rewarding the sexually responsible youth who are prime
Guidance and counseling on reproductive health.
Setting good examples.
Advising them on the mode of dressing.
Advising them to pray and follow God’s commandment.
Teach them that sex is only good for married people.
The bible teaches against adultery, fornication rape etc.

7 x 1 = 7mks

(c) Features of a Traditional African family.

Procreation
Polygamy
Women were surbonate to their husbands
Divorce was rare
Relationships between in-laws were maintained well.
Marriage promoted one’s status.
Marriage was a covenant relationship and was not to be broken unless there were extreme problems.

7 x 1 = 7mks

(a) How the birth of Jesus fulfils the prophecies of Isaiah

Jesus mother Mary was a virgin as foretold by Isaiah
He was to inherit the throne of his father David
He was born of the Holy Spirit to fulfill the prophecy that the messiah will be filled by the Holy Spirit.
Jesus birth is in a humble and poor setting to justify the prophecy that he humble & simple.
Jesus was a descendent of King David
Jesus’ rule was to last for ever from the message of the Angle to Mary.
In the magnificat, the humble were to be uplifted to show that he was to be a just ruler.
He was to be named Immanuel to fulfill Isaiah’s’ prophesy.

6 x 1 = 6mks

(b) Identify the Activities that took place following the birth of John the Baptist

Relatives and friends gathered at Elizabeth’s home
They celebrated birth of a baby boy.
There was giving of gift and presents.
On the eight day the child was circumcised.
There was an argument over the name of the baby/ The mother gave the name John.
Zechariah wrote down the name
Zechariah regained his speech
He praised God for what He had done
The baby was given the name John.
Zachariah song the Benedicts in praise f God. 7×1=7 marks

(c) How a Christian couple should respond to the problems of childlessness

They should accept their state
They should consult medical experts for advice
They should pray for God to open their ways
They may think of adopting children.
They can visit children’s Homes to offer their services
Attend guidance and counseling sessions on family life Education
Read literature on childlessness as a way of getting the solution.

SAMPLE PAPER 6

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CHRISTIAN RELIGIOUS EDUCATION PAPER 2

a) Identify incidences in the life of Jesus that fulfilled Old Testament prophecies about the messiah. (7 mks)
b) What were the Jewish expectations of the Messiah? (6 mks)
c) Give ways in which John the Baptist was an outstanding prophet. (7 mks)

a) Relate the healing of the paralytic in Luke 5:18 – 26. (7 mks)
b) Identify ways in which the disciples of Jesus supported Him. (7 mks)
c) Jesus encountered many problems as a result of performing miracles. Identify those problems.

(6 mks)

a) Describe the triumphant entry of Jesus into Jerusalem according to Luke 19:28 – 40.(7 mks)
b) Identify the signs of the end of times as taught by Jesus. (7 mks)
c) How should Christians prepare themselves for the second coming of Jesus Christ?(6 mks)

a) Relate what Peter said about Jesus on the day of Pentecost. (7 mks)
b) Identify the fruits of the Holy Spirit as taught by St. Paul. (6 mks)
c) What are some of the ways in which Christians in Kenya fulfil Paul’s teaching on believers as “The body of Christ” ? (7 mks)

a) Explain the effects of incest. (7 mks)
b) Explain why violence against women is rampant in Kenya. (6 mks)
c) Identify some of the strategies put in place by the government to ensure justice and fairness for (7 mks)

a) Outline ways in which you can make the work of your employees enjoyable. (7 mks)
b) Give factors that may lead to misuse of leisure in modern society. (7 mks)
c) What are the advantages of genetic engineering to mankind? (6 mks)

MARKING SCHEME

SAMPLE PAPER 6

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1 a) Incidences in Jesus life that fulfilled O.T prophecies about the messiah

Prophecies that he would be born in linage of David fulfilled when the Angel was sent to Mary who was engaged to Joseph who come from David’s lineage
Nathan prophecy that the messiah would establish an everlasting dynasty is fulfilled when Angel Gabriel told Mary that the child to be born would reign over the house of Jacob forever
Isaiah prophecy of a virgin conception is fulfilled in Angel Gabriel’s message to mary that she was going to conceive by the power of the Holy spirit
Isaiah prophecy that he will be called Immanuel is fulfilled in Angel’s message to Joseph that Mary’s son would be called Emanuel
Jesus was born in Bethlehem also the birth place of king David a fulfillment of prophet Micah’s prophecy about a rule of Israel whom God will raise up from the town of Bethlehem
Jesus was hailed as the messiah , descendent from David by the blind beggar of Jericho-fulfillment of Nathan’s and Jeremiah’s prophecy that the Messiah would come from the family of David
Jesus quoted from Isaiah 61: 11 that the messiah would set the captives free e.t.c an affirmation that He was the messiah that Isaiah had talked about.
Isaiah’s prophecy about a messiah who would work miracles was fulfilled when Jesus turned water into wine in Cana, raised the dead , cast out evil spirits etc
Messiah would be betrayed by a close friend is fulfilled when Judas, one of this disciple betrayed Jesus
Isaiah prophecy about a suffering servant is fulfilled through the passion and death of Jesus Christ

Any 7×1= 7 mks

1 b) Jewish expectations of the messiah

Jews expect any one who had a specific mission to fulfill from God eg priests, prophets or kings
Expected one who would be a universal messiah from the lineage of David
Expected a political / military messiah / liberator/ conqueror who would drive out the Romans
One who would establish an everlasting kingdom for them
Messiah who would bring God’s blessings and complete peace and joy to Israel
One who would lead Israel into a time of great national power and prosperity
One who would come in future with more power/ authority than David / Moses etc 6×1= 6 mks

c) Ways in which John the Baptist was an outstanding prophet

He was filled with the Holy spirit since birth
His birth was foretold by an angel
Had a prophetic ministry
Invited people to baptism/ repentance
He was in the spirit/ power of Elijah
Led an austere life
Denounced the Pharisees for their self sufficiency castigates Herod for incest
Announced the coming of the kingdom of God 7×1 = mks

2 a) Relate the healing of the paralytic man Lk 5: 18- 26

One day when Jesus was teaching is a house among his audience were scribes and Pharisees
Some people brought to him a man who was paralysed , but finding no way to bring him in because of the crowd , they went up on the roof and let him down with his bed thought the tiles , into the midst before Jesus
When Jesus saw the great faith of those who brought him, told the paralytic “ your sins are forgiven my friend”
The Pharisees who heard him grumbled, saying. it was only God who could forgive sins
To prove to the Pharisees and scribes that he had divine power to forgive sins, he challenged them, `is it easier to say your sins are forgiven you or to say get up and walk?’
Jesus then ordered the paralytic to rise up, pick up his bed and go home.
Immediately the paralytic rose up and departed to his house glorifying God
The people were amazed and they glorified God and were filled with fear. 7×1= 7 mks

2 b) How Jesus’ disciples supported him

Joined him in prayer
They left their jobs , families to follow him
Invited him to their homes
Served him e.g. prepared the Passover meal
Kept him company
Helped him spread the gospel
Accepted the miracles of Jesus and those performed in his name
They put their faith in him
Gave him physical protection e.g. peter
Peter was prepared to die with Jesus
Sought advice from him
Some financed his ministry e.g. Joan
They performed miracles in his name 7×1= 7 mks

2 c)

Was accused of using the power of the devil to perform miracles
Accused of violating the Sabbath
Crowd followed him some hoping to receive a miracle from him eg after feeding the five thousand
Jewish religious leaders plotted to have him killed as he was very popular
Jesus was accused of blasphemy (claiming to be the messiah)
Jesus was mocked / ridiculed
He came under great scrutiny from Pharisees
He was rejected in some places where his miracle brought loss e.g. casting out demons from pigs
At times, he/ his disciple were forced to take refuge to escape from the crowd.
He became exhausted. 6×1 = mks

3 a) Jesus triumphant entry into Jerusalem

When Jesus came near Beth phage , Bethany he sent two of his disciples
They were to get him a colt on which nobody had ridden
They were to unite it and take it to Jesus
If the owner asked, they were to say that the master needed it
When they got the colt, they threw their garments on it and helped Jesus to sit on it
As Jesus rode along they spread their garments on the road
The crowds following him rejoiced and praised God for all the wonderful works Jesus had done ie healing & feeding the hungry
The Pharisees objected to the crowds singing and asked Jesus to silence them
Jesus responded that if his followers were silent the stones would cry out 7×1= 7 mks

3 b) Signs of the end times as taught by Jesus

People would come claiming to be Jesus, the messiah , the son of god
Wars between nations would arise
Natural calamities like earthquakes , famine and plagues would occur
Strange and celestial beings would fall from the sky
Jesus’ disciples would be arrested , persecuted and imprisoned
The disciples would also be betrayed to the authorities by close relatives and even put them to death
Disciples would be hated on Jesus account
There will be disruption in the sky and in the sea
People will faint from fear as they witness theses signs 7×1 = 7 mks

c) How Christians should prepare for the second coming

Not losing hope in the face of trials and tribulations because the kingdom of God with its promise of a new life will definitely come.
Leading a righteous life, they are to avoid too much feasting & drunkenness
Avoiding being pre- occupied with worries and cares of this world
Being watchful by praying so that God will give them courage & strength to stand firm in their faith
Preaching the word of God to those who have not heard it
Being obedient to God’s commandments
Through helping the needy
Reading the bible
Fellowship 6×1= 6 mks

4a) What peter said about Jesus on the day of Pentecost

Jesus was accredited by God to do miracles, wonders and signs for Gods purpose
He was put to death by the evil plans of the Jews
God raised him from the dead & disciples were witnesses
God raised Jesus to life and he is exalted at the right hand was what the people had witnessed
That God had made Jesus both lord and Christ
That David foresaw the resurrection of Jesus
That David recognized the divinity of Jesus as lord
God through Jesus had poured the Holy Spirit and that was what the people had witnessed.

7×1 = 7 mks

4 b) Identify the fruits of the Holy Spirit as taught by St Paul

Love
Peace
Patience
Kindness
Goodness
Faithfulness
Gentleness
Self control
Generosity

6×1= 6 mks

4 c) Ways in which Christians in Kenya fulfill Paul’s teaching on believers as “the body of Christ “

(7 mks)

Believers belong to the universal church(the body of Christ)
Christians are one in Christ since all have been baptized into one body by the spirit , whether Jews, Greeks, slaves or free etc
There are many Christians in the church but every member plays a vital role for the common good of the church i.e. preachers, teachers, miracles, worker , healers, ushers, counselors, administrators, pastors etc
Christians are honoured by God who distributes spiritual gifts to them as each requires
Christians are members of the same body since they share the same rights and privileges in Christ
Christians work harmoniously for the benefits of the church just like all body parts must be in harmony for the human body to function properly
Christians should be responsible for one another for if one part of the body suffers the whole body suffers too
Christians practice cooperation in all fields, social economical and political, for the benefits of the society.

7×1 = 7 mks

5) Effects of incest

Brings shame & guilt among parties involved / psychological problems
Destroys the relationship within the family and leads to divorce
Destroys dignity, self respect & self esteem of victims
Leads to pregnancy/ unwanted children
May lead to abortion which may be a health hazard
Victims may contract sexually transmitted diseases
Undermines the healthy relationship between members of the family
Destabilizes the kinship system
Brings Gods judgment
Chances of abnormality in children is high
Boys or girls who are abused may never establish a health relationship with members of the opposite sex 7×1 = 7 mks

b) Why violence against women is rampant in Kenya

Male chauvinism / superiority
Poverty
Women are vulnerable
Ignorance of laws/ women do not report cases of violence
Attitudes towards women
Cultural beliefs/ norms
Lack of laws/ legislation guarding women against violence
Lanient punishment by the court of law against offenders/ corruption legal procedures
Drug abuse/ alcohol
Male dominated society in leadership 6×1= 6 mks

c) Strategies for ensuring justice and fairness for all

Enacting relevant laws/ legislatives to safeguard from injustice
Equal distribution of resources to all parts of the country/ citizens
Educating the citizens about their rights
Guiding and counseling
Through rehabilitation
Stamping out corruption in our community /nation condemning corrupt practice
Upright/ moral , law enforcement officers
Establishment of code of conduct for all those serving in various offices
Punishing wrong doers/ offenders
Creation of employment /balanced opportunities
The support of government for human rights activities and implementation of international conventions on human rights 7x 1= 7 mks

6 a) Outline ways in which you can make the walk of your employees enjoyable

When you pay them on time/ a just wage/ equivalent wage for work done
When you pay them according to the labour contract
Extra work , extra pay
Enable them to rest/ have leisure time
Equal work, equal pay / no discrimination
Provide favourable and safe working conditions
When you enable them to undergo personal development and training which can give them an opportunity for promotion
When you allow them , join or form associations or trade unions
Treat them with respect and dignity / avoid mistreatment of employees/ avoid over taxation
Be considerate to the employees’ grievance
Give them mobiration to the employees e.g. salary incremets and promotions 7×1= 7 mk s

6 b) Give factors that may lead to misuse of leisure in the modern society

Ignorance on proper use of leisure
Expensive and inappropriate forms of leisure activities
Failure to provide facilities
Bad company
Lack of variety of the leisure activities
Failure to collect and balance the types (passive and active)
Misunderstanding in the family
Poor planning may lead to over indulging
Poverty / too much wealth
Feelings of insecurity
Social oppression / injustice may interference with people’s freedom to involve in leisure
Influence of mass media 7×1 = 7 mks

6 c) Advantages of genetic engineering to mankind (6 mks)

It increases the rate of growth and maturity of livestock – beneficial to human beings in livestock products
It has aided research in the manufacture of human growth harmone
It increases disease resistance in crops
To originate generic finger printing for forensic work
To produce genetically engineered bacteria
Helps to identify / determine the biological parents of a child
Increase plant & animal yields for the benefits of an ever increasing human population

(6 x 1 = 6mks)

SAMPLE PAPER 7

313/2

CHRISTIAN RELIGIOUS EDUCATION PAPER 2

a) Narrate the message of the Angel of the Lord to Zachariah in the annunciation of John’s birth. (8mks)
b) Examine the evidence in the Luke’s Gospel that Jesus was a messiah sent to the poor. (7mks)
c) What do Christians learn from Simeon’s prophecy when baby Jesus was presented in the Temple? (5mks)

a) Describe Jesus healing of the Centurion’s servant. (8mks)
b) Outline the parable of the Great Feast. (6mks)
c) Highlight the role of Christians in a burial Ceremony. (6mks)

a) Highlight on the injustices that characterized the trials of Jesus. (8mks)
b) Show how Joseph of Arimathea acknowledged the Lordship of Jesus. (7mks)
c) What is the relevance of Jesus teachings on the cost of discipleship to Christians? (5mks)

a) Outline the events that took place on the day of Pentecost. (8mks)
b) How did Paul address the abuse of the Lord’s supper at Corinth? (7mks)
c) Identify the manifestation of the Holy Spirit in our churches today. (5mks)

a) Compare the Christian and Traditional African understanding of human sexuality. (8mks)
b) Why should a Christian prefer to live a celibate life as an alternative to marriage? (6mks)
c) Outline the social effects of HIV/AIDS in the modern family. (6mks)

a) Show ways in which Jesus upheld the dignity of work during His life and ministry. (8mks)
b) Identify the Christian criteria for spending leisure time. (7mks)
c) How has over involvement in alcohol affected the growth of our country? (5mks)

MARKING SCHEME

SAMPLE PAPER 7

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a) Message of Angel to Zachariah
i) Told him not to be afraid
ii) That God had answered his prayers

His wife would bear him a child / son

The child was to be called John
Many would rejoice with Johns birth
The child would be great before Lords eyes

The child will not take wine nor strong
John will be filled with holy spirit from birth.

John will bring many back to God.
He will go before the Lord strong like Elijah.
He will bring the fathers and children together.

He would get the Lords people ready for Him. ( 1 x 8 mks)

Evidences form St. Luke that Jesus came for the poor.
1. The Parents of Jesus were from poor background
2. He lived in Nazareth, a town of the poor people

He was born in a cattleshed/ laid in a manger.

His birth was revealed by the shepherds
At dedication, a pair of doves / pegions were offered
During ministry he associated more with the poor /suffers

Some of His Apostles were from poor background.
He taught that God’s Kingdom was for the poor

He told John’s Apostles that Good News was being preached to the poor.
He challenged the rich to share their wealth with the poor ( 7 x 1 mks)

Lessons learnt from Simon’s prophecy
1. God keeps His promises
2. Jesus is the salvation that people were waiting for.

He said that Jesus’ salvation is universal

All Christians are witnesses to the salvation of Christ
By faith in Christ, we escape God’s Judgment
Should be ready for ridicule and persecution in the name of Christ

Christians learn to receive Christ with glory / Christ has brought glory to Christians. ( 5 x1 mks)

a) Jesus healing of the centurions servant in universal.
i) Roman centurions servant was sick / about to die
ii) He requested Jewish leaders to go and tell Jesus to heal his servant.

Jewish leaders went and told Jesus how the centurion had built them a synagogue

They said that the centurion needs to be helped and Jesus went with them.
Centurion sent messenger to meet them.
He said that he was not worthy to receive Christ

He requested Jesus just to order and His servant would be healed.
Jesus was surprised to hear this.

He/ Jesus told the crowd that He had not found such faith in Israel.
The messenger went back and found the servant well / healed.

( 1 x 8 mks)

Outline the Parable of the Great Feast.
i) A man gave a feast and invited many guests.
ii) When it was time, he sent his servants for quests.
- The invited gave excuses and did not come.

One said he had bought a field & must attend to it.
Another said he bought oxen and had to try them out in the field.
One had just married and could not leave his new bride.

The servants reported these to their master.
The master sent them to the streets to bring the poor/ lame / blind / cripples to the feast.

The master said that non of the invited guests would taste his dinner.

( 1 x 6 mks)

c) Role of Christians in Burial ceremony
i) Read scriptures / from the Bible
ii) Officiate the burial ceremony

Sing Gospel songs / relevant Hymns

Preaching about death and God’s Kingdom.
Provide counsel / hope to bereaved members
Give company to the bereaved family

Participate in the reception / serving visitors
Offer prayers / intercede for the bereaved and the deceased.

Provide burial necessities casket, food, attire.

a) Injustices that characteristics Jesus Trials.
i) Jesus was not told the cause of His arrest.
ii) He was mocked & beaten even before trial

False accusations were brought against Him

The Sanhedrin insisted on His death after Pilate found Him innocent.
The crowd intimidated Pilate to put Him to death
Though innocent, Pilate was to have Him whipped

A criminal /Barnabbas was released to have way for His death.
Herod treated Him with contempt / ridicule at the trial.

Peter who followed Him was treated with threats / intimidations on night before trials ( 1 x 8 mks)

How Joseph of Arimathea acknowledged Jesus’s Lordship
1. He believed in the Kingdom of God.
2. Disagreed with the Sanhedrin about crucify Jesus.

He went to pilate and asked for Jesus body.

He removed the body of Jesus from the cross
He wrapped Jesus body with expensive linen
He buried Jesus in a personal tomb

He buried Jesus the very day / Friday in order for him to attend and obey Sabbath day/ Saturday.
He recognized the righteousness of Jesus at cross.

( 1 x 7 mks)

Relevance of Jesus teachings on cost of discipleship today.
1. Motivates Christians to sacrifice comfort in order to spread the Gospel.
2. Reminds Christians to have strong faith.

Discourages Christians from desiring any aspect of their past sinful lives.

Guides Christians to forsake anything that is an obstacle to salvation.
Teaches Christians to give priority to Jesus over all other things.
Hardens the Christians to accept suffering and rejection for Jesus sake.

Gives Christians the spirit of spreading the Gospel.
Makes Christians to be ready to serve the poor.

Enables Christians to convert persecution into glory for Christ.

( 1 x 5 mks)

a) Outline of events of Day of Pentecost
i) All the Apostles gathered in one place.
ii) Suddenly a noise like strong wind filled the house.

Tongues of fire landed on each of their head

Disciples talked in tongues understood by the crowd
The crowd wondered in amazement and confusion at the preaching about Jesus.
Part of the crowd accused disciples for being drunk

Peter stood up and explained the meaning of the event / quoted from the book of Joel about Holy Spirit.
Peter preached to the crowd about the Gospel of Jesus ministry, death and resurrection

They asked Peter what they should do to save themselves
Many accepted to be baptized and about 3,000 people repented.
The crowd dispersed and each went away / home talking about Jesus and the promised Holy Spirit.

( 1 x 8 mks)

How Paul address abused of Lords supper at Corinth
1. Told them that it was for spiritual not physical satisfaction
2. Told them to wait for one another in the celebration

Advised them to eat and drink in their homes before partaking Lords supper.

Told them that they were to perform all the events that took place at last supper.
Reminded them that the bread and wine were the body and blood of Jesus Christ.
Reminded them to re- examine self before partaking Lords supper.

Talked of God’s judgment upon those who are not worthy to receive Lords Supper.

( 1 x 6 mks)

Manifestations of the Gifts of Holy Spirit Today.
1. Translations of tongues are common
2. Christians tell the future events through prophesy

Church leaders are elected based on their wisdoms.

Faith healing is evident in Churches today.
Preaching has been enhance by gift of knowledge
Christians have felt wonders in life by the gift of miracles e.g. found innocent in courts of law.

Christians have condemned others found to leading others astray by gift of discerning
Gift of love is expressed by Christians helping the needy in the society

( 1 x 6 mks)

a) Compare Human sexuality in Christian and African Contexts
i) Sexuality is sacred / gift from God. / initiated by God.
ii) Virginity is up help and only broken in marriage

Procreation is allowed at marriage only

Marriage is enhanced by unity, co-operation among the couple.
In both set ups, there are laid down rules and regulations that controlled sexual feelings.
Have similar sexual offenses such as incest, adultery, unfaithfulness, fornication.

Marital / sexual offences are punishable by murder, fines etc.
In both, male – female relationships are controlled especially among the youths by either instilling moral values or supervision.

Sex education is taught to the children and youth to equip themselves with knowledge and life skills in sexual issues.
The sexually up right act as good role models and are rewarded for their good morals e.g. by words of praise ( 1 x 8mks)

b) Reasons for celibate life in Christians
i) It is a gift and blessing from God
ii) Promotes total commitment and service to God

One of the vocations that strengthens self control toward sexual feelings

Creates ample time to serve God and His creations
Ensures that the Body, mind and soul of the Christians concentrated on heavenly ideal
Helps to Christians to avoid worldly problems i.e. becoming a slave of another person

It enables one to be a good role model for other Christians to emulate benefits in God’s Kingdom than marriage.
Celibacy reduces ones vulnerability to commit sins ( 1 x 6 mks)

c) Social effects of HIV and AIDS on modern family .
i) Increases frequent visits to medical facilities
ii) Death of the infected creates a gap never to be filled in the family

hence inspiration

Leads to creation of other social groupsg. orphans, widow, widower in the family

Changes the social roles in family e.g. child – headed family when parents are dead.
Social amenities g. schools experience high drop out rates due to orphaned learners or as caretakers of sick parents.
Discrimination in social places in family meetings.

Stigmatization by family members or the general society to the infected and affected family.
The infected members tend to withdraw, isolate themselves from other family members hence fail to make decisions on family matters. ( 1x 6 mks)

a) Ways in which Jesus upheld dignity of work
i) He was a worker / was a carpenter.
ii) Established God’s Kingdom on earth by doing the work of preaching, forgiving, miracles etc.

He called different workers to be His disciples e.g. fishermen, tax collectors e.t.c

Used work to make His teachings clearer e.g. parables of the sower, vineyard etc.
He solved situations that reduced work / healed Simon mother in-law for him to be effective at work.
Gave advice that promoted work e.g. “Pay to Caesar (taxes) what belongs to him.

He promoted division of labour by commissioning the twelve disciples to different parts and perform various works of God.
He discouraged overworking / He rested and worshipped on a Sabbath day.

Jesus called adults to be His disciples to ridiculed idea of child labour.
He found fulfillment in His work / Expressed His work with compassion and never complained at work. ( 1x 8 mks)
Christians guidelines to spend leisure time.
1. To serve God for it was ordained by Him
2. To have fellowship with other people.

Should come after the daily routine works.

Be used to do good /show love to others.
Be used for worship and praise to God.
Be used to break monotony from the routine work / make work interesting.

It does not mean laziness /lazing around.
Leisure activities be in accordance to God’s laws and wills.

Should be done with moderation for relaxation of body, mind and soul.

( 1 x 7 mks)

Effects of alcoholism among civil servants
1. Frequent absentism reducing quantity of production
2. May kill the morale to work among fellow workers due to quarrels, abuses, fights e.t.c

Lateness and hangovers reduces efficiency at work

Working while drunk may lead to mishandling of machinery, breakdowns and accidents
Ill health of the worker may lead to high cost of treatment and low productivity
Poverty due to misuse of salary hence none to invest in the national sectors for growth.

Retrenchment / dismissed from work leads to incontinous production due to difficulty in getting suitable replacement.
Death of the worker that lead to loss of qualified and experienced servant, hence reduces the efficiency and effectiveness of production.

( 1 x 5 mks)

SAMPLE PAPER 8

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CHRISTIAN RELIGIOUS EDUCATION PAPER 2

a) With reference to St. Luke’s Gospel, state the mission of John the Baptist as prophecised by his father in the Benedictus. (8mks)
b) What did Angel Gabriel reveal to Mary concerning Jesus in the annunciation story? (5mks)
c) Give reasons why children should be introduced to the worship of God at an early age. (7mks)

a) With reference to the story of the woman caught in adultery, explain the teachings of Jesus on forgiveness. (8mks)
b) What is the New testament teaching on the Jewish attitude towards sin? (7mks)
c) State the actions the church members would take in handling cases of dishonesty. (5mks)

a) Describe the arrest of Jesus as narrated in (Luke 22:47-53) (9mks)
b) Why were Jesus’ disciples reluctant to listen to the Holy Women’s testimony concerning the resurrection of Jesus. (6mks)
c) Give reasons why modern Christians should accept to suffer in Christ’s name. (5mks)

a) Explain the role of the Holy Spirit as taught by Jesus. (8mks)
b) Relate the message of Peter on the day of Pentecost. (7mks)
c) How do Christians misuse Spiritual gifts in the church today? (5mks)

a) identify irresponsible sexual behaviour condemned by Christians. (6mks)
b) Give reasons why domestic violence is widespread in Kenya. (7mks)
c) Explain the role of the church in controlling abortion in the society today. (7mks)

a) Identify the life skills that are useful to personal development. (5mks)
b) Give reasons why Christians are against Euthanasia (7mks)
c) Explain how modern media technology has enhanced evangelization. (8mks)

MARKING SCHEME

SAMPLE PAPER 8

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a) With reference to St. Luke’s gospel state the mission of John the Baptist as prophecied

by his father in the Benedictus.(7mks)

He would be the prophet of the most high God.
He would go before the Lord to prepare the way for him.
His duty would involve imparting knowledge of salvation to the people
He would call people to repentance and forgiveness in order to restore the right relationship with God.
He would give right to those people living in darkness and guide them into the path of peace.
He would tell of God’s mercy and care on human beings
He would preach God’s love and justice on earth.

1 x 7 = 7mks

b) What did the Angel reveal to Mary concerning Jesus in the annunciation story (6mks)

– Jesus was going to be great

– He would be called the son of the most high God.

– He would be given the throne of his father David.

– He would rule forever.

– Jesus was going to be an everlasting King and his Kingdom would have no end.

– He was to bring salvation to mankind.

– His conception was to be affected by the power of the Holy Spirit.

1 x 6 = 6mks

c) Give reasons why children should be introduced to the worship of God at an early age(7mks)

– To create in them an awareness of God / creation of the universal man.

– To learn the true teachings of their church.

– To familiarize themselves with the religious practices of their church / sacramental

life of the church.

– To learn to trust / obey God/ Give themselves to God.

– To learn the prayers / songs of worship.

– To enable them become members of the church.

– To initiate the example of Christ

– To give them sense of belongings to the Christian community. 1 x 7 = 7mks

a) With reference to the story of the woman caught in adultery, explain the teachings of

Jesus on forgiveness (8mks).

God is merciful towards sinners / he gives a chance to repent / come back to him.
God forgives all types of sin / according to the Jews such a sin is too big to be forgiven.
We should not judge others / all are sinners and no one has a right to judge others.
Jesus does not condemn sinners / Jesus loves sinners but hates sin.
We should avoid sinning after being forgiven / those whose sins are forgiven, should live according to the law of God/ live righteously.
We are all sinners no one is righteous everyone needs forgiveness.
We should not kill sinners / life is sacred and only God can take it away.
We should be tolerant – sinners / we should love sinners / pray for them / help there to change and take them to God. 8 x 1 = 8mks

What is the New Testament teaching on the Jewish attitude towards sin?(7mks)
- Sinners are outcasts, they are not to be mixed with.
- Sinners are stoned killed executed
- Sinners are despised / rebuked
- Sinners are hated.
- Sinners are grouped / classified
- Sin is inherited
- Sin causes ill health /sickness / suffering God punish sin.
- Only God can forgive sin/ no human being can forgive sin. 1 x 7 = 7mks

State the actions the church members would take in handling cases of dishonesty (5mks)
- Find out what problems that members have/ reasons for dishonesty.
- Help the people to meet their basic needs
- Offering guidance and counseling
- Forgive them.
- Encourage tem to pray.
- Report the matter to the authorities / discipline them.
- Help the people to set up income generating projects
- Fellowship with them
- Preach to them
- Setting good example for them to emulate
- Delegating duties responsible persons.
- Teaching them to live responsibly
- Encourage them to avoid situations which could lead them to dishonesty.
- Pray for them (5 x 1=5mks)

a) Describe the arrest of Jesus as narrated in Lk 22: 47- 53 (9mks)
- Jesus was arrested at olives at evening while with his disciples
- He was arrested by a team led by Judas Iscariot.
- The team included the Chief priest, temple officials , Romans soldiers.
- Judas moved forward and kissed Jesus.
- Jesus asked Judas are you betraying the son of man with a kiss.
- The team was heavily armed.
- One of the discples struck of the ear of the chief priest’ servant.
- Jesus quickly ordered for peace stopped resistance.
- He touched the ear of the servant of the Chief priest and it got healed
- He asked why they had came to arrest him while he had been in the temple with them daily
- They arrested Jesus and took him to the house of the Chief priest. ( 9 x 1 = 9mks)
Why were Jesus disciples reluctant to listen to the Holy women’s testimony concerning the resurrection of Jesus ? ( 6mks)
- The women were looked down upon/ despised in the Jewish community / seen as gossip
- The disciples had witnessed Jesus’ death and burial.
- Jesus had not appeared to them therefore they though it was a lie.
- It had never happened before / it was an extraordinary event.
- They had little faith / weak faith in Jesus as the messiah.
- They were still in a state of shock having lost their master / state of fear.
- There were confliting information about the resurrection.
- They knew the tomb was heavily guarded.
- They had forgotten Jesus teachings about destroying the temple and rebuilding after three days.He had hinted about his coming suffering and death in Jerusalem. 6 x 1 = 6mks
- They were influenced by their historical background on resurrection .(Lk 20: 27 – 40)
Give reasons why modern Christians should accept to suffer in Christ’s name(5mks)
- To imitate Christ / because Christ suffered to bring salvation.
- Suffering strengthens Christian faith
- To have the experience/ feeling of the suffering
- To protect the unfortunate / defend the rights of the weak.
- To save and to support the needy with basic needs
- To act as a role model to the young Christians
- To harden their bodies against temptations of the flesh.
- Jesus came for the poor and the suffering
- To demonstrate the glory of Christ over evil forces. ( 5 x 1 = mks)
a) Explain the role of the Holy Spirit as taught by Jesus ( 8mks)
- He would be a counselor / advocate/ a comforter i.e. aiding believers / helper.
- He would convict people about sins, righteousness and the coming judgment
- He would remind the believers everything that Jesus had taught them.
- He would reveal the truth concerning God the father and Jesus the son.
- He would pass judgment to the sinners
- He would declare things to come through the believers
- He will glorify Jesus by declaring what belongs to Jesus Christ from the father.
- He would teach the believers of all things and reveal the sins of the world.
- He would give authority to the believers to forgive sin.
- He would give the believers power to become witness of Jesus Christ.
- He would enable the believers to discern and expose the secret heart of sinful people.
- He would affirm the right of Jesus as the son of God. ( 8 x 1= 8mks)
Relate the message of Peter on the day of Pentecost (7mks)
- What was happening was the fulfillment of Joel’s prophecy about the outpouring

Of God’s spirit/ disciples were not drunk

The outpouring of the Holy spirit was a positive proof that the messianic age had arrived through Jesus Christ.
Jesus was from Nazareth through who God worked miracles.
Jesus suffered and was crucified by sinful people in accordance with God’s plan.
God raised Jesus up to fulfill the prophesy if King David.
The apostles are living witnesses to the resurrection of Jesus.
God had exalted Jesus and made him both Lord and Saviour
Jesus is a descendant of David.
Peter told the people to repent so that they could be forgiven and receive the gift of

the Holy Spirit. ( 7 x 1 = 7mks)

How do Christians misuse spiritual gifts in the church today? ( 5mks)
- Rivalry about superiority because of spiritual gifts e.g. those who speak in tongues belief that God is closer to them. Thus brings division
- Some preachers /Christians give false prophecy in order to gain favour e.g They prophecy peace when there is no peace.
- The are never interpretted, hence they do not benefit or edify the listeners
- Some Churches completely ignore spiritual gifts, they claim that the gifts ended with the apostolic church.
- Preachers without the gift of wisdom preach wayward messages to please the audience.
- Pretence of possession of gifts e.g. claim to have gifts of healing / miracle performance
- Commercialisation of the gift of healing / miracle performance e.g. planting the seed.

( 5 x 1 = 5mks)

a) Irresponsible sexual behaviours condemned by Christians (5mks)
- Homosexuality / Lesbianism
- Prostitution
- Incest
- Fornication
- Adultery
- Beastiality (6 x 1=5mks)

b) Reasons why Domestic violence is widespread in Kenya (7mks)

– Low moral standards

– High cost of living

– Negative attitude towards women.

– The struggle for equality between men and women.

– lack of laws protecting women and children from domestic violence.

– Irresponsible sexual behaviours

– Unemployment

– Poverty

– Drugs and alcohol abuse.

– Lack of guidance and counseling. 7 x 1 = 7mks

c) The role of the church in controlling abortion in society today(7mks)

– Christians showed be role models and guide the youth on the issue of chastity as

well as upholding Christian principles.

Should organize massive campaign against attempts to legalize abortion.
Abortion is murder and is condemned in the Bible Christians should strongly condemn the practice.
Should preach against pre-marital and extra marital sex.
Should sensitise the society on the need to respect human life as it is special gift from God.
Doctors who carry out abortion showed be punished heavily.
Christians should apply the principles of critical thinking, decision making, wisdom in dealing with abortion.
Pray God to help.
Emphasise moral teachings in Churches. ( 7 x 1 = 7mks

a) The life skills that are useful for personal development (5mks)
- Critical thinking
- Creative thinking
- Decision making
- Self-esteem
- Assertiveness 5 x 1 = 5mks

b) Reasons why Christians are against euthanasia (7mks)

– Life is sacred and holy God has the right to take it away.

– Euthanasia is against the ethnics and ethos of medical profession.

– Euthanasia infringe on other people’s rights Christians have bear suffering and

not end their lives or that of others which is the cost of discipleship.

Accepting Euthanasia weakened the societies respect to life.
Some patients have recovered after years of deep coma, thus switching off life- supporting machines is morally wrong.
Euthanasia is contrary to the teachings and works of Jesus Christ in the restoration of human life.
Accepting euthanasia discourages medical research on vaccines that may cure various diseases
Those who practice euthanasia render life meaningless by distorting its sanctify
Euthanasia is equivalent to modes or suicide which is strongly condemned in the Bible.
Every human being is created in the image of God. 7 x 1 = 7mks

c) How modern media technology has enhance evangelization (8mks)

– use of electronic media such a television and radio to teach the God News has

enabled preachers to read large number of people.

Use of public address systems during crusader has allowed clearer delivery of messages.
Use of modern means of communication has shorten the distances of evangelists
Gospel music is recorded in cassettes and distributed globally.
Modern technology such as DVDs VCDs and Cassettes are used to record preaching and Gospel music which can be listened for a long time after the actual events.
The mass media has allowed people to listen to preaching from wherever they are be it at home, in office or while traveling.
Print media has allowed the message to be printed and distributed. 8 x 1= 8mks

SAMPLE PAPER 9

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CHRISTIAN RELIGIOUS EDUCATION PAPER 2

(a) Identify the roles of the Messiah according to prophet Isaiah. (6mks)

(b) Explain the Jewish expectations of the Messiah. (7mks)

(a) Identify five occasions in St Luke’s gospel when Jesus was tempted. (7mks)

(b) Outline the functions of the temple of Jerusalem at the time of Jesus. (7mks)

What do Christians learn about Jesus from his temptations? (6mks)

(a) Relate the parable of the widow and the unjust judge. (Luke 18: 1-8) (6mks)

(b) State the occasions when Jesus prayed. (8mks)

(a) Describe Peter’s message on the day of Pentecost. (9mks)

(b) In what ways was unity demonstrated by Christians in the early church? (6mks)

(a) Describe ways in which science and technology challenge the Christian

understanding of human dignity. (7mks)

(b) Give reasons that make Christians oppose the use of artificial contraceptives. (6mks)

(a) Under what circumstances is strike action justified. (5mks)

(b) What are the obstacles to maintenance of law and order in modern society? (8mks)

MARKING SCHEME

SAMPLE PAPER 9

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(a) Identify the roles of the Messiah according to prophet Isaiah. (6mks)(Isaiah 61: 1 – 3)

To bring / teach good news to the poor.
To heal the broken hearted.
To announce release / liberate the captives and offer freedom to those in prison.
To proclaim the accepted year of the lord.
To save his people and defeat their enemies.
To bring joy and gladness instead of grief /mourning.
He would restore sight to the blind. (any 6 points x 1 = 6mks)

(b) Explain the Jewish expectations of the Messiah.(7mks)

He would conquer the enemies of Israel as a military ruler.
He would rule from the throne of David.
He would avenge the enemies of Israel.
He would be born from a royal family.
He would lead Israel into political and economic prosperity.
One who would appear in Jerusalem full of glory
He would come after the return of Elijah.
Perform miracles and mighty deeds.
Not to associate himself with the poor sinners and gentiles. (any 7 points x1=7mks)

(c) State ways in which Christians are preparing for the Parousia.

Be watchful/ vigilant in prayer’s
Forgiving others
Repenting sin/ living righteous lives
Baptism
Preaching the gospel/ good news/ evangelise/ witnessing.
Attending church / fellowships
Studying / reading the bible / bible study.
Obeying God’s commandments
Live in unity as one body of Christ.
Enduring sufferings/ tribulations/ temptations. (any 7 points x 1 = 7mks)

(a) Identify the occasions in St Luke’s gospel when Jesus was tempted.

On the cross by the unrepentant thief.
In the wilderness after His baptism
The Pharisees tempted Jesus on the issue of tax payment to caeser.
In Nazareth when He was told to perform a miracle that he had performed in Capernaum.
In the garden of Gethsemane when he asked God to” remove the cup” from Him.
By the soldiers during the trials when they asked Him to prophesy who had hit Him.
During the transfiguration when Peter requested the construction of the three tents/ thus trying to stop Jesus mission to save man.
From the two sons of Zebeddee when they asked Him to destroy by divine fire the inhospitable Samaritan village.
When he predicted his coming suffering and death and Peter objected. (Any 7 points x 1 = 7mks)

Outline the functions of the temple of Jerusalem at the time of Jesus.
it was a house / of place of prayer/ worship.
All first born male children were dedicated in the temple.
The priests offered sacrifices / burnt incense in the temple.
The teaching of the law was conducted in the temple.
Religious festivals were held here
Circumcision/ Naming took place here.
The animal and birds for sacrifices were sold/ exchanged here.
The priests were dedicated / in the temple.
It was a major religious centre for the Jews/ unifying factor for the Jews.
The Ark of the Covenant was kept in the temple/ sign of God’s presence among his people.
It was a dwelling place for the priests.
It was the seat for the Sanhedrin / acted as a court.
It was a place of pilgrimage for Jews of Diaspora.(any 7 points x 1= 7mks)

(c) What do Christians learn about Jesus from his temptations?

Jesus is human
Jesus has power over Satan / devil / evil
Jesus had complete faith in his father.
Jesus was well versed / knowledgeable in Jewish scriptures.
Jesus mission to destroy/ overcome Satan’s kingdom/ save mankind.
Jesus was ready for his mission
Jesus was a humble messiah
Jesus is divine /son of God
Jesus was obedient to his father. (any 6 points x 1 =6mks)

(a) Relate the parable of the widow and the unjust judge. (Luke 18: 1 -8)

Jesus told the disciples the parable to teach them that they should always pray.
In a certain town there was a judge who neither feared God nor respected man.
In the same town there was also a widow who kept coming to him pleading for justice/ her rights against her opponents.
For a long time the judge refused to act.
Finally the judge said to himself though he did not fear God or respected man because of the widow’s persistence he would see that she got justice.
Jesus said if the unjust judge finally gives justice how much more will God be willing to give them help. (Any 6 points x 1=6mks)

(b) State the occasions when Jesus prayed.

During his baptism
During his temptations in the wilderness.
Before choosing of the 12 disciples.
At the feeding of the five thousand men.
When his disciples asked him to teach them to pray.
Before his arrest in the Mount of Olives.
During the transfiguration
During the last supper.
After the return of the seventy two from their mission.
When he was on the cross.
After Peter confessed his true identity as the Christ.
At the table in Emmaus after resurrection.
After the healing of the leper at Capernaum. Any 8 points x 1 = 8mks

(c) Why do Christians find it difficult to pray? (6mks)

Lack of faith
Laziness
Frustrations at personal/ family level
Permissiveness
Discouragement from peers
Lack of role models
Too bussy in their work
Trust in wealth/ education/ materialism other than in God.
Lack of scriptural knowledge
Lack of training from an early age.
Devil’s attacks/ humiliation.

(a) Describe Peter’s message on the day of Pentecost. (9mks)

He defended the disciples that they were not drank.
He told the crowd that what they had seen is fulfillment of Joel’s prophecy.
Jesus had been sent by God to save mankind but was rejected by the Jews.
Jesus was raised from the dead as a fulfilment of old testament prophecies (Joel 2 : 28 – 32)
Jesus was innocent and they killed him.
Jesus would forgive them if they repent their sins.
Jesus had conquered death / had victory over death.
He told them that God had raised Jesus from death.
God had made Jesus both lord and messiah.
He was attested by God to work miracles.
Jesus death and crucifixion was according to God’s plan.
Peter told the people to repent so they would be forgiven and receive the gift of the Holy Spirit.
David had prophesied about the resurrection of Jesus. (Any 9 points x1=9mks)

(b) In what ways was unity demonstrated by Christians in the early church. (6mks)

They met for prayers/ fellowship
They prayed for each other
They shared meals together
They shared belongings/ property together
They showed concern for the less privileged.
They welcomed each other in their homes
Celebrated Holy Communion together/ breaking of bread/ agape meal.
They removed devisive traditions from the church e.g circumcision of the gentiles.
Helped in winning new convents / preached the same gospel (of salvation through the risen Christ)
They met together for apostolic teachings/ instruction.
Helped in solving problems in the church.
They sold their property and distributed their money among themselves.

(Any 6 points x 1 = 6mks)

(c) How do Christians reflect the example of being the “new people of God? (5mks)

Praying and praising God
Devoting their lives to obedience to God.
Forgiving each other.
Meeting the needs of each other by sharing.
Repenting their sins
Teaching the gospel/ good news
Living exemplary lives. (Any 5 point x 1 = 5mks)

(a) Describe ways in which science and technology challenge the Christian understanding of human dignity. (7mks)

Evolution theory reduces the dignity of a person according to Christian teachings which asserts that man was created by God.
Science and technology has sometimes been used to destroy God’s creation (e.g. through experiments of animals and human beings)
Science and technology have been used to create things in thus taking the place of God.
Some nations use their scientific and technological advances to humiliate others yet all human being are equal.
Science and technology have created divisions in the society making some people richer than others (economical imbalance)
Science and technology has been used to deny people employment e.g. tea picking machines and computers.
Science and technology has been used to make items that interfere with the normal functions e.g. contraceptives drugs and alcohol etc.
Science and technology has led to environmental degradation e.g. pollution that threaten human existence.
Science and technology have made devise that expose man to danger through accidents.

(Any 7 points x 1 = 7mks)

(b) Give reasons that make Christians to oppose the use of contraceptives. (6mks)

Too much use of contraceptives has led to widespread extra-marital sex.
Contraceptives used have made couples become suspicious of each other leading to quarrel/ divorce.
Contraceptives have increased prostitution.
At times sexually transmitted infections become prevalent among those on contraceptives.
God has made life sacred hence there should be no man’s interference.
Some contraceptives can cause actual abortion.
Use of contraceptives by unmarried
People are condemned as sinful.
God has endowed man with ability to exercise self- control.
The virtue of chastity is highly esteemed in the bible. (any 6 points x 1= 6mks)

(c) How do Christians use the print media to promote the spread of Christianity? (7mks)

Using posters/ pictures.
Printed Christian messages teachings.
Christian publications / magazines.
Pamphlets where write religious and family life issues are written.
Printing cards which carry Christian messages / verses.
Use of newspapers to educate people on the work of the church/ how they should behave in certain situations.
Rising funds through advertisements for charity.
Use of the bible dictionary / encyclopedia to explain Christian messages.
Use of bible Atlas/ charts to illustrate geographical spread of the gospel.
Selling books / magazines with Christian messages. (any 7points x 1=7mks)

(a) Under what circumstances is strike action justified.

As a last resort
The good to be achieved is greater than the negatives/ evil effect of the strike.
To be just a strike must have a very serious cause.
If it does not lead to mass sackings/ victimization.
If it is peaceful action
If it is well supported by all the workers.
Means used in carrying out the strike must be lawful.
There must be reasonable hope of success. (any 5 points x 1 = 5mks)

(b) What are the obstacles to maintenance of law and order in modern society? (8mks)

Poverty
Increasing rate of crime especially in urban areas.
Political instability.
Totalitarians / dictatorial governments
Greed for power and position
High level of unemployment.
Corruption/ bribery
Due to people’s ignorance on the law
Tribalism / clanism / nepotism / sexism.
Inequal distribution of resources / wealth.
Permissiveness in the society.
Hypocrisy– where leaders mislead the society through propaganda ( siasa ya pesa nane)

(Any 8 points x 1 = 8mks)

(c) Suggest the factors that can hinder Christians from taking part in leisure

activities (7mks)

Excessive poverty
Ignorance on leisure activities.
Leisure is expensive
Lack of social amenities
Work pressure
Greed for wealth
Sicknesses
Physical disability
Insecurity/ fear of attacks by mobs / Roundy youths.
Some churches discourage certain leisure activities e.g. gambling. (any7 points x 1 =7mks)

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CHRISTIAN RELIGIOUS EDUCATION PAPER 2

a) Describe Zachariah’s vision in the temple Luke1:5-80. (5mks)
b) Explain the message contained in the Benedictus (8mks)
c) State how the infancy narratives of Jesus show that he was the son of God. (6mks)

a) Give an account of the sinful woman Luke 7: 36 – 50. (8mks)
b) Explain why Jesus criticized the Jewish religious leaders. (8mks)
c) Give reasons why Jesus healed the sick. (4mks)

a)Who were the Sadducees? (8mks)
b) What was Jesus’ response to the Sadducees question on resurrection? (6mks)
c) State the dangers of wife inheritance. (6mks)

a) Explain how the symbolic expression of the vine and the branches is used to express the unity of believers in the New Testament. (8mks)
b) What are the characteristics of the New people of God according to the New Testament? (7mks)
c) State the reasons why Christians in Kenya should work in unity. (5mks)

a) Explain the basis of Christian ethics. (10mks)
b) Outline the Christian teachings on fornication. (5mks)
c) What are the reasons why young people are seeking church weddings? (5mks)

a) Explain the Christian understanding of the use of Science and Technology. (8mks)
b) What are the disadvantages of plastic surgery? (6mks)
c) Show how modern family planning methods have affected Kenyan families. (6mks)

MARKING SCHEME

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1 a) -Zachariah’s vision in the Temple.

(i) Zachariah was a priest in the temple belonging to the division of Abijah.

(ii) One day Zachariah was offering incense in the temple.

Angel, Gabriel appeared to him and announced that his wife Elizabeth would

give birth to a son whom he was to name John.

The angel went a head to describe the qualities of the son to be born.
Zachariah was afraid but the angel re assured him that he had come with Good news.
Zachariah expressed doubt regarding the angel’s message and insisted & be given a proof because he and his wife were old.
Zachariah was punished for disbelieving the angel’s message.
The angel told him that he would be dureb until the day of John’s birth.

(5×1=5mks)

b) – Message contained in the Benedictines.

(i) Describes God as redeemer because He saved His people from sins.

(ii) God has brought salvation through Jesus from the house of David as He promised

through His prophets.

God fulfils the promises that He made to Abraham and through the prophets in

the old testament.

Zachariah outlines the mission of John as the prophet of the most high.
John would go before the lord to prepare the way for him. He was the forerunner of the messiah.
John’s duty would involve imparting the knowledge of salvation & the people.
He would call people to repentance and forgiveness in order to restore the right relationship with God.
He would give light to those living in darkness and guide them into the path of peace. (4×2=8mks)

c) –How the infancy narratives about the birth of Jesus show that he was the son of

God.

His birth had been foretold long before by the old testament – prophets.
It was a virgin birth.
He was conceived through the power of the Holy spirit.
His name Jesus (saviour) was given by the angel.
The angel announced & the shepherds about His birth.
A bright star shown from the East.
A host of angels sang praises “ a long to God in the highest and on earth peace among men.”
During his presentation is the temple, simeon recognized Him as the messiah. (7×1=7mks)

2 a) – Account of the sinful woman. Lk.7:36:8:3,

(i) On one occasion, Jesus was invited by Simon the Pharisee to his house to

eat with him.

A woman who was known & be sinner came to Jesus weeping on His feet’s and

wiping Him with her hair.

This action drew a negative response from Simon the host. Simon wondered how

Jesus could allow a sinner to touch Him. He was a prophet. He would have

known that she was a prostitute.

She kissed his feet and anointed them with an expensive ointment.
Jesus perceived Simon’s thoughts and told him the parable of the debtors. One owned five hundred denarii and the other fifty. When they could not pay, he forgave them. Then Jesus asked Simon a question, now which of them will love more? Simon answered,” the one he forgave more”.
Jesus told Simon that the woman loved more by her actions which Simon had not done i.e. she wet his feet with tears and wiped them with her hair.
He forgave her sins though many since she loved more. Those present wondered who Jesus was “… who even forgave sins?” Jesus told the woman, “Your faith has saved you, go in peace”. (5×1=5mks)

b) – Why Jesus criticized the religious leaders.

(i) They were jealous about. His popularity Lk.20;1-8.

(ii) They were hypocrites.

They neglected justice and love of God.
They offered long prayers.
They lacked faith. Lk.17:20.
They were boastful e.g. parable of pherises & tax collector.
They never accepted him as the messiah (parable of the tenant and great feast)
They took advantage of weak i.e widows Lk.20.
They loved greetings at market places.
They were dishonest i.e attached to money matters.
They were a bad influence to the people e.g. wheat and tares.
They loved to seat in front seats so as to be recognized. (7×1=7mks)

c) – Why Jesus healed the sick.

(i) To demonstrate God’s love for human beings.

(i) To take a way their pain / suffering.

(iii) People believed in his healing power / people had faith in Him.

To show that physical healing sometimes symbolized spiritual hearing.
As a way of destroying the work of Satan.
To glorify God / show God’s power.
To fulfill Old Testament prophecies.
To demonstrate that God’s kingdom had come.
To show that he was the son of God / Messiah.
To show that his mission was universal.
To show that he had power over nature.
To support / strengthen his teachings.
To show that works of mercy must go on even on the Sabbath / He was loved for the Sabbath. (8×1=8mks)

3 a) – Characteristics of the Sadducees.

(i) They were a consenstive group / sect who accepted the Mosaic law (Torah)

(ii) They rejected their own traditions of elders.

Most of them were priests from wealthy families.
They did not believe in angels, demons etc.
They did not believe in resurrection of the body.
They did not believe in judgment.
They did not believe in the messiah.
They supported the Roman rule.
Some of them were members of the Sanhedrin. (8×1=8mks)

b) – Jesus response to the Sadducees question on resurrection. Lk. 20;27-40.

(i) They entered an argument with Jesus based on the Mosaic teaching. Deut:25:5-6.

The law made provision for levivate marriage.

A woman married to seven brothers who died yet left no children. Then who shall be the woman’s husband.
Jesus responded:

There was no marriage after resurrection / no husband-wife relationship.

A resurrected body would have an angelic form/ heavenly form.
There was no death after resurrected. The righteous will live eternally.
God is for those who are alive and not for those who are dead. He is the God of Abraham, Isaac and Jacob who are a live. (6×1=6mks.)

c) – Dangers of wife inheritance.

(i) The spread of HIV/AIDS resulting to orphans suffering.

(ii) Jealousy from other wives – conflicts.

Children suffer from fear / insecurity.
Fear of losing family property.
May lead to poverty.
Rivaling among wives.
Psychological problems.
Rejection of the woman defies the cultural stand on being inherited. (6×1=6mks)

4 a) – Explain how the symbolic expression of the vine and the branches is used to

express the unity of believers in the new Testament. (8marks)

The vine refers to Jesus Christ, the son of God.
The branches are Christian believers who are united through the vine tree.
The branches are attached to the vine, just like Christians are attached to Jesus Christ.
The gardener / vine dresser / or care taker of the vine is God creator.
The fruits are acceptable Christian behaviours such as love for one another .
The vine (Jesus) therefore links all the branches (Christians) to the vine dresser (God) so, all branches are the same vine tree and have a single caretaker.
The fruitless branches refer to those unfaithful Christians who do not live up to the Christian standards. Those who will be destroyed God.
The branches that bear fruits represent faithful Christian.
Christians must therefore abide by Jesus teachings and love others to be fruitful.
Christians must also remain united in Christ.
Jesus used vine to show the close relationship between Him and His followers. (8×1=8mks).

What are the characteristics of the New people of God according to the New testament?
1. God’s own people choosen by Him to worship Him 1 peter 2:9
2. They are baptized born again

The proclaim the wonderful deeds of Christ.

They share in the resurrection of Christ
They are determined to inherit the Kingdom of God.
They abandon the peace and prosperity of this world.

They are set aside by their moral uprightness. They posses the values of love kindness, humility generosity and caring.
They have a common destiny that is the Kingdom of God.

They are guided by the Holy Spirit
They are under the umbrella of Jesus Christ Titus 2:14.
They are a royal priest hood.

They are a holy nation.
They are a people called out of darkness. Any 7 x 1 = 7mks

State reasons why Christians in Kenya should work in unity (5mks).
i) In order to promote oneness in Christ
ii) To promote the teaching of Jesus Christ
- To share the scarce resources e.g. minimize expenses

In order to achieve effective evangelism or ministry
In order to adopt a common attitude to the integration of African culture in worship of achieve ecumenism.
To prevent the formation of splinter groups and cults

To reduce internal wrangling Any 5 x 1 = 5mks

a) Explain the basis of Christian ethnic. (10mks)

– The Bible which is universally accepted as the word of God.

– Conscience / the individual sense of good and bad in thought and action.

– Faith in God which makes Christians to trust in God and make an effort to

live according to the teachings of Jesus Christ.

The teachings of Jesus Christ especially on how to behave towards those in problems and enemies.
The church community which guides Christians on moral behaviour.
Rituals like marriage where partners vow to remain faithful to one another.
The authoritative Christian literature like those written by theological / papal or pastoral/ directives.
The secular law/ constitution of the nation which guides the behaviour of Christians.
God’s revelation, through people like prophets, sermons the Bible, dreams visions which guide people on how to approach certain issues. 5 x 2 = 10mks

b) Outline the Christian teaching on fornication .

– It is immoral and condemned in the Bible

– It is against the divine purpose of sex because sex is exclusively for

married couples.

It defiles the body which is the temple of the Holy Spirit
It can lead to pregnancy which can cause complications and schools drop- out.
It can lead to transmission of HIV/ AIDS and STIs
It is a misuse of God’s gift of sex / parties are only out to satisfy their sexual desires without a sense of responsibility.
It creates guilt feelings as those involved already know that they are sinning before God.
If perfected it can lead to unfaithfulness in marriage in later life.
It is against the church teachings such as abstinence and chastity. 5 x 1 = 5mks

c) What are the reasons why young people are seeking church weddings? (5mks)

– Marriage is God ordained.

– So that people can learn and accept teachings of the church marriage

– It is a respectable way to publicize marriage.

– The two will be obliged to honour the vows taken in the church.

– They discourage young people from entering into unchristian marriages

– The wedding gives the couples the opportunity to celebrate their marriage

with others.

Christian marriage is legal it legalizes marriage.
It helps the new couple to realize that they are members of the wider community.
Helps in strengthening of the faith of the couples.
To stress the salvation of the couple through becoming one body in Christ .
The church provides guidance and counseling to the couple. Any 5 x 1 = 5mks

a) Explain the Christian understanding of the use of science and technology 8mks

– Science should not replace man in jobs e.g. use of computers.

– It should not destroy human life.

– It should not destroy values about the dignity of man and sanctity of life.

– It should not be used to destroy the environment

– It should help man to appreciate the splendour of God / The beauty of

God’s creation.

It should be made to serve the needs of man for material and spiritual happiness .
It should help man in solving problems
It should be used to glorify God the creator rather the selfish ends. 8 x 1 = 8mks

b) What are the disadvantages of plastic surgery? 6mks

– It may lead to a lot of suffering to the patient incase the tissues as “foreign”

and amounts an immune response against it, which gradually destroys it.

May lead to death or permanent damage on the patient incase of failure of the operation.
Some of the operations are very expensive and many people may not afford them therefore remain with their defects for life.
Although cosmetic surgery may make someone look young for sometime. This will not stop the ageing
There is a danger of people not appreciating what God has created
There is danger of transmitting certain diseases especially in case where tissues are detached from a “donor’
It may lead people to idolize beauty. 6 x 1 = 6mks

c) Show how modern family planning methods have affected Kenyan families (6mks)

– Have led to misunderstanding in the family due to disagreements on what methods to use.

– Have led to sterility

– Have led to unfaithfulness in marriage

– Led to health problems e.g. breast cancer.

– Have led to miscarriages and premature births

– Have enabled people to have the size of family they can manage

hence improvement of the living standards.

Some of the methods are very expensive
Incase of failure of the methods used, the mother and child tend to be rejected / frustrated.
Some methods are irreversible e.g. Tubal ligation may lead to psychological and emotional effect on those who may decide to have children later.

END

Teachers' Resources

ELEMENTARY PROGRAMMING PRINCIPLES COMPUTER NOTES

October 5, 2025 Maverick John Leave a comment

ELEMENTARY PROGRAMMING PRINCIPLES

Definition of terms

Computer Program:

A computer program is a set of coded instructions given to the computer, and represents a logical solution to a problem. It directs a computer in performing various operations/tasks on the data supplied to it.

Computer programs may be written by the hardware manufacturers, Software houses, or a programmer to solve user problems on the computer.

Programming:

Programming is the process of designing a set of instructions (computer programs) which can be used to perform a particular task or solve a specific problem.

It involves use of special characters, signs and symbols found in a particular programming language to create computer instructions.

The programming process is quite extensive. It includes analyzing of an application, designing of a solution, coding for the processor, testing to produce an operating program, and development of other procedures to make the system function.

The program created must specify in detail the logical steps to be taken & the method of processing the data input into the computer in order to carry out the specified task.

A computer program performs the following:

Accepts data from outside the computer as its input.
Carries out a set of processes on the data within the computer memory.
Presents the results of this processing as its output, and
Stores the data for future use.

Programming Languages:

A programming language is a set of symbols (a language) which a computer programmer uses to solve a given problem using a computer.

The computer must be able to translate these instructions into machine-readable form when arranged in a particular sequence or order.

TERMS USED IN COMPUTER PROGRAMMING

Source program (source code)

The term Source program refers to program statements that the programmer enters in the program editor window, and which have not yet been translated into machine-readable form.

Source code is the code understood by the programmer, and is usually written in high-level language or Assembly language.

Object code (object program).

The term Object code refers to the program code that is in machine-readable (binary) form.

This is the code/language the computer can understand, and is produced by a Compiler or Assembler after translating the Source program into a form that can be readily loaded into the computer.

LANGUAGE TRANSLATORS

A computer uses & stores information in binary form, and therefore, it cannot understand programs written in either high-level or low-level languages. This means that, any program code written in Assembly language or high-level language must be translated into Machine language, before the computer can recognize & run these programs.

A Translator is special system software used to convert the Source codes (program statements written in any of the computer programming languages) to their Object codes (computer language equivalents).

The Translators reside in the main memory of the computer, and use the program code of the high-level or Assembly language as input data, changes the codes, and gives the output program in machine-readable code.

In addition, translators check for & identify some types of errors (e.g., Syntax/grammatical errors) that may be present in the program being translated. They will produce error messages if there is a mistake in the code.

Each language needs its own translator. Generally, there are 3 types of language translators:

Note. Interpreters & Compilers translate source programs written in high-level languages to their machine language equivalents.

Assembler

An assembler translates programs written in Assembly language into machine language that the computer can understand and execute.

Functions of an Assembler.

It checks whether the instructions written are valid, and identifies any errors in the program.

The Assembler will display these errors as well as the complete source and object programs. If the program has no errors, the job control will let it run immediately, or save the object program so that it may run it later without translating it again.

It assigns memory locations to the names the programmer uses.

E.g., the Assembler keeps a table of these names so that if an instruction refers to it, the Assembler can easily tell the location to which it was assigned.

It generates the machine code equivalent of the Assembly instructions.

Usually, the Assembler generates a machine code only when no errors are detected. Some of the errors include;

Typing mistakes.
Using the wrong format for an instruction.
Specifying a memory location outside the range 0 2047.

Note. The Assembler cannot detect Logic errors. The programmer knows of these errors only when the program is run & the results produced are incorrect (not what the programmer expected). The programmer must therefore, go through the program & try to discover why an incorrect result was being produced.

Interpreter

An interpreter translates a source program word by word or line by line. This allows the CPU to execute one line at a time.

The Interpreter takes one line of the source program, translates it into a machine instruction, and then it is immediately executed by the CPU. It then takes the next instruction, translates it into a machine instruction, and then the CPU executes it, and so on.

The translated line is not stored in the computer memory. Therefore, every time the program is needed for execution, it has to be translated.

Compiler

A compiler translates the entire/whole source program into object code at once, and then executes it in machine language code. These machine code instructions can then be run on the computer to perform the particular task as specified in the high-level program.

The process of translating a program written in a high-level source language into machine language using a compiler is called Compilation.

For a given machine, each language requires its own Compiler. E.g., for a computer to be able translate a program written in FORTRAN into machine language; the program must pass through the FORTRAN compiler (which must know FORTRAN as well as the Machine language of the computer).

The object code file can be made into a fully executable program by carrying out a Linking process, which joins the object code to all the other files that are needed for the execution of the program. After the linking process, an executable file with an .EXE extension is generated. This file is stored on a storage media.

Points to note.

The job of a Compiler is much more difficult than that of an Assembler in that, a single statement in a high-level language is equivalent to many machine instructions.

The format of an Assembly instruction is fairly fixed, while high-level languages give a lot of freedom in the way the programmer writes statements.

Functions of a compiler.

A Compiler performs the following tasks during the compilation process:

It identifies the proper order of processing, so as to execute the process as fast as possible & minimize the storage space required in memory.

It allocates space in memory for the storage locations defined in the program to be executed.

It reads each line of the source program & converts it into machine language.

It checks for Syntax errors in a program (i.e., statements which do not conform to the grammatical rules of the language). If there are no syntax errors, it generates machine code equivalent to the given program.

It combines the program (machine) code generated with the appropriate subroutines from the library.
It produces a listing of the program, indicating errors, if any.

Differences between Compilers and Interpreters

Interpreter

Compiler

1. Translates & executes each statement of the source code one at a time.

The source code instruction is translated & immediately obeyed by the computer hardware before the next instruction can be translated.

(Translation & execution go together).

2. Translates the program each time it is needed for execution; hence, it is slower than compiling.

3. Interpreted object codes take less memory compared to compiled programs.

4. For an Interpreter, the syntax (grammatical) errors are reported & corrected before the execution can continue.

5. An Interpreter can relate error messages to the source program, which is always available to the Interpreter. This makes debugging of a program easier when using an Interpreter than a Compiler.

1. Translates all the source code statements at once as a unit into their corresponding object codes, before the computer can execute them.

A Compiler translates the entire source program first to machine code, and then the code is executed by the CPU.

(Translation & execution are separate phases)

2. Compiled programs (object codes) can be saved on a storage media and run when required; hence executes faster than interpreted programs.

3. Compiled programs require more memory as their object files are larger.

4. For a Compiler, the syntax errors are reported & corrected after the source code has been translated to its object code equivalent.

5. Once the source program has been translated, it is no longer available to the Compiler, so the error messages are usually less meaningful.

Linkers & Loaders

Computer programs are usually developed in Modules or Subroutines (i.e., program segments meant to carry out the specific relevant tasks). During program translation, these modules are translated separately into their object (machine) code equivalents.

The Linker is a utility software that accepts the separately translated program modules as its input, and logically combines them into one logical module, known as the Load Module that has got all the required bits and pieces for the translated program to be obeyed by the computer hardware.

The Loader is a utility program that transfers the load module (i.e. the linker output) into the computer memory, ready for it to be executed by the computer hardware.

Syntax

Each programming language has a special sequence or order of writing characters.

The term Syntax refers to the grammatical rules, which govern how words, symbols, expressions and statements may be formed & combined.

Semantics

These are rules, which govern the meaning of syntax. They dictate what happens (takes place) when a program is run or executed.

Review Questions.

Define the following terms:

Computer program.
Programming language.

With reference to programming, distinguish between Source program and Object code.
What is the function(s) of: Assemblers, Interpreters and Compilers in a computer system?
(a). What are the main functions of a Compiler?

(b). Differentiate between a Compiler and an Interpreter.

LEVELS OF PROGRAMMING LANGUAGES

There are many programming languages. The languages are classified into 2 major categories:

Low-level programming languages.

High-level programming languages.

Each programming language has its own grammatical (syntax) rules, which must be obeyed in order to write valid programs, just as a natural language has its own rules for forming sentences.

LOW-LEVEL LANGUAGES

These are the basic programming languages, which can easily be understood by the computer directly, or which require little effort to be translated into computer understandable form.

They include:

Machine languages.
Assembly languages.

Features of low-level languages

They are machine hardware-oriented.
They are not portable, i.e., a program written for one computer cannot be installed and used on another computer of a different family.
They use Mnemonic codes.
They frequently use symbolic addresses.

Machine languages (1^st Generation languages)

Machine language is written using machine codes (binary digits) that consist of 0s & 1s.

The computer can readily understand Machine code (language) instructions without any translation.

A programmer is required to write his program in strings of 0s & 1s, calculate & allocate the core memory locations for his data and/or instructions.

Different CPUs have different machine codes, e.g., codes written for the Intel Pentium processors may differ from those written for Motorola or Cyrix processors. Therefore, before interpreting the meaning of a particular code, a programmer must know for which CPU the program was written.

A machine code instruction is made up of 2 main parts;

An Address (operand):

It specifies the location (address) of the computer memory where the data to be worked upon can be found.

A Function (operation) code:

It states to the Control Unit of the CPU what operation should be performed on the data/item held in the address, e.g., Addition, Subtraction, Division, Multiplication, etc.

Note. The computer can only execute instructions which are written in machine language. This is because; it is the only language which the computer can understand. Therefore, any program written in any other programming language must first be translated into machine language (binary digits) before the computer can understand.

Assembly language (2^nd Generation Languages).

Assembly languages were developed in order to speed up programming (i.e., to overcome the difficulties of understanding and using machine languages).

The vocabulary of Assembly languages is close to that of machine language, and their instructions are symbolic representations of the machine language instructions.

Assembly language programs are easier to understand, use & modify compared to Machine language programs.

Assembly language programs have less error chances.

To write program statements in Assembly language, the programmer uses a set of symbolic operation codes called Mnemonic codes.

The code could be a 2 or 3 shortened letter word that will cause the computer to perform specific operation. E.g., MOV move, ADD – addition, SUB subtraction, RD – read.

Example;

RD PAT, 15 (read the value 15 stored in the processor register named PAT)

SUB PAT, 10 (subtract 10 from the value in register PAT)

A program written in an Assembly language cannot be executed/obeyed by the computer hardware directly. To enable the CPU understand Assembly language instructions, an Assembler (which is stored in a ROM) is used to convert them into Machine language.

The Assembler accepts the source codes written in an Assembly language as its input, and translates them into their corresponding computer language (machine code/ object code) equivalent.

Comments are incorporated into the program statements to make them easier to be understood by the human programmers.

Assembly languages are machine-dependent. Therefore, a program written in the Assembly language for a particular computer cannot run on another make of computer.

Advantages of Low-level languages

The CPU can easily understand machine language without translation.
The program instructions can be executed by the hardware (processor) much faster. This is because; complex instructions are already broken down into smaller simpler ones.
Low-level languages have a closer control over the hardware, are highly efficient & allow direct control of each operation.

They are therefore suitable for writing Operating system software & Game programs, which require fast & efficient use of the CPU time.

They require less memory space.
Low-level languages are stable, i.e., they do not crash once written.

Disadvantages of Low-level languages

Very few computer programs are actually written in machine or Assembly language because of the following reasons;

Low-level languages are difficult to learn, understand, and write programs in them.
Low-level language programs are difficult to debug (remove errors from).
Low-level languages have a collection of very detailed & complex instructions that control the internal circuiting of the computer. Therefore, it requires one to understand how the computer codes internally.

Relating the program & the problem structures is difficult, and therefore cumbersome to work with.
The programs are very long; hence, writing a program in a low-level language is usually tedious & time consuming.
The programs are difficult to develop, maintain, and are also prone to errors (i.e., it requires highly trained experts to develop and maintain the programs).

Low level languages are machine-dependent (specific), hence non-portable.

This implies that, they are designed for a specific machine & specific processor, and therefore, cannot be transferred between machines with different hardware or software specifications.

It is not easy to revise the program, because this will mean re-writing the program again.

HIGH-LEVEL PROGRAMMING LANGUAGES

High-level languages were developed to solve (overcome) the problems encountered in low-level programming languages.

The grammar of High-level languages is very close to the vocabulary of the natural languages used by human beings. Hence; they can be read and understood easily even by people who are not experts in programming.

Most high-level languages are general-purpose & problem-oriented. They allow the programmer to concentrate on the functional details of a program rather than the details of the hardware on which the program will run.

High-level language programs are machine-independent, (i.e., they do not depend on a particular machine, and are able to run in any family of computers provided the relevant translator software is installed).

Programs written in a high-level language cannot be obeyed by the computer hardware directly. Therefore, the source codes must be translated into their corresponding machine language equivalent. The translation process is carried out by a high-level language software translator such as a Compiler or an Interpreter.

Features of high-level programming languages.

They contain statements that have an extensive vocabulary of words, symbols, sentences & mathematical expressions, which are very similar to the normal English language.

Example;

Read (TaxablePay);

IF TaxablePay<1000 THEN

Tax: =0;

ELSE

Tax: =TaxRate * TaxablePay;

Write (Tax: 6:2);

Allow modularization (sub-routines).
They are user-friendly and problem-oriented rather than machine-based. This implies that, during a programming session, the programmer concentrates on problem-solving rather than how a machine operates.
They require one to be obey a set of rules when writing the program.
Programs written in high-level languages are shorter than their low-level language equivalents, since one statement translates into several machine code instructions.
The programs are portable between different computers.

Purpose of High-level languages.

To improve the productivity of a programmer. This is because; the source programs of high-level languages are shorter than the source programs of low-level languages, since one statement translates into several machine code instructions.

To ease the training of new programmers, since there is no need to learn the detailed layout of a procession/sequence.

To speed up testing & error correction.

To make programs easy to understand & follow.

Advantages of High-level languages.

They are easily portable, i.e., they can be transferred between computers of different families and run with little or no modification.

High-level language programs are short, and take shorter time to be translated.

They are easy to lean, understand and use.

They are easy to debug (correct/remove errors), & maintain.

High level language programs are easy to modify, and also to incorporate additional features thus enhancing its functional capabilities.

They are user-friendly & problem-oriented; hence, can be used to solve problems arising from the real world.

They enable programmers to adapt easily to new hardware. This is because; they dont have to worry about the hardware design of the computer.

High-level language programs are self-documenting, i.e., the program statements displays the transparency of purpose making the verification of the program easy.

High level languages are more flexible; hence, they enhance the creativity of the programmer and increase his/her productivity in the workplace.

Disadvantages of using High-level languages

High-level languages are not machine-oriented; hence, they do not use of the CPU and hardware facilities efficiently.

The languages are machine-independent, and cannot be used in programming the hardware directly.

Each high-level language statement converts into several machine code instructions. This means that, they use more storage space, and it also takes more time to run the program.

Their program statements are too general; hence, they execute slowly than their machine code program equivalents.

They have to be interpreted or compiled to machine-readable form before the computer can execute them.

The languages cannot be used on very small computers.

The source program written in a high-level language needs a Compiler, which is loaded into the main memory of the computer, and thus occupies much of memory space. This greatly reduces the memory available for a source program.

TYPES OF HIGH-LEVEL LANGUAGES.

High-level languages are classified into five different groups:

Third generation languages (Structured / Procedural languages).
Fourth generation languages (4GLs).
Fifth generation languages (5GLs)
Object-oriented programming languages (OOPs).
Web scripting languages.

The various types of high-level languages differ in:

The data structures they handle.
The control structures they support.
The assignment instructions they use.
Application areas, e.g., educational, business, scientific, etc.

STRUCTURED LANGUAGES

A structured (procedural) language allows a large program to be broken into smaller sub-programs called modules, each performing a particular (single) task. This technique of program design is referred to as structured programming.

Structured programming also makes use of a few simple control structures in problem solving. The 3 basic control structures are:

Sequence
Iteration (looping).

Advantages of structured programming.

It is flexible.
Structured programs are easier to read.
Programs are easy to modify because; a programmer can change the details of a section without affecting the rest of the program.
It is easier to document specific tasks.
Use of modules that contain standard procedures throughout the program saves development time.
Modules can be named in such a way that, they are consistent and easy to find in documentation.
Debugging is easier because; each module can be designed, coded & tested independently.

Examples of Third generation programming languages include:

BASIC (Beginners All-purpose Symbolic Instructional Code).

BASIC is a simple general-purpose high-level language used in most computer processing tasks such as developing business and educational applications.

It is easy to learn & use; hence, suitable for students who wish to easily learn programming.

Translation in most versions of BASIC is carried out by an Interpreter.

Disadvantages of BASIC.

BASIC is available in so many versions with different dialects/languages & therefore, it has no standard.
Some dialects are limited to data & control structures they support.
Some versions of BASIC offer limited facilities in terms of structured programming & meaningful variable names.

PASCAL

PASCAL is a general-purpose, high-level programming language, which was named after a French mathematician called Blaise Pascal.

It was developed as an academic tool to help in the teaching and learning of structured programming.

PASCAL supports structured programming, i.e., it uses procedures & functions, which allow a top-down approach to solving problems.

It is not easy to learn because; it has strict rules in its grammar on typing of variables (data names) & declarations.

It is poor (has limited ability) on handling of data files.

COBOL (COmmon Business Oriented Language)

COBOL is designed for developing programs that solve business problems, e.g., can be used to develop commercial data processing applications such as computer-based inventory control systems.

COBOL is mostly used where large amounts of data are to be handled, because it supports powerful data & control structures.

COBOL programs are semi-compiled, and the intermediate code is interpreted.

A program written in COBOL language consists of 4 divisions: –

Identification division: Where the programmer & the program details are specified, e.g., program ID, programmer name, etc.

Environment division: Where the equipments to be used by the source & the object programs are defined, e.g., the computer hardware.

Data division: Where the various files to be used by the program are described, e.g., a description of the input files.

Procedure division: Where all the procedures required to manipulate/interrelate the data into information are defined.

Advantages of COBOL.

It is easy to read.
It is portable, i.e., can be used on different types of computers. This is because; it has an American National Institute

American National Standards Institute (ANSI): An international organization that devised/ invented the group of standardized symbols used in flowcharting.

It is widely used, and has a pool of skilled programmers.

Disadvantage of COBOL.

The structure of a COBOL program is too long even for simple programs.

E.g., consider the following assignment statement:

DIVIDE A into B giving C.

This statement when used in BASIC language can much short C=A/B. However, notice that the COBOL statement above is more self defining.

FORTRAN (FORmula TRANslator)

It was developed for mathematicians, scientists and engineers. It provides an easier way of writing scientific & engineering applications.

FORTRAN statements are mostly in form of mathematical expressions; hence, it is useful in writing of programs that can process numeric data.

FORTRAN programs are compiled.

Advantages of FORTRAN.

It is portable, i.e. it can be used on different types of computers.

Disadvantage of FORTRAN.

It is not suited for business applications.

Ada

This language was named after the first lady programmer Ada Lovelace.

It is suitable for developing military, industrial and real-time systems.

C is mainly used for developing system software such as the operating system as well as developing the application packages.

It has powerful commands that permit the rapid development of programs, and allows direct control over the hardware.

Disadvantage of C

It is difficult to read & learn because of its strict dialect rules.

LOGO

LOGO was designed for educational use in which children can explore & develop concepts through programming the movement of a pen.

COROL

COROL is used in Real-time processing.

COROL programs are compiled.

RPG (Report Program Generator)

RPG is used in report generating applications, (i.e. it is designed to facilitate the output of reports of business data).

A Report generator is a software tool that extracts stored data to create customized reports that are not normally/usually produced by existing applications.

SNOBOL (String Oriented Symbolic Language).

It is a high-level language designed to manipulate strings of characters. It is therefore used for non-numeric applications.

FOURTH GENERATION LANGUAGES (4GLS).

4GLs make programming even easier than the 3GLs because; they present the programmer with more programming tools, such as command buttons, forms, textboxes etc. The programmer simply selects graphical objects called controls on the screen, and then uses them to create designs on a form by dragging a mouse pointer.

The languages also use application generators (which in the background) to generate the necessary program codes; hence, the programmer is freed from the tedious work of writing the code.

4GLs are used to enquire & access the data stored in database systems; hence, they are described as the Query languages.

Purpose of fourth generation languages.

The 4GLs were designed to meet the following objectives: –

To speed up the application-building process, thereby increasing the productivity of a programmer.
To enable quick & easy amendments and alteration of programs.
To reduce development & maintenance costs.
To make languages user-friendly. This is because, the 4GLs are designed to be user-oriented, unlike the 3rd generation languages which are problem & programmer oriented.

To allow non-professional end-users to develop their own solutions.

`To generate bug-free codes from high-level expressions of requirements.

Examples of 4GLs are:

visual Basic
Delphi Pascal
Visual COBOL (Object COBOL)
Access Basic

Advantages of fourth generation languages.

They are user-based, and therefore, easy to learn & understand.

The grammar of 4GLs is very close to the natural English language. It uses menus & prompts to guide a non-specialist to retrieve data with ease.

Very little training is required in order to develop & use 4GL programs.

They provide features for formatting of input, processing, & instant reporting.

FIFTH GENERATION LANGUAGES (5GLS).

The 5GLs are designed to make a computer solve a problem by portraying human-like intelligence.

The languages are able to make a computer solve a problem for the programmer; hence, he/she does not spend a lot of time in coming up with the solution. The programmer only thinks about what problem needs to be solved and what conditions need to be met without worrying about how to implement an algorithm to solve the problem.

5GLs are mostly used in artificial intelligence.

Examples of 5GLs are:

PROLOG
LISP
Mercury

LISP (LISt Processing)

In LISP, both programs & data are arranged (structured) as lists.

It is used in artificial intelligence. However, it is not suitable for commercial data processing applications.

PROLOG (PROgramming in LOGic)

PROLOG was developed from LISP by the Japanese.

It is designed for use with Expert systems & Artificial Intelligence. It is mostly used for solving problems, which involve objects and relationships between objects.

Like LISP, it is not suitable for commercial data processing applications.

OBJECT-ORIENTED PROGRAMMING LANGUAGES (OOPs)

Object-Oriented Programming is a new approach to software development in which data & procedures that operate on data are combined into one object.

OOPs use objects. An Object is a representation of a software entity such as a user-defined window or variable. Each object has specific data values that are unique to it (called state) and a set of the things it can accomplish called (functions or behaviour).

Several objects can be linked together to form a complete program. Programs send messages to an object to perform a procedure that is already embedded in it. This process of having data and functions that operate on the data within an object is called encapsulation.

The data structure & behaviour of an object is specified/described by a template (called a class). Classes are hierarchical, and it is possible to pass the data & behaviour of an object in one class down the hierarchy.

Object-Oriented programming enables rapid program development. Every object has properties such as colour, size, data source, etc, which can be set easily without much effort. In addition, every object has events associated with it that can be used to trigger certain actions, e.g. remove the window from the screen on clicking the Close button.

OOP has contributed greatly to the development of graphical user interface operating systems and application programs.

Examples of Object-oriented programming languages are: –

Simula
C++
SmallTalk
Java

Java is sometimes associated with development of websites, but it can be used to create whole application programs that do not need a web browser to run.

JAVA

Java is an OOP language that resembles Object C (a simplified form of C++).

The code of Java displays graphics, accesses the network, and interfaces with users via a set of capabilities known as classes. Classes define similar states & common methods for the behavior of an object.

JAVA programs are not compiled into machine code; instead, they are converted into a collection of bytes that represent the code for an abstract Java Virtual machine (VM). A Java interpreter running on a physical machine is then used to translate those bytes into local actions, such as printing a string or drawing a button.

WEB SCRIPTING LANGUAGES.

Web scripting languages are mostly used to create or add functionalities on web pages.

Web pages are used for creating Web sites on the Internet where all sorts of advertising can be done.

Web pages are hypertext (plain-text) documents written using a language called HyperText Markup Language (HTML). HTML documents have a file extension of .Html or .Htm.

Note. HTML doesnt have the declaration part and control structures, and has many limitations. Therefore, to develop functional websites, it must be used together with other web scripting languages like JavaScript, VBScript and Hypertext Preprocessor.

Comparison of Programming languages.

Machine language

Assembly language

High-level languages

1. Instruction set is made up of binary digits (0s & 1s).

2. Instruction is made of 2 parts: operation code & operand.

3. No translation is needed. (This is the computer language; hence, the computer understands it directly).

4. Executed by the hardware directly & is faster.

5. Difficult to learn, develop & maintain.

6. Programs are lengthy & tedious.

7. It is time-consuming to develop machine code programs.

8. Used in applications where efficient use of the CPU time is necessary, e.g., developing Operating systems & other Control programs that coordinate the working of peripherals.

1. Instruction set is made up of Mnemonics & labels.

2. Instruction is made up of 2 parts: operation code & operand, but comments can be added.

3. Uses an Assembler to convert the assembly language source codes to their object code equivalents

4. Executed faster than High-level, but slower than the machine code programs.

5. Its easier to learn, develop & maintain as compared to machine code programs.

6. Like machine code language, the programs are lengthy & tedious.

7. They take a shorter time to develop as compared to machine code programs, but take longer than High-level language programs.

8. Like machine language, Assembly language programs are used in applications where efficient use of the CPU time is necessary.

1. Instruction set is similar to English language statements & mathematical operators.

2. The instruction varies depending on the particular language.

3. Uses compiler or interpreter

Compiler translates all the source code at once into object code; Interpreter translates line by line.

4. Translation & execution is very slow.

5. Easy to learn, develop, maintain and use.

6. Programs are shorter & simpler than Machine & assembly lang. programs.

7. Developing High-level language programs takes very short time.

8. Most High-level languages are general-purpose, & can be used to do almost all computer-processing tasks.

Factors to consider when choosing a Programming language.

The following factors should be considered when choosing a Programming language to use in solving a problem:

The availability of the relevant translator
Whether the programmer is familiar with the language
Ease of learning and use
Purpose of the program, i.e., application areas such as education, business, scientific, etc.
Execution time

Applications that require quick response are best programmed in machine code or assembly language. High-level languages are not suitable for such application because, they take long to be translated & executed.

Development time

Development time is the time a programmer takes to write and run a program.

High-level languages are easy to read, understand and develop; hence, they require less development time. Machine code & Assembly languages are relatively difficult to read, understand and develop; hence, they are time-consuming.

Popularity

The language selected should be suitable and/or successful in the market with respect to the problems to be solved.

Documentation

It should have accompanying documentation (descriptions) on how to use the language or maintain the programs written in the language.

Maintenance

Programs are developed to solve specific problems, and the problems keep on changing; hence, the programs are also changed to perform the new functions.

Program maintenance is the activity of incorporating more routines onto the program, modifying the existing routines or removing the obsolete routines to make the program adapt to a functionally enhanced environment.

The maintenance is made easier if the language used is easy to read and understand.

Availability of skilled programmers

The language selected should have a pool of readily available programmers to ease the programming activity, and reduce development time.

Review Questions

(a). What is a Programming language?

(b). Explain the two levels of programming languages.

(a). What is meant by Machine language?

(b). Explain why machine language programming is so error-prone.

(c). Show the difference between Machine language and Assembly language.

(d). Give two advantages & three disadvantages of Machine language programming.

(a). What are High-level languages?

(b). Give the features/characteristics of high-level programming languages.

(c). Describe briefly how a program written in high-level programming language becomes a machine code program ready for operational use.

(d). Explain the advantages and disadvantages of using a High-level programming language for writing a program.

(e). List four examples of high-level programming languages. Indicate the application of each language in computing.

(a). What is meant by program portability?

(b). Why are low-level languages not considered to be portable?

List 8 factors that need to be considered when selecting a programming language.

PROGRAM DEVELOPMENT.

Stages involved in the program development cycle.

The process of program development can be broken down into the following stages:

Problem recognition (Identification of the problem).
Problem definition.
Program design.
Program coding.
Program testing & debugging.
Program Implementation and maintenance.
Program documentation.

Problem recognition.

Problem recognition refers to the understanding and interpretation of a particular problem.

The programmer must know what problem he/she is trying to solve. He/she must also understand clearly the nature of the problem & the function of the program.

In order to understand a problem, look for the keywords such as compute, evaluate, compare, etc.

Usually, a programmer identifies problems in the environment and tries to solve them by writing a computer program.

There are 3 situations that cause the programmer to identify a problem that is worth solving:

Problems or undesirable situations that prevent an individual or organizations from achieving their purpose.

Opportunity to improve the current program.

A new directive given by the management requiring a change in the current system.

Sample problem: Develop a program that can be used to calculate/find the area of a circle. Use the equation A = π * r².

Problem definition (Problem Analysis).

In Problem definition, the programmer tries to define (determine) the:

Output expected from the program.
Inputs needed to generate the output information.
Processing activities (requirements), and
Kind of files which may be needed.

The programmer should write a narrative on what the program will do, and how it is meant to achieve the intended purpose. Within this narrative, he/she is required to determine what data is to be input & what information is to be output.

For example:

In calculating the area of any circle, the parameters needed to determine the area of any circle are:

Input: (a) Pie (π) which is a constant.

(b) The radius of the circle.

Process: The formula for calculating area of a circle, which is π * radius * radius.
Output: The area of the circle (A).

At the end of the problem definition, the programmer is required to write a requirements report/document for the new program. This document will enable the programmer to come up with a program design that meets the needs at hand.

Note. Problem definition should be done thoroughly to ensure user satisfaction, and to facilitate the subsequent stages in the program development cycle. A failure at this stage usually results in a system that will not work as intended, or that may not work at all.

Program design

Program design is the actual development of the programs process or problem solving logic called the Algorithm.

It involves identifying the processing tasks required to be carried out in order to solve the problem.

The design stage enables the programmer to come up with a model of the expected program (or a general framework (outline) of how to solve the problem, and where possible, break it into a sequence of small & simple steps.

The models show the flow of events throughout the entire program from the time data is input to the time the program gives out the expected information.

The processing tasks must be in order & systematic. Therefore, the programmer identifies the processing tasks required, and the exact order in which they are to be carried out.

The design process does not take account of the programming language to be used in the final product, since it only defines program logic.

Program design provides for easy maintenance.

Note. It is important to design programs before entering them into the computer. The programmer should only attempt to covert a design into a program code after ensuring that it is logically correct. If possible, check the logical order on the desk.

Some programmers produce rough & ready solutions at a Keyboard, and continue to amend the programs until eventually the program appears to do what was expected. This is not recommended in programming because of the following reasons:

The final code may not be easy to follow, since it was just cobbled together.

Variable names & specific items of code may not be documented.

Programs produced by continuous amendments & changing of codes mostly lead to unforeseen side effects.

E.g., there may not have been plan for testing the program or procedures, hence, the program may easily fail.

A programmer may be asked to modify the code at a later date. Without sufficient documentation, the programmer will be forced to trace through the program in order to gain an insight into how the program functions.

Modular programming

Many programs are non-monolithic (i.e., they are not usually made up of one large block of code). Instead, they are made up of several units called modules, that work together to form the whole program with each module performing a specific task.

This approach makes a program flexible, easier to read, and carry out error correction.

Program coding

Program coding is the actual process of converting a design model into its equivalent program.

Coding requires the programmer to convert the design specification (algorithm) into actual computer instructions using a particular programming language.

For example;

The programmer may be required to write the program code either in Pascal, C++, Visual Basic or Java, and develop (invent) suitable identifiers, variable names, & their data types. However, remember that, at this stage the coding is still a Pencil & paper exercise.

The end result of this stage is a source program that can be translated into machine readable form for the computer to execute and solve the target problem.

Rules followed in coding a program.

Use the standard identifiers or reserved words.
Make the program more readable by using meaningful identifiers.
Dont use similar variables.
Keep spellings as normal as possible.
Use comments to explain variables & procedures. This makes the program readable.
Avoid tricks write the program using straightforward codes that people can readily understand.
Modularize your program.

Sample programs written in Pascal language.

Example 1:

Develop a program code that would be used to solve the equation of a straight line given by the expression: Y = mx + c

Program StraighLine (input, output);

VAR

y, m, x, c: INTEGER;

BEGIN

Writeln (Input the value of M);

Readln (M);

Writeln (Input the value of X);

Readln (X);

Writeln (Input the value of C);

Readln (C);

Y: = (m * x) +c;

Writeln (The value of y is:, Y);

END.

Pascal code	Explanation
Program StraightLine (input, output);	This is the program Header. The word Program indicates the beginning of the program whose name is StraightLine. The (input, output) statements shows that the program expects some input from the Keyboard and display the output on the Screen.
VAR	VAR is short form for Variable. A variable is a location for data in the computer memory. This statement tells the computer that variables are about to be declared. When a variable is declared, the computer sets aside some memory space to store a value in the variable.
y, m, x, c: INTEGER;	Four variables of type Integer have been declared. This means that, the memory spaces that will be set aside can only hold values that are whole numbers.
BEGIN	The Begin statement marks the start of the program body. Statements in this section are executed by the computer. E.g., execution starts by asking the user to input the value of m.
Writeln (Input the value of M);	The Writeln statement displays whatever is between the inverted commas in the brackets. The statements will be sent to the screen exactly the way they appear in the brackets. This is because; the inverted commas are meant to make the output readable on the screen. To display the value held in a variable on the screen, remove the inverted commas and write the name of the variable in the brackets, e.g., Writeln (y) will display the value held in the variable y.
Readln (M);	The Read or Readln statement reads a value and stores it in a variable. When the program is running, a Read/Readln statement in the code will displays blinking cursor that indicates to the user where to type the input.
Y: = (m * x) +c;	Calculates the value of y. in Pascal, the symbol : = is called the *Assignment statement*. The values on the right are calculated then the answer stored in the variable y which is on the left of the assignment symbol.
Writeln (The value of y is:, Y);	The Writeln displays the value held in the variable y on the screen. Note. Y is not within the inverted commas.
END.	The END. statement shows the end of a program.

Example 2:

Program AreaCircle (input, output);

CONST

Pi = 3.142;

VAR

Radius, Area: REAL;

BEGIN

Writeln (Enter the radius);

Readln (Radius);

Area: = Pi * Radius * Radius;

Writeln (The Area is, Area);

END.

Pascal code	Explanation
Program AreaCircle (input, output);	The Header of the program. The statements in ( ) shows that the user inputs data via Keyboard and the program display information on the Screen.
CONST Pi = 3.142;	A constant has been declared with a name Pi and value 3.142.
VAR Radius, Area: REAL;	Variables with fractional parts have been declared.
BEGIN	Marks the beginning of the program body.
Writeln (Enter the radius);	Displays on the screen the string between the inverted commas.
Readln (Radius);	Displays a blinking cursor that tells the user that an input is needed before the program can continue.
Area: = Pi * Radius * Radius;	Calculates the Area. An assignment statement (: =) has been used.
Writeln (The Area is, Area);	Displays the value stored in the variable Area.
END.	Marks the end of the program.

Revision Questions.

State the rules followed in coding a program.

Program Testing and Debugging

After designing & coding, the program has to be tested to verify that it is correct, and any errors detected removed (debugged).

TESTING:

Testing is the process of running computer software to detect/find any errors (or bugs) in the program that might have gone unnoticed.

During program testing, the following details should be checked;

The reports generated by the system.
The files maintained in connection to the systems information requirements.
The input to the system.
The processing tasks.
The controls incorporated within the system.

Note. The testing process is a continuous process, and it ends only when the Programmer & the other personnel involved are satisfied that when operational, the program will meet the objectives and the growing demands of the organization.

Types of program errors

There are 5 main types of errors that can be encountered when testing a program. These are:

Syntax errors.
Run-time (Execution) errors.
Logical (arithmetic) errors.
Semantic errors.
Lexicon errors.

Syntax errors

Every programming language has a well-defined set of rules concerning formal spellings, punctuations, naming of variables, etc. The instructions are accepted only in a specified form & and must be obeyed by the programmer.

Syntax errors are therefore, programming errors/mistakes that occur if the grammatical rules of a particular language are not used correctly.

Examples:

Punctuation mistakes, i.e., if the programmer does not use the right punctuations & spaces needed by the translator program, e.g., omitting a comma or a semicolon.
Improper naming of variables.
Wrong spellings of user defined and reserved words.

Reserved words are those words that have a special meaning to the programming language, and should not be used by the programmer for anything else.

Syntax errors are committed by the programmer when developing, or transcribing the program, and can be detected by the language translators, such as the Compiler as it attempts to translate a program. Such errors must be corrected by the programmer before the program runs.

Logical (arithmetic) errors.

These are errors in the program logic.

Logical errors relate to the logic of processing followed in the program to get the desired results. E.g., they may occur as a result of misuse of logical operators.

Logical errors cannot be detected by the translator. The programmer will detect them when the program results are produced.

The program will run, but give the wrong output or stop during execution.

Run-time (Execution) errors.

These errors occur during program execution.

Run-time (execution) errors occur when the programmer introduces new features in the program, which are not part of the translators standards.

For example; they may occur if:

The computer is asked to divide a number by zero.
The number generated as a result of an instruction is too large to fit in a memory location.
When you raise a number to a very big power that cannot be accommodated in the Registers structure of the computer.
In case of a closed loop in the program, leading to a set of instructions being executed repetitively for a long time.

Execution errors are not detected by the translator programs, but are detected by the computer during execution. Sometimes, execution errors may lead to premature end of a program.

To detect and eliminate Execution errors, a test run should be performed on the program after it has been translated.

Semantic errors.

These are meaning errors. They occur when the programmer develops statements, which are not projecting towards the desired goal. Such statements will create deviations from the desired objectives.

Semantic errors are not detected by the computer. The programmer detects them when the program results are produced.

Example;

IF GP>=1500 OR 2200 THEN

TAX: = GP – (GP * 13%)

IF GP>=1500 AND GP<= 2200 THEN

TAX: = GP – (GP * 13%)

In the 1^st statement, if the selection is between 1500 & 2200, the computer will pick only 1500 & 2200, and the other values will not be touched.

In the 2^nd statement, the computer will be able to pick all the values between 1500 & 2200 because of the AND operator.

Lexicon errors.

These are the errors, which occur as a result of misusing Reserved words (words reserved for a particular language).

Revision Questions.

State the three types of errors that can be experienced in program testing, and how each can be detected.
Syntax errors can be detected by the help of translators while logical errors are detected differently. Explain FIVE methods which can be used to detect Logical errors.

DEBUGGING:

The term Bug is used to refer to an error in a computer program.

Most programming errors often remain undetected until an attempt is made to translate a program.

The most common errors include:-

Improperly declared Constants and Variables.
A reference to undeclared variable.
Incorrect punctuation.

Debugging is therefore, the process of detecting, locating & correcting (removing, eliminating) all errors (mistakes or bugs) that may exist in a computer program.

TYPES OF TESTING (Methods of error detection)

For the program to be assumed as correct, several testing needs to be conducted by the programmer to ascertain/establish their validity.

There are several methods of testing a program for errors. These include:

Dry running (Desk checking).
Translator system checking.
Functional testing.
Use of Test data.
Use of debugging utilities.
Diagnostic procedures.
System test with actual data.

Dry Running (Desk checking):

Dry running is a method of checking a program for errors by making the corrections on a paper before entering it in the program editor.

It involves going through the program while still on paper verifying & validating its possible results. If the final results agree with the original test data used, the programmer can then type the program into the computer and translate it.

Dry running helps the programmer to identify the program instructions, detect the most obvious syntax and logical errors, & the possible output.

Dry running is much faster. This is because; it involves the use of human brain as the processor, which has got a well inbuilt common sense.

Translator system checking:

This is a type of testing, which involves the computer & the translator programs.

After entering the program, it is checked using a translator to detect any syntax errors. The translator can be a Compiler or an Interpreter, which goes through the set of instructions & produces a list of errors, or a program/statement listing which is free from errors.

Functional testing (White-box testing):

This type of testing is based upon examining the internal structure of a program & selecting test data, which give rise to the alternative cases of control flow.

Use of Test data.

The accuracy of a program can be tested by inputting a set of values referred to as Test data. The test data is designed to produce predictable output.

There are 2 types of test data;

Real data (live data): – test data obtained from the real problem environment (practical applications).

Dummy data: – assumed test data.

The programmer invents simple test data, which he/she uses to carry out trial runs of the new program. At each run, the programmer enters various data variations including data with errors to test how the system will behave. For example, if the input required is of numeric type, the programmer may enter alphabetic characters. The programmer will then compare the output produced with the predicted (actual) output.

Notes.

Where possible, the program should be tested using the same test data that was used for desk checking. More strict/rigid tests should be applied on the program in order to test the program to its limits.

Only Logical errors & Semantic errors can be corrected by the programmer using test data.

A good program should not crash due to incorrect data entry but should inform the user about the irregularity and request for the correct data to be entered.

Use of debugging utilities.

After the program has been entered in the program editor, debugging utilities which are built in the computer can be run during translation to detect any syntax errors in the program.

The errors are corrected and the debugging process is repeated again to find out more errors, before the program is executed.

Diagnostic procedures.

For complex programs, diagnostic procedures, such as Trace routines, may be used to find logical errors.

A Trace prints out the results at each processing step to enable errors to be detected quickly.

System Test with actual data.

This is whereby the new program is run in parallel with the existing system for a short time so that results can be compared and adjustments made. In such cases, the system test is made using actual data.

Review Questions.

Differentiate between Testing and Debugging.
What is Dry running?

Implementation and Maintenance.

IMPLEMENTATION

Implementation refers to the actual delivery, installation and putting of the new program into use.

The program is put into use after it is fully tested, well documented, and after training the staff who will be involved in the running of the new program.

Structured Walk Through:

It is an organized style of evaluating/reviewing a program by a team of other programmers, which then reports to the programming team.

REVIEW AND MAINTENANCE.

Once the program becomes operational, it should be maintained throughout its life, i.e., new routines should be added, obsolete routines removed, & the existing routines adjusted so that the program may adapt to enhanced functional environments.

The main objective of maintenance is to keep the system functioning at an acceptable level.

Program maintenance mainly involves: –

Correcting errors that may be encountered after the program has been implemented or exposed to extensive use.
Changing procedures.
Hardware and software maintenance.
Changing parameters and algorithms used to develop the original programs.
Making any adjustments as new technology comes.

Note. Program maintenance runs parallel to the maintenance of the program documentation, i.e., any time maintenance is carried out on the program, the documentation should also be updated to convey the right image of the system.

Program documentation.

After writing, testing, and debugging a program, it must be documented. In other words, the programmer should describe all what he was doing during the program development stages.

Program documentation is the writing of supportive materials explaining how the program can be used by users, installed by operators, or modified by other programmers.

Note. All the program development activities (i.e., from the initial stage up to the complete program) should be documented/recorded in order to assist in the development of the program, future modification of the program, general maintenance, machine & software conversion at a later date, and program changeover.

Documentation can either be; Internal or External.

Internal documentation is the writing of non-executable lines (comments) in the source program that help other programmers to understand the code statements.

External documentation refers to reference materials such as user manuals printed as booklets.

Types of program documentation.

There are 3 target groups for any type of documentation:

User-oriented documentation.

This enables the user to learn how to use the program as quickly as possible, and with little help from the program developer.

Operator-oriented documentation:

This is meant for computer operators such as the technical staff. It is used to help them install & maintain the program.

Programmer-oriented documentation:

This is a detailed documentation written for skilled programmers. It provides the necessary technical information to help in future modification of the program.

Some documents used in program documentation.

User guide/ manual.

This is a manual provided for an end-user to enable him/her use or operate the program with minimal or no guidance.

A User guide is used in user-oriented documentation.

Reference guide.

It is used by someone who already knows how to use the program but needs to be reminded about a particular point or obtain more detailed information about a particular feature.

Quick Reference guide.

This could be a single sheet or card small enough to fit into a pocket. It is used by the user to get help for the common tasks carried out within the program.

Technical manuals.

They are intended for System analysts & Programmers. They assist in maintaining & modifying the program design and code.

Contents in a program document.

Documentation includes:

Title of the program.
Function of the program.
Language used.
Hardware & Software required to support the processing of the system.
File specifications (details of the data structures used, & details of how data files are to be organized, accessed, and kept secure).
Limitations of the program.
Format of the input & the output expected.
Design of the program using the design tools (i.e., detailed algorithms & procedures used).
A listing of the Source program and the program flowcharts.
A carefully devised set of Test data, and a table of expected results.
Detailed instructions on how to run the program.

Review Questions.

What is program designing?
(a). Define program documentation.

(b). What does a program documentation contain?

Briefly explain how each of the following documents are useful in programming?
- User manual / guide.
- Reference guide.
- Quick reference guide.
Program documentation is different from Implementation. Explain.
Outline and briefly explain the stages involved in program development.

DEVELOPING OF ALGORITHMS

After carefully analyzing the requirements specification, the programmer usually comes up with the algorithm.

Definition of an Algorithm:

An Algorithm is a limited number of logical steps that a program follows in order to solve a problem.

A step-by-step (a set of) instructions which when followed will produce a solution to a given problem.

Algorithms take little or no account of the programming language.

They must be precise/ accurate, unambiguous/clear and should guarantee a solution.

Program design Tools.

Algorithms can be illustrated using the following tools:

Decision Tables.
Decision Trees.

Note. For any given problem, the programmer must choose which algorithm (method) is best suited to solve it.

PSEUDOCODES.

A pseudocode is a method of documenting a program logic in which English-like statements are used to describe the processing steps.

These are structured English-like phrases that indicate the program steps to be followed to solve a given problem.

The term Code usually refers to a computer program. This implies that, some of the words used in a pseudocode may be drawn from a certain programming language and then mixed with English to form structured statements that are easily understood by non-programmers, and also make a lot of sense to programmers.

However, pseudocodes are not executable by a computer.

Guidelines for designing a good pseudocode.

The statements must be short, clear and readable.
The statements must not have more than one meaning (i.e., should not be ambiguous).
The pseudocode lines should be clearly outlined and indented.
A pseudocode must have a Begin and an end.

i.e., a pseudocode should show clearly the start and stop of executable statements and the control structures.

The input, output and processing statements should be clearly stated using keywords such as PRINT, READ, INPUT, etc.

Example 1:

Write a pseudocode that can be used to prompt the user to enter two numbers, calculate the sum and average of the two numbers and then display the output on the screen.

START

PRINT Enter two numbers

INPUT X, Y

Sum = X + Y

Average = Sum/2

PRINT Sum

PRINT Average

STOP

Example 2:

Write a structured algorithm that would prompt the user to enter the Length and Width of a rectangle, calculate the Area and Perimeter, then display the result.

Solution

Step 1: Draw the rectangle of Length (L) and Width (W).

Step 2: Write down the Pseudocode.

START

PRINT Enter Length and Width

READ L, W

Area = L * W

Perimeter = 2 (L + W)

PRINT Area

PRINT Perimeter

STOP

Example 3:

Write a pseudocode that can be used to calculate the Diameter, Circumference and Area of a circle and then display the output on the screen.

START

Set π to 3.14

Prompt the user for the Radius (R)

Store the radius in a variable (R)

Set Diameter to 2 * Radius

Set Circumference to π * 2 * Radius

Set Area to π * Sqr (Radius)

PRINT Diameter

PRINT Circumference

PRINT Area

STOP

Example 4:

Write a pseudocode for a program that would be used to solve equation: E = MC².

START

Enter values from M to C

E = M * C * C

Display E

STOP

It is important to use program control structures when writing Pseudocodes. The most common constructs are:

Looping (Repetition / Iteration)  used where instructions are to be repeated under certain conditions.
Selection  used when choosing a specified group of instructions for execution. The group chosen depends on certain conditions being satisfied.

Example 5:

Write a pseudocode for a program that can be used to classify people according to age. If a person is more than 20 years; output Adult else output Young person.

START

PRINT Enter the Age”

INPUT Age

IF Age > 20 THEN

PRINT Adult

ELSE

PRINT Young person

STOP

Note. Pseudocodes make an algorithm easier to understand. This is because; the algorithm can be read from top to bottom without the need for jumping backwards or forwards to follow the logic of the algorithm as in flowcharts.

FLOWCHARTS.

A Flowchart is a diagrammatic or pictorial representation of a programs algorithm.

It is a chart that demonstrates the logical sequence of events that must be performed to solve a problem.

Types of Flowcharts.

There are 2 common types of Flowcharts:

System flowchart.

A System flowchart is a graphical model that illustrates each basic step of a data processing system.

It illustrates (in summary) the sequence of events in a system, showing the department or function responsible for each event.

Program flowchart.

This is a diagram that describes, in sequence, all the operations required to process data in a computer program.

A program flowchart graphically represents the types of instructions contained in a computer program as well as their sequence & logic.

PROGRAM FLOWCHARTS.

A Flowchart is constructed using a set of special shapes (or symbols) that have specific meaning. Symbols are used to represent operations, or data flow on a flowchart.

Each symbol contains information (short text) that describes what must be done at that point.

The symbols are joined by arrows to obtain a complete Flowchart. The arrows show the order in which the instruction must be executed.

SYMBOLS USED IN PROGRAM FLOWCHARTS.

Below is a standard set of symbols used to draw program flowcharts as created by American National Standard Institute (ANSI).

Terminal symbol.

Ellipse (Oval in shape)

It is used to indicate the point at which a flowchart, a process or an algorithm begins & ends.

All Flowcharts must have a START & STOP symbol. The START/BEGIN symbol is the first symbol of a flowchart, & identifies the point at which the analysis of the flowchart should begin. The STOP/END symbol is the last symbol of a flowchart, & indicates the end of the flowchart.

The words Begin & End (or Start & Stop) should be inserted in the Terminal symbol.

Input or Output symbol.

(Parallelogram)

– It is used to identify/specify an input operation or output operation.

For example;

Input operation Output operation

Note. The words mostly associated with I/O operations are READ & PRINT. READ describes the entry of computer data, while PRINT relates to the printed output of information.

Process symbol.

(Rectangle)

– Process symbol is used to indicate that a processing or data transformation is taking place.

The information placed within the process symbol may be an algebraic formula or a sentence to describe processing.

SUM = A + B

Commission is computed at 20% of Total Sales

Processing defined as a Formula Processing defined as a Sentence

Decision symbol.

NO (Rhombus)

YES

– It is used to indicate/ specify a condition or to show the decision to be made.

There are 2 main components of a Decision symbol:

A question asked within the Decision symbol, that indicates the comparison / logical operation.
The results of the comparison (which are given in terms of YES or NO).

The arrows labeled YES or NO lead to the required action corresponding to the answer to the question.

Flow lines.

Flow lines with arrowheads are used to indicate the direction of processing of the program logic, i.e., they show the order in which the instructions are to be executed.

The normal flow of a flowchart is from Top to Bottom, and Left to Right.

Note. Flow lines should never cross each other.

Connector symbol.

Sometimes, a flowchart becomes too long to fit in a single page, such that the flow lines start crisscrossing at many places causing confusion & also making the flowchart difficult to understand.

The Connector symbol is used as a connecting point for arrows coming from different directions.

A Connector symbol is represented by a Circle, and a letter or digit is placed within the circle to indicate the link.

Note. Connectors do not represent any operation. They are used to connect two parts of a flowchart, indicating that the flow of data is not broken.

General guidelines for drawing a program flowchart.

A flowchart should have only one entry/starting point and one exit point (i.e., ensure that the flowchart has a logical start and finish).
The flowchart should be clear, neat and easy to follow.
Use the correct symbol at each stage in the flowchart.
The flowchart should not be open to more than one interpretation.
Avoid overlapping the lines used to show the flow of logic as this can create confusion in the flowchart.
Make comparison instructions simple, i.e., capable of YES/NO answers.
The logical flow should be clearly shown using arrows.

Note. A flowchart should flow from the Top to Bottom of a page, and from the Left to the Right.

Where necessary, use Connectors to reduce the number of flow lines.

Connectors are helpful when a flowchart is several pages long, and where several loops are needed in the logic of the flowchart.

Check to ensure that the flowchart is logically correct & complete.

Example 1:

Draw a flowchart for a program that can be used to prompt the user to enter two numbers, find the sum and average of the two numbers and then display the output on the screen.

Example 2:

Draw a flowchart for a program that would prompt the user to enter the Length and Width of a rectangle, calculate the Area and Perimeter, then display the result.

Example 3:

Draw a flowchart for a program that can be used to calculate the Diameter, Circumference and Area of a circle and then display the output on the screen.

Example 4:

Design a flowchart for a program that can be used to classify people according to age. If a person is more than 20 years; output Adult else output Young person.

Yes

Example 5:

Draw a flowchart for a program that would be used to classify animals according to sex. If a letter M is input, the program should display Male otherwise it should display Female.

Yes

Example 6:

Write a program using a flowchart to convert temperature from ⁰C to ⁰F.

Fahrenheit =32 + (9^o x C/5).

Notes.

A flowchart must have a Start and an end.

A flowchart is useful when the algorithm is short & the flowchart can fit conveniently on a single page. If the flowchart is too large, it is recommended to use Pseudocodes for long & complicated programs.

Advantages of using Flowcharts.

The following are the benefits of Flowcharts:

Quicker understanding of relationships.

They assist programmers to understand procedures more quickly.

A programmer can represent a lengthy procedure more easily with the help of a flowchart than describing it by means of written notes.

Effective synthesis.

Flowcharts may be used as working models in the design of new programs and systems.

Proper program documentation.

Program flowcharts serve as good program documentation, which is needed for the following reasons:

If programs are modified in future, the flowcharts will direct the programmer on what was originally done.
When staff changes occur, the flowcharts may help new employees understand the existing programs.
Flowcharts assist in program conversion when new hardware/software are acquired.

Effective coding.

Program flowcharts act as a guide during the program preparation stage. Instructions coded in a programming language may be checked against the flowchart to ensure that no steps are omitted.

Orderly debugging and testing of programs.

Flowcharts help in detecting, locating and removing mistakes.

The programmer can refer to the flowchart as he/she re-checks the coding steps, & the logic of the written instructions.

Efficient program maintenance.

Flowcharts facilitate the maintenance of operating programs. They help the programmer to concentrate on the part of the information flow which is to be modified.

Limitations of using Flowcharts.

Flowcharts are complex, clumsy & become unclear, especially when the program logic is complex.

If changes are to be made, the flowchart may require complete re-drawing.

Reproduction of flowcharts is usually a problem, since the flowchart symbols cannot be typed.

No uniform practice is followed for drawing flowcharts as it is used as an aid to the program.

Sometimes, it becomes difficult to establish the link between various conditions, and the actions to be taken upon a particular condition.

Revision Exercise.

Define the following:
(a). State the various types of flowcharts.

(b). Discuss the advantages and disadvantages of flowcharts.

PROBLEM SOLVING

Problem Identification (problem recognition).

Write a program which:

Requests the user to enter a temperature in ^o
Calculates the corresponding temperature in ^o
Outputs the given temperature and the converted value.

Problem definition & Problem Analysis.
- Determine the general requirements, i.e., the main inputs to the program, the main outputs from the program, & also the kind of files which may be needed.

Find out how to convert the given temperature. If the given temperature is in ^oC, then convert it to ^oF.

F = 32 + (9^oC/5)

The Keyboard will be used to enter the Centigrade temperature, and display the output on the Screen.

Design the program.

Develop an Algorithm (a method) for solving the problem.

An Algorithm is a set of instructions which when followed will produce a solution to a given problem.

Write the instructions in such a way that they can be easily converted into a form which the computer can follow.

Computer instructions fall into 3 main categories:

Input instructions used for supplying data to a program inside the computer. The data supplied is stored in the memory of the computer.

Processing instructions used for manipulating data inside the computer.

These instructions allow us to Add, Subtract, Multiply, & Divide. They also allow us to compare two values, and act according to the result of the comparison.

Output instructions used to get information out of the computer.

Note. The programmer must choose which algorithm (method) is the best suited to solve it. This may involve drawing a Flowchart or writing Pseudocode.

Algorithm.

STEP 1: [Prompt the user to enter temperature in ^oC ]

STEP 2: [Store the value in memory]

STEP 3: [Calculate the corresponding temperature in ^oF]

STEP 4: [Store the result in memory location]

STEP 5: [Output the values in^oC & ^oF]

STEP 6: [Stop]

After the algorithm is developed, it must be checked by use of appropriate data values to make sure it is doing its job correctly. This process is called Dry running or Desk checking the algorithm, & is used to pin-point any errors in logic before the program is actually written.

Note. You should never start writing programming codes unless you are absolutely sure that the algorithm is correct.

Algorithms do not depend on any particular language.

Flowchart.

Flowcharts are useful for specifying small algorithms.

A flowchart consists of a set of flowchart symbols connected by arrows. Each symbol contains information about what must be done at that point & the arrows show the order in which the instructions must be executed.

Write a computer program corresponding to the algorithm.

Choose a suitable language & write the computer program using the algorithm.

Use the appropriate language statements & control structures which are found in that programming language.

Test the program to find out whether it is doing its intended job.

Testing involves supplying data values (Test data) to the program for which the answer is known. Some values we can use are 0, 10, 100.

If the program does not give us the correct answers, then we know the program contains some errors.

Debug the program.

I.e., find & correct any errors in the program.

Document the program.

Write out an explanation of how the program works, and how to use it. This includes:

The statement of the problem.
The Algorithm for solving the problem.
The program listing.
Test data, and the results produced by the program.

Note. Documentation should be done at the same time as all the other activities. Each activity produces its own items of documentation which will be part of the final program documentation.

User documentation (User guide)

It enables a non-technical person to use the program without the need to know about the internal workings of the program.

User guides are intended to help the user to use the program (to operate it) with minimal or no guidance.

Example 1:

Assume that the program is called TEMPCONV, and resides on a disk named CONVERT.

Program Name: TempConv

Machine:

This program is designed for use on IBM Microcomputers.

Purpose:

This program simulates the conversion of temperature in ^oC to the corresponding temp. in ^oF. (Attach a description of how the game is played).

The strategy used is for the program to request the user to enter a temperature in ^oC. The user specifies this temperature, and the program calculates the corresponding temperature in ^oF, printing the given temperature and the converted value.

Location:

On the disk named CONVERT

How to Use:

Start PASCAL in drive C:

Insert the disk CONVERT into the disk drive.

Press ALT + ‘F’ to open the File menu, then choose Open.

Type A:TEMPCONV, then press ENTER

Input:

The program requests the user to enter a temp. in ^oC. When the message ENTER TEMP. IN DEGREES CENTIGRADE appears on the screen, you must type the value of the temperature you want converted.

Only digits 0 to 9, a possible sign, and a possible decimal point must be typed. E.g., 25, -20, or 30.5. Typing 25C, for instance, is invalid and will result in an error.

Output:

The program will print a heading followed by the two temperatures. For example, if 40 is entered as the Centigrade temp., the output will be;

Centigrade Fahrenheit

40 104

Example 2:

Program Name: OneZero

Machine:

This program is designed for use on an IBM Microcomputer.

Purpose:

This program simulates the playing of the One zero game. (A description of how the game is played should be attached).

The strategy used is to limit a turn to a maximum number of throws. The user specifies this maximum (as well as the number of turns to play), and the program simulates a game, printing the score obtained.

The program can be used to enable the user to select the maximum number of throws to make on each turn, in order to increase the chances of winning an actual game.

Location:

On the disk named GAMES

How to Use:

Start PASCAL in your computer.

Insert the disk GAMES into the disk drive.

Choose Open from the File menu.

Type A:OneZero, then press ENTER

Press ALT + R

When the prompt: ENTER NUMBER OF TURNS PER GAME appears, type the number of turns for which you want a game to last. Thus, if you want the game to consist of 20 turns, type 20.

When the prompt: ENTER MAXIMUM THROWS PER TURN appears, type a number, say, 5.

The program will simulate one game of 20 turns; each turn will consist of a maximum of 5 throws. When it is finished, it will print the results as in the following sample:

The game consisted of 20 turns

Each turn consisted of a maximum of 5 throws

The score obtained for the game was 156

It will then display the prompt: ENTER MAXIMUM THROWS PER TURN.

Again, you can enter another number (or the same one, if you wish), and the computer will simulate another game of 20 turns, using the new maximum that you have given. If you dont wish to continue, you can enter 0 at this stage and the program will end.

Restrictions:

If you wish to vary the number of turns in a game, the program has to be rerun. You can type: ALT + R, and enter the required number when the prompt: ENTER NUMBER OF TURNS PER GAME appears.

Technical documentation (Technical manuals)

Technical manuals are intended for the systems analysts or programmers.

This documentation is useful to a programmer & can help them in maintenance & modification of the program design and code at a later stage.

Example

Program Name: DiceGame

Purpose:

This program simulates the playing of a dice game. (Attach a description of how the game is played).

Program Structure:

The program consists of three modules;

Main

SimulateOneGame

SimulateOneTurn

The following diagram shows the relationship between these modules:

(Note. Here give the documentation for each individual module. The documentation given is only for the module SimulateOneTurn. Therefore, develop the documentation for the other modules along similar lines).

Module Name: SimulateOneTurn

Parameters: MaxThrowsPerTurn, ScoreThisTurn

Purpose:

Given MaxThrowsPerTurn, this module simulates one turn and returns (in ScoreThisTurn), the score obtained for that turn.

Variables used:

MaxThrowsPerTurn – a parameter representing the maximum number of throws per turn. This value is supplied to the module.

ScoreThisTurn – a parameter used to return the score for the turn to the calling module.

NumberOfThrows – used to count the number of throws made. If the count reaches the maximum, the turn ends.

Modules Called:

Only the standard module RANDOM is called to simulate the throwing of the dice. RANDOM (1, 6) produces a random number in the range 1 to 6, inclusive.

The Algorithm:

MODULE SimulateOneTurn (MaxThrowsPerTurn, ScoreThisTurn)

Set ScoreThisTurn to 0

Set NumberOfThrows to 0

WHILE NumberOfThrows < MaxThrowsPerTurn DO

Set ThrowValue to RANDOM (1, 6)

Add 1 to NumberOfThrows

IF ThrowValue = 1 THEN

Set ScoreThisValue to 0

Set NumberOfThrows to MaxThrowsPerTurn {force loop exit}

ELSE

Add ThrowValue to ScoreThisTurn

ENDIF

ENDWHILE

ENDMODULE

Explanation Notes:

If a 1 is thrown, the turn ends. In this case, a forced exit of the WHILE loop is made. This is done by setting NumberOfThrows to MaxThrowsPerTurn.

Note. If the module was tested individually; a program listing, the test data used, and the results obtained should be included in the documentation of the module.

After each module has been documented, sample runs of the entire program should be added. This should include:

The complete program listing.
Test data used.
Results obtained.

Review Exercise

Why is documentation an essential part of the program development process?
Name four items which the user documentation of a program must contain.
What items make up the technical documentation of a program?

PROGRAM CONTROL STRUCTURES

Control structures are blocks of statements that determine how program statements are to be executed.

Control statements deal with situations where processes are to be repeated several number of times or where decisions have to be made.

There are 3 control structures used in most of the structured programming languages:

Iteration (looping).

SEQUENCE CONTROL STRUCTURES

In Sequence control, the computer reads instructions from a program file line-by-line starting from the first line sequentially towards the end of the file. This is called Sequential program execution.

Start End

Note. Sequential program execution enables the computer to perform tasks that are arranged consecutively one after another in the code.

SELECTION (DECISION) CONTROL STRUCTURES

Selection involves choosing a specified group of instructions/statements for execution.

In Selection control, one or more statements are usually selected for execution depending on whether the condition given is True or False.

The condition must be a Boolean (logical) expression, e.g., X >= 20

In this case, the condition is true if x is equal to or greater than 20. Any value that is less than 20, will make the condition false.

Generally, there are 4 types of selection control structures used in most high-level programming languages:

IF THEN
IF THEN ELSE
Nested IF
CASE – OF

Note. These control structures are used in a program based on a sequence of instructions, which require that a choice (decision) be made between two or more alternatives.

In such a situation, the computer must be programmed to compare data, and take action depending on the outcome of the comparison.

IF THEN

IF THEN structure is used if only one option is available, i.e., it is used to perform a certain action if the condition is true, but does nothing if the condition is false.

The general format of the IF-THEN structure is:

IF < Condition > THEN

Program statement to be executed if condition is true;

ENDIF

If the condition is TRUE, the program executes the part following the keyword THEN. If the condition is FALSE, the statement part of the structure is ignored, and the program continues with the statements below the ENDIF.

The diagrammatic expression of the IF-THEN structure is:

FALSE

TRUE

Continuation of program

Example 1;

In a school, the administration may decide to reward only those students who attain a mean mark of 80% and above.

Flowchart

Pseudocode

IF Mark > 80 THEN

Print Give reward Yes

ENDIF

Example 2;

A user is asked to enter a set of positive numbers, one at a time. She enters a 0 (zero) to indicate that she has no more numbers to enter.

Develop an algorithm to print the largest number entered.

Pseudocode

START

Prompt the user for a number, Largest

Prompt the user for another number, NewNumber

IF NewNumber > Largest THEN

Set Largest to NewNumber

ENDIF

Prompt the user for a number, NewNumber

Output (The largest number entered is, Largest)

STOP

As each number is entered, the algorithm checks if the number entered is larger than the previous ones. If it is larger, it is saved as the largest. If it is smaller, it is ignored, and holds onto the largest number so far.

Example 3;

PROGRAM AgeTalk (Input, Output);

VAR Age: INTEGER;

BEGIN {program}

Writeln (How old are you?); Readln (Age);

IF Age >= 18 THEN

Writeln (You are old enough to join the army.);

END. {program}

Note. Compound statements can also be used with the IF THEN structure.

Example 4;

PROGRAM Service; {*This program displays a message depending on the number of years you have worked for a company*}

VAR Years: INTEGER;

BEGIN

CLRSCR

Writeln (How long have you been with the company?); Readln (Years);

IF Years > 20 THEN

Writeln (Get a Gold watch);

IF (Years > 10) AND (Years <= 20) THEN

Writeln (Get a Paper weight);

IF Years <= 10 THEN

Writeln (Get a pat on the back );

END.

IF THEN -ELSE

The IF-THEN-ELSE structure is suitable when there are 2 available options to select from.

The general format of the IF-THEN-ELSE structure is:

IF < Condition > THEN

Statement 1; (called the THEN part)

ELSE

Statement 2; (called the ELSE part)

ENDIF (indicates the end of the control structure)

The diagrammatic expression of the IF-THEN-ELSE structure is:

TRUE FALSE

Continuation of program

When the IF-THEN-ELSE structure is encountered:

The Condition is tested.

If the Condition is TRUE, the statements between THEN & ELSE (i.e., the THEN part) are executed.

The ELSE part is skipped, and execution continues with the statement following ENDIF.

If the Condition is FALSE, the THEN part is skipped. The statements between ELSE & ENDIF (i.e., the ELSE part of the structure) are executed, and execution continues with the statement following ENDIF.

After either group of statements has been executed, the program will then continue executing the program statements after the last ENDIF.

Note. Using IF-THEN-ELSE, for any given test of the condition, only one set of statements is selected for execution (not both statements).

Example 1;

In a football match, if a player makes a mistake which is considered serious by the rules of the game, he/she is given a Red card. Otherwise, he/she is given a Yellow card.

Flowchart

Pseudocode

IF Fault = Serious THEN

Print Give red card No Yes

ELSE

Print Give Yellow card

ENDIF

Example 2;

Write an algorithm which asks a user for two numbers; A and B, and calculates the value of A divided by B. However, if B is 0, a message is printed which says that division by 0 is not allowed.

Pseudocode

START

Prompt the user for the two numbers, A and B

IF B = 0 THEN

Writeln (Division by 0 is not allowed) ELSE

Set C to A/B

ENDIF

Output A, B, and C

STOP

Explanation.

Suppose the user enters 1 for A and 0 for B in response to the prompt.

The algorithm will test if B=0. Since B is 0, the condition is True. Therefore, the THEN part is executed printing the message: Division by 0 is not allowed.

Suppose the user enters 20 for A and 5 for B in response to the prompt.

The algorithm will test if B=0. Since B is not 0, the condition is False. Therefore, the statements between ELSE & ENDIF are executed (i.e., A is divided by B, and the result is stored in C).

NESTED IF

Nested IF structure is used where 2 or more options have to be considered to make a selection.

The general format of the Nested IF structure is:

IF < Condition 1 > THEN

Statement 1

ELSE

IF < Condition 2 > THEN

Statement 2

ELSE

IF < Condition 3 > THEN

Statement 3

ELSE

Statement 4;

ENDIF

Example;

In an Olympics track event, medals are awarded only to the first three athletes as follows:

Position 1: Gold medal
Position 2: Silver medal
Position 3: Bronze medal

The pseudocode and flowchart below can be used to show the structure of the Nested IF selection.

Pseudocode

IF Position = 1 THEN

Medal = Gold

ELSE

IF Position = 2 THEN

Medal = Silver”

ELSE

IF Position = 3 THEN

Medal = Bronze

ELSE

Medal = nil

ENDIF

Flowchart

No No No

Yes Yes Yes

When IF statements are embedded within one another, they are said to be Nested.

Note. Each IF-THEN or IF-THEN-ELSE is terminated with the comment {ENDIF}. The number of {End Ifs} must be equal to the number of ELSEs.

The CASE structure

CASE-OF allows a particular group of statements to be chosen from several available groups.

It is therefore used where the response to a question involves more than two choices/alternatives.

The general format of the CASE structure is:

CASE Expression OF

Label 1: statement 1

Label 2: statement 2

Label 3: statement 3

Label n: statement n

ELSE

Statement m

ENDCASE

The Boolean expression for the CASE structure can only be expressed using Integers or alphabetic characters Hence;

CASE Integer OF or CASE Char OF

A statement is executed only if one of its corresponding labels matches the current value of the expression. This implies that, the current value of the expression determines which of the statements will be executed.

Example 1;

Write a pseudocode of a program that requests the user to type a number from 1 to 7. The program then prints the corresponding day of the week.

Pseudocode

START

Prompt the user for a number from 1 to 7, Day

CASE Day OF

1: Writeln (Sunday);

2: Writeln (Monday);

3: Writeln (Tuesday);

4: Writeln (Wednesday);

5: Writeln (Thursday);

6: Writeln (Friday);

7: Writeln (Saturday);

ENDCASE

STOP

The CASE structure consists of:

The word CASE.
A Control variable (e.g., Day).
The word OF.
A group of one or more statements, each group labeled by one or more possible values of the control variable.
The word ENDCASE, indicating the end of the construct.

When a CASE statement is encountered, the value of the control variable is used to determine which group of statements is executed, e.g., if the value of Day is 5, then the group of statements labeled 5 is selected for execution, and the statement; Thursday is printed.

After executing this group of statements, execution continues at the statement following ENDCASE.

NOTES:

The programmer should ensure that the value of the control variable appears as a label. g., suppose the value entered for Day was 9. Since 9 does not label any statement within the CASE construct, an error will result.

A given label can be used on only one group of statements. g., 5 cant be used to label two groups of statements. If this is done, the computer will not know which group to select & unpredictable results can occur.

Example 2;

Write a pseudocode of a program that requests the user to type a number from 1 to 7. Depending on the number entered, print the message, It is a School day or It is on a Weekend.

Pseudocode

Prompt the user for a number from 1 to 7, Day

IF (Day < 1) OR (Day >7) THEN

Print (Invalid number entered —, Day)

ELSE

CASE Day OF

2, 3, 4, 5, 6: Writeln (It is a School day);

1, 7: Writeln (It is on a Weekend);

ENDCASE

ENDIF

STOP

In this pseudocode, the IF statement has been used to validate the value of Day. This ensures that, only valid data gets processed by the CASE statement.

Otherwise, if the ELSE part is executed, we are sure that the value of Day will lie between 1 and 7 inclusive.

Example 3;

Pseudocode

CASE Average OF

80 .. 100: Grade = A

70 .. 79: Grade = B

60 .. 69: Grade = C

50 .. 59: Grade = D

40 .. 49: Grade = E

ELSE

Grade = F

ENDCASE

Flowchart

No No No No

Yes Yes Yes Yes

Example 4;

PROGRAM CaseSample (Input, Output);

VAR Grade:CHAR;

BEGIN {Program}

Writeln (What grade did you get?); Readln (Grade);

CASE Grade OF

A, B : Writeln (Very Good);

C : Writeln (Pass);

D, F : Writeln (Wake up);

End; {Case}

Readln;

End. {Program}

ITERATION (LOOPING / REPETITION) CONTROL STRUCTURES

Looping refers to the repeated execution of the same sequence of statements to process individual data. This is normally created by an unconditional branch back to a previous/earlier operation.

The loop is designed to execute the same group of statements repeatedly until a certain condition is satisfied.

Note. Iteration is important in situations where the same operation has to be carried out on a set of data many times.

The loop structure consists of 2 parts:

Loop body, which represents the statements to be repeated.
Loop control, which specifies the number of times the loop body is to be repeated.

Types of loops:

Conditional loop: – This is where the required number of repetitions is not known in advance.

Pseudocode

STEP 1: [Prompt the user for temperature in ^oC]

STEP 2: [Store the value in memory]

STEP 3: IF C = 0 THEN Stop

STEP 4: [Calculate temperature in ^oF]

F: = 32 + (^oC * 9/5)

STEP 5: [Output temperature in ^oC &^oF]

STEP 6: [GOTO Step 1]

Flowchart

YES

This algorithm illustrates Conditional execution. Conditional execution is a situation that requires that a logical test be carried out, and then a particular action be taken depending on the outcome of that test.

In this case, going to Step 4 will depend on whether the condition is True or False. E.g., If C = 10 then the condition C = 0 is False, and the program goes to Step 4. But if C = 0, then the condition is True, and the program stops.

Unconditional loop: – This is where the execution of the instructions is repeated some specified number of times.

Continuous (infinite/unending) loop: – This is where the computer repeats a process again and again, without ending.

Example:

STEP 1: [Prompt the user for temperature in ^oC]

STEP 2: [Store the value in memory]

STEP 3: [Calculate temperature in ^oF]

F: = 32 + (^oC * 9/5)

STEP 4: [Output temperature in ^oC &^oF]

STEP 5: [GOTO Step 1]

As long as a number is entered for^oC, the algorithm does not stop when it reaches STEP 5 but rather transfers control to STEP 1, causing the algorithm/process to be repeated.

However, a zero (0) can be used to stop the program because; the program cannot give the Fahrenheit equivalent to 0 ^oC.

Requirements for loops:

Control variable (Counter): – it tells/instructs the program to execute a set of statements a number of times.
Initialization: – allocating memory space, which will be occupied by the output.
Incrementing: – increasing the control variable by a certain number before the next loop.

Generally, there are 3 main looping controls:

The WHILE loop
The REPEATUNTIL loop.
The FOR loop.

The FOR loop

The FOR loop is used in situations where execution of the chosen statements has to be repeated a predetermined number of times.

The general format of the FOR loop is:

FOR loop variable = Lower limit TO Upper limit DO

Statements;

END FOR

The flowchart extract for a FOR loop that counts upwards is:

YES

Example;

Consider a program that can be used to calculate the sum of ten numbers provided by the user. The FOR loop can be used to prompt the user to enter the ten numbers at most 10 times. Once the numbers have been entered, the program calculates and displays the accumulated sum.

Pseudocode Flowchart

FOR count = 1 TO 10 DO

PRINT Enter a number (N)

Sum = Sum + N

END FOR

Display SUM

YES

Explanation

The loop variable (Count) is first initialized/set to the Lower limit whose value is 1.

The lower limit is then tested against the Upper limit whose value is set at 10.
If the lower limit is less than or equal to 10, the program will prompt the user to enter a number N, otherwise the computer will exit the loop.

After the last statement in the loop has been executed, the loop variable (count) is incremented by a 1 and stored in the lower limit, i.e., Lower limit = Count + 1.

The lower limit is again tested, and if it is less than or equal to 10, the loop is repeated until the time the lower limit will equal the upper limit.

NOTE:

The FOR loop can also be used to count downwards from the upper limit to the lower limit.

E.g., FOR count = 10 DOWN TO 1DO

In this case, the upper limit 10 is tested against the lower limit 1.

Pseudocode for a FOR loop that counts from upper limit down to the lower limit:

FOR loop variable = Upper limit DOWN TO Lower limit DO

Statements;

END FOR

The flowchart extract for a FOR loop that counts downwards is:

YES

The WHILE loop

The WHILE loop is used if a condition has to be met before the statements within the loop are executed.

E.g., to withdrawal money using an ATM, a customer must have a balance in his/her account.

Therefore, it allows the statements to be executed zero or many times.

Pseudocode Flowchart

WHILE Balance > 0 DO

Withdraw cash

Update account

ENDWHILE

YES

Exit loop

Explanation

The condition balance > 0 is first tested.
If it is TRUE, the account holder is allowed to withdraw cash.
The program exits the loop once the balance falls to zero.

The general representation of the WHILE loop is:

Pseudocode segment Flowchart extract

WHILE Condition DO

Statements;

ENDWHILE

TRUE

FALSE

Exit loop

The REPEATUNTIL loop

In REPEATUNTIL, the condition is tested at the end of the loop. Therefore, it allows statements within it to be executed at least once.

E.g., if REPEATUNTIL is used in case of the ATM cash withdrawal, the customer will be able to withdraw the cash at least once since availability of balance is tested at the end of the loop.

Pseudocode Flowchart

REPEAT

Withdraw cash

Update account

UNTIL balance <= 0;

Yes

Exit loop

The general format of the REPEATUNTIL loop is:

Pseudocode segment Flowchart extract

REPEAT

Statements;

UNTIL Condition;

True

False

Exit loop

DEVELOPING COMPLEX ALGORITHMS

Example 1:

With aid of a pseudocode and a flowchart, design an algorithm that:

Prompt the user to enter two numbers X and Y.
Divide X by Y. However, if the value of Y is 0, the program should display an error message Error: Division by zero.

Pseudocode

START

PRINT Enter two numbers X and Y

INPUT X, Y

IF Y = 0 THEN

PRINT Error: Division by zero

ELSE

Quotient = X/Y

PRINT X, Y, Quotient

ENDIF

STOP

Flowchart

Yes

Example 2:

In an athletics competition, an athlete is rewarded as follows:

1st position: Gold

2nd position: Silver

3rd position: Bronze

Draw a pseudocode and a flowchart for a program that would be used to determine the type of medal to be rewarded to each athlete.

Pseudocode

START

PRINT Enter athlete Name and Position

INPUT Name, Position

IF Position = 1 THEN

Medal = Gold

ELSE

IF Position = 2 THEN

Medal = Silver

ELSE

IF Position = 3 THEN

Medal = Bronze

ELSE

Medal = None

ENDIF

Flowchart

No No No

Yes Yes Yes

Yes

Example 3:

The class teacher of Form 3S in a secondary school requested a programmer to design for her a simple program that would help her do the following:

Enter the names of students and marks obtained in 8 subjects Mathematics, English, Kiswahili, Biology, Chemistry, Business studies, Computer studies, and History.

After entering the mark for each subject, the program should calculate the total and average marks for each student.

Depending on the Average mark obtained, the program should assign grade as follows:
- Between 80 and 100 A
- Between 70 and 79 B
- Between 60 and 69 C
- Between 50 and 59 D
- Below 50 E
The program should then display each students Name, Total marks and the Average grade.

Using a pseudocode and a flowchart, write an algorithm that shows the design of the program.

Pseudocode

START

REPEAT

PRINT Enter student Name and subject marks”

INPUT Student name, Maths, Eng, Kisw, Bio, Chem, Business, Computer, History

SUM = Maths + Eng + Kisw + Bio + Chem + Business + Computer + History

AVG = SUM/8

IF (AVG => 80) AND (AVG <= 100) THEN

Grade = A

ELSE

IF (AVG => 70) AND (AVG <= 79) THEN

Grade = B

ELSE

IF (AVG => 60) AND (AVG <= 69) THEN

Grade = C

ELSE

IF (AVG => 50) AND (AVG <= 59) THEN

Grade = D

ELSE

Grade = E

ENDIF

PRINT Student name, Sum, AVG, Grade

UNTIL Count = Number of students

STOP

Flowchart

No No No No

Yes Yes Yes Yes

Yes

Example 4:

The gross salary of employees in ZAG BOOKS ENTERPRISE is based on basic salary and additional benefits as follows:

Employees who have worked for the company for more than 10 years receive an additional pay of 10% to their basic salary.

Monthly salary bonus based on monthly sales of books as follows:

Monthly sales	Bonus Rate (%)
Above 500,000	15
Between 250,000 and 500,000	10
Below 250,000	5

Draw a flowchart for a program that would be used to calculate the gross salary then output each employees basic salary, gross salary and all benefits.

Yes

Yes No

Example 5:

A lady deposits 2,000 shillings in a Microfinance company at an interest rate of 20% per annum. At the end of each year, the interest earned is added to the deposit and the new amount becomes the deposit for that year.

Write a pseudocode for a program that would track the growth of the deposits over a period of seven years.

START

INPUT Initial Deposit

INPUT Interest Rate

SET Deposit to Initial deposit (i.e., 2000)

SET Year to 0

WHILE Year <= 7 DO

Interest = Deposit x Interest rate

Total = Deposit + Interest

Deposit = Total {the new deposit}

Year = Year + 1

ENDWHILE

PRINT Deposit, Year

STOP

Example 6:

Draw a flowchart for a program that is to prompt for N numbers, accumulate the sum and then find the average. The output is the accumulated totals and the average.

YES

Example 7:

Mutuku took a loan of Ksh. 400,000 from a local bank at an interest rate of 10% payable in four years. Assuming you wish to develop a computer program that will keep track of monthly repayments:

Identify the input, processing and output requirements for such a program.
Design the algorithm for the program using a simple flowchart and pseudocode.

Requirements:

Input – Initial amount borrowed

– Interest rate

– Number of years

Processing – equation to calculate Yearly repayments and Monthly repayments.

Output – Monthly repayments calculated by the process

Pseudocode:

START

INPUT Initial amount borrowed

INPUT Interest rate

INPUT Number of years

Calculate Yearly repayments

Monthly repayments = (Yearly repayments / 12)

OUTPUT Monthly repayments

STOP

Flowchart:

Teachers' Resources

A STUDY GUIDE TO A DOLL’S HOUSE BY HENRIK IBSEN LATEST

October 5, 2025 Maverick John Leave a comment

A STUDY GUIDE TO

A DOLL’S HOUSE

BY HENRIK IBSEN

ABOUT THE AUTHOR

Henrik Ibsen was born on March 20, 1828, in Skien, Norway. He was the second son in a wealthy family that included five other siblings. When he was about 8 years old, his family was thrown into poverty due to complications with his father’s business. It was after this when Ibsen started to invest his time reading, writing, painting, and doing magic tricks.

Ibsen wrote his first play, Catiline, in 1850 which generated little interest. His second play, The Burial Mound, however, was performed at the Christiania Theatre on September 26, 1850.

Later, he wrote a series of plays which included Lady Inger (1855), The Feast at Solhoug (1856), Olaf Liljekrans (1857), The Vikings at Helgeland (1858), The Pretenders (1863), Peter Gynt (1867), The League of Youth (1869), Emperor and Galilean (1873), Pillars of Society (1877), A Doll’s House (1879), Ghosts (1881), An Enemy of the People (1882), The Wild Duck (1884), Hedda Gabler (1890), The Master Builder (1892), Little Eyolf (1894), John Gabriel Borkman (1896) and When We Dead Awaken (1899). He also wrote a dramatic epic poem, “Brand” (1866).

He married Suzannah Thoresen in 1858 and their only child, Sigurd, was born the following year. In 1900, Ibsen suffered his first of several strokes and poor health ended his writing career. He died on May 23, 1906.

GENRE

A Doll’s House is a family drama for the obvious reason that it concerns a family. It is a “drama” because it is a play—a piece of literature that is never fully realized until it is put on stage in front of an audience.

It is also a modern tragedy because it focuses on the trials and tribulations that face women in a patriarchal society. The play explores not only the status of women, but how they are victims of social forces to the extent that they are left with the role of a “doll-wife.” In this tragedy, we don’t get blood and death at the end; we get the death of a marriage and of the characters’ old selves. Ibsen shows Nora, and maybe all the other characters, trapped in a society defined by restrictive gender roles. In order to become more than a doll, Nora must shatter the cornerstone that her entire society is based on: marriage.

The play can also be categorized as a realist drama. In a realist drama, the characters talk in a close approximation of everyday speech. The speeches are straightforward, conversational and concerned with normal, everyday things; which makes the play really easy for a modern audience to associate with.

Example

MRS. LINDE: “You must not forget that I had a helpless mother and two little brothers. We couldn’t wait for you, Nils; your prospects seemed hopeless then.” (Pg 86)

The vast majority of modern plays, TV shows, and movies are written in a similar style.

ABOUT THE TITLE

Just before Nora leaves her husband and children at the end of the play, she has the following to say to her husband, Torvald: “Our home has been nothing but a playroom. I have been your doll-wife.”(Pg 111-112)

It is therefore not too hard to guess where the play’s title might have come from. Torvald has never treated Nora as anything more than a plaything. He admires her beauty. He gets her to dance for him. He even dresses her up in costumes. In effect, she is his doll. The home they live in seems perfect and picturesque, but in reality it is just like the Helmers’ marriage: all for show.

Nora adds, “at home I was papa’s doll-child.”(Pg 112) She has never been anything but a man’s plaything. Every house she’s ever lived in has been just as artificial; first her father’s house, and now her husband’s house.

No wonder the play is titled A Doll’s House!

TONE

In the beginning, the play seems to be biased toward Nora. We are definitely inclined to sympathize with her. It is very hard to be on Torvald’s side. From his reaction toward Nora for eating macaroons, we know that he is overbearing. His demeaning little pet names for Nora further confirm this.

Torvald, however, redeems himself in the end with the last line, “The most wonderful thing of all?”(Pg 120) The line seems to indicate that he is heading toward the same spiritual awakening as Nora.

This makes us move from seeing Nora as Torvald’s prisoner to seeing that all the characters, Torvald included, have been prisoners in some way.

In the end, the tone of the play becomes more objective. Sympathy can be found for all its characters. Hence the play can be said to end with a serious, intense and somber tone.

SETTING

Setting can be discussed from three dimensions: Geographical, historical and social setting.

Geographical setting

This refers to the place or location where the events in the play are taking place. In the play A Doll’s House, the events take place in The Helmers’ Living Room. The dwelling contains comfortable and stylish furniture and such items as a china cabinet, a bookcase with well-bound books, and a piano on carpeted floor—all of which demonstrate a stable financial situation. On a broader level, it is assumed that the events take place in Norway in Europe; however there are no references to anything specially Norwegian. This assumption is made because that is where Ibsen was born and raised.

Historical setting

This refers to the time in history when the events in the play took place. The events in A Doll’s House took place in The Victorian Era, presumably around the late 1870s. During this time, gender roles were very stiff and clearly outlined. Women were expected to be submissive to their husbands; husbands were expected to dominate. Women raised the children; men went out to work. Anyone who challenged these deeply entrenched values faced some serious consequences. This charged atmosphere of gender division was the reason that the play became such a phenomenon.

Social setting

Social setting refers to the kind of a society in which the events in the play are taking place. The play involves a middle-class society of family and friends who are reeling under the pressure of strict Victorian values which eventually result to conflicts.

STRUCTURE

Henrik Ibsen’s A Doll’s House is divided into three Acts. Ibsen followed the form of a well-made play. Features of a well-made play include increasing suspense by methodical plotting, introducing past events early on and unraveling a secret, which leads to the climax of the play.

The play circumvents through four major stages:

Major conflict – This comes in the form of Nora’s struggle with Krogstad, who threatens to tell her husband about her past crime, incites Nora’s journey of self-discovery and provides much of the play’s dramatic suspense. Nora’s primary struggle, however, is against the selfish, stifling, and oppressive attitudes of her husband, Torvald, and of the society that he represents.
Rising action – This comes in Nora’s first conversation with Mrs. Linde; Krogstad’s visit and blackmailing of Nora and Krogstad’s delivery of the letter that later exposes Nora.
Climax – This is reached when Torvald reads Krogstad’s letter and erupts angrily.
Falling action – This finally comes in Nora’s realization that Torvald is devoted not to her but to the idea of her as someone who depends on him and her decision to abandon him to find independence.

CHARACTER LIST

Nora Helmer

Nora is the play’s protagonist and the wife of Torvald Helmer. She is the central character, who is a “doll” for her husband to dress up, show off, and give direction to. She is childlike and romps easily with her three children. She has never lived alone, going immediately from the care of her father to that of her husband. Inexperienced in the ways of the world as a result of this sheltering, Nora is impulsive and materialistic. She takes a loan from Krogstad to make her husband’s holiday possible. Later, she emerges as a fully independent woman who rejects both the false union of her marriage and the burden of motherhood.

Torvald Helmer

Torvald Helmer is Nora’s husband of eight years. At the beginning of the play, he has been promoted to manager of the bank. He was once gravely ill and needed to go to a southern climate to improve his health. He has built his own legacy through his own work and not from family money. He lives his life according to society’s norms – both professionally and personally. He spends a great deal of his time at home in his study, avoiding general visitors and interacting very little with his children. In fact, he sees himself primarily as responsible for the financial welfare of his family and as a guardian for his wife. Torvald is particularly concerned with morality. He also can come across as stiff and unsympathetic. Still, the last Act of the play makes it very clear that he dearly loves his wife.

Dr. Rank

Dr Rank is a friend of the family of Torvald as well as his physician. He is sick from consumption of the spine (tuberculosis of the spine) as a result of a venereal disease contracted by his father. He confesses his desire for Nora in the second Act and dies in the third Act.

Mrs. Christine Linde

Mrs Linde is an old schoolmate of Nora’s. She is a widow. She comes back into Nora’s life after losing her husband and mother. She successfully asks Nora to help her secure a job at Torvald’s bank. Ultimately, she gets married to Krogstad.

Nils Krogstad

Nils Krogstad is a man from whom Nora borrows money to pay for her family’s trip to Italy. He is an acquaintance of Torvald’s and an employee at the bank which Torvald has just taken over. He is also a lawyer and moneylender. Krogstad was involved in a work scandal many years previously; as a result, his reputation is tainted because he once committed a forgery. When his job at the bank is threatened by Torvald, he blackmails Nora to ensure that he does not lose it. Dr. Rank calls Krogstad “morally diseased.”(Pg 25)

Ivar, Bob, and Emmy

These are Nora’s young children. They spend little time with their mother or father: they are mostly with their nurse, Anne. In the play, the children speak no individualized lines; they are “Three Children.” Ibsen facilitates their dialogue through Nora’s mouth.

Anne

Anne is the family nurse. She raised Nora too after she (Nora) lost her mother to death. She stayed on to raise Nora’s children. Nora is confident that she can leave her children in Anne’s care. She gave up her own daughter to “strangers.”

Helen

Helen is a housemaid employed by the Helmers.

Porter

The porter brings in the Christmas tree at the very beginning.

Nora’s father

Although he never makes a physical presence during the play, Nora’s father’s influence is felt throughout its course. Torvald repeatedly brings up his loose morals and past scandals to compare them to Nora.

SYNOPSIS

Nora Helmer once secretly borrowed a large sum of money so that her husband, Torvald Helmer, could recuperate from a serious illness in Italy. She never told him of this loan and has been secretly paying it back in small installments by saving from her household allowance. Her husband thinks her careless and childlike, and often calls her his doll.

When he is appointed bank director, his first act is to relieve a man who was once disgraced for having forged his signature on a document. This man, Nils Krogstad, is the person from whom Nora has borrowed her money! It is then revealed that she forged her father’s signature in order to get the money.

Krogstad threatens to reveal Nora’s crime and thus disgrace her and her husband unless Nora can convince her husband not to fire him. Nora tries to influence her husband, but he thinks of Nora as a simple child who cannot understand the value of money or business. Thus, when Torvald discovers that Nora has forged her father’s name, he is ready to disclaim his wife even though she had done it for him.

Later when all is solved, Nora sees that her husband is not worth her love and she leaves him.

PLOT SUMMARY AND ANALYSIS

ACT I

SUMMARY

The play opens on the day before Christmas. Nora returns home from shopping; although her husband is expecting a promotion and payrise, he still criticizes her excessive spending. In response, Nora plays around with her husband as a child might, and, indeed, Torvald addresses her as he might a child. He hands her more money but only after having criticized her spending. Their relationship compares with that of a daughter and father and, indeed, is exactly like the relationship Nora had with her father. Early in this act the audience is aware that the relationship between the Helmers is based on dishonesty when Nora denies that she has eaten macaroons, knowing that her husband has forbidden her to do so.

Nora is visited by an old friend, Christine Linde. Mrs. Linde tells Nora that she has had some difficult problems and is looking for employment. Nora confesses to Mrs. Linde that she, too, has been desperate and reveals that she had been forced to borrow money several years earlier when her husband was ill. The money was necessary to finance a trip that saved her husband’s life, but Nora forged her father’s signature to secure the loan and lied to Torvald that her father had given them the money. Thus, she has been deceiving her husband for years as she worked to repay the loan. She tells this story to Mrs. Linde to demonstrate that she is an adult who is capable of both caring for her family and conducting business. Unfortunately, Nora’s secret is known by Krogstad, an employee at Torvald’s bank. After a confrontation with Krogstad, Torvald decides to fire Krogstad and hire Mrs. Linde in his place.

Krogstad threatens Nora, telling her that if he loses his job he will expose her earlier dishonesty. For her part, Nora cannot believe that forging her father’s signature – an act that saved her husband’s life – could lead to a serious punishment. Still, she is concerned enough to plead with Torvald on behalf of Krogstad. Torvald refuses to reconsider firing Krogstad and forbids Nora to even mention his name.

ANALYSIS

The Helmers’ house is decorated tastefully, showing they are relatively well-off. Nora’s happiness as she returns with the Christmas shopping reveals that she enjoys both spending money and doing nice things for her husband and children. At the same time, it will soon become clear that eating the macaroons is an act of deceit and disobedience, as she has been forbidden by Torvald.

Torvald’s nicknames for Nora suggest that he thinks of her almost as a child or a pet. This impression is emphasized when Nora hides the macaroons, like a mischievous child afraid of getting caught. Torvald’s parent-like attitude is highlighted by the way he talks to Nora about money, implying that he thinks she’s not intelligent enough to be financially responsible.

Nora’s happiness shows she enjoys performing the role of a wife and mother. At the same time, her request for money to buy something for herself suggests she wants to be allowed to make decisions for herself. But Torvald clearly doesn’t trust Nora with the money.

Even though Torvald and Nora appear to be in love, Torvald does not trust her, and Nora on her part doesn’t hesitate to lie to him; she was eating macaroons earlier.

Money is central to Torvald and Nora’s happiness. Torvald’s emphasis on their new prosperity suggests how important money is to him as well.

Mrs. Linde has been visibly changed by her life experiences. Nora’s happiness in the last eight years has left her remaining girlishly innocent and naive, whereas Mrs. Linde seems much older. Mrs. Linde’s decision to travel alone was unusual for women at the time, and Nora’s admiration of her “courage” suggests a desire for independence. Mrs. Linde’s status as a widow adds to the impression that she is much older than Nora.

In this part of the play Nora is childishly impolite. Mrs. Linde is obviously in a bad situation following the death of her husband, yet instead of listening to her Nora begins to insensitively boast about her and Torvald’s good fortune. Her speech also shows that she believes money leads to freedom and happiness.

Mrs. Linde’s story shows how difficult it was for women to survive without the financial support of men. The need for money effectively forced her to marry her husband, and after his death her struggle to support her family highlights the obstacles women faced in earning a reasonable income.

Both Mrs. Linde and Nora have strange and suspicious reactions to Krogstad’s arrival. Thus when Krogstad claims he is here on “routine” business matters, we are tempted to believe there is more to the story.

Here, Krogstad reveals more about Nora’s deceitful nature; not only did she lie to Torvald (and everyone else) about where the money for the trip to Italy came from, but she also committed forgery, an illegal act. He threatens to reveal the secret unless she does him a particular favour. Nora is terrified to the point that she even seems to be going mad.

ACT II

SUMMARY

Mrs. Linde stops by to help Nora prepare for a costume ball. Nora explains to Mrs. Linde that Krogstad is blackmailing her about the earlier loan. After Nora again begs Torvald not to fire Krogstad, her husband sends Krogstad an immediate notice of his dismissal. Nora is desperate and decides to ask help from Dr. Rank, a family friend, for a loan, to clear Krogstad. Before she can ask him for his help, Dr. Rank makes it obvious that he is in love with her and Nora decides that because of this it would be unwise to ask his help. Krogstad visits Nora once again and this time leaves a letter for Torvald in which Nora’s dishonesty is revealed. To divert Torvald’s attention from the Krogstad’s letter in the mailbox, Nora engages him to help with her practice of the dance she is to perform, the tarantella. Finally, Nora asks Torvald to promise that he will not read the mail until after the party.

ANALYSIS

In the opening of the second act, the stripped Christmas tree not only shows that time has passed, but also symbolizes a negative shift from the joy of Christmas to a sense of ruin and chaos. Nora’s obsession in checking to see if any person or letter has arrived and assurances that no one will come for two days gives a sense of time running out and impending disaster.

Nora cannot think of anything else but her secret and the possibility of someone finding out. She tries to occupy herself with the clothes but is unable to.

As the play progresses, it becomes more and more clear how possessive Torvald is. Nora’s pride at saying Dr. Rank is “her” friend suggests she doesn’t really have many friends now that she is married. Nora believes that the reason that Torvald is so controlling is because he is so in love with her.

Nora seems increasingly desperate and crazed. Her mutterings to herself when she is alone show the effect that concealing her secret in front of others is having on her. She lies easily to Dr. Rank, showing how natural lying has become to her.

Nora flirts with Dr. Rank in a very provocative manner. When she teases him with the stockings, this is a very explicit sexual gesture. Her promise to dance for him likewise betrays a disregard for the boundaries of her marriage and a delight in displaying her femininity and sexuality.

Nora is almost asking Dr. Rank to help with keeping the secret of the debt from Torvald, but she is stopped by his confession of love. The confession changes her view of Dr. Rank completely. Where before she perhaps thought flirtation was harmless, the fact that Dr. Rank seems to genuinely love her becomes too much to handle, and she retreats in a rather childlike way.

Krogstad is determined to keep his position at the bank, to the extent of lacking etiquette for Nora, which shows he is desperate. Meanwhile, Nora must cover her tracts in front of everyone—even the maid—hence increasing her isolation.

ACT III

SUMMARY

In this act, it is revealed that Krogstad had years earlier been in love with Mrs. Linde. At the beginning of this act they agree to marry, and Krogstad offers to retrieve his letter from Torvald. However, Mrs. Linde disagrees and thinks that it is time that Nora is forced to confront the dishonesty in her marriage. After the party, the Helmers return home and Torvald opens the letter from Krogstad. While Torvald reads it in his study, Nora pictures herself as dead, having committed suicide by drowning in the icy river. Torvald interrupts her fantasy by demanding that she explains her deception.

However, he refuses to listen and is only concerned with the damage to his own reputation. Torvald’s focus on his own life and his lack of appreciation for the suffering undergone by Nora serve to open her eyes to her husband’s selfishness. She had been expecting Torvald to rescue her and protect her, and instead he only condemns her and insists that she is not fit to be a mother to their children.

At that moment another letter arrives from Krogstad telling the Helmers that he will not take legal action against Nora. Torvald is immediately excited and is willing to forget the entire episode. But having seen her husband revealed as self-centered, egoistic and hypocritical, Nora tells him that she can no longer live as a doll and expresses her intention to leave the house immediately. Torvald begs her to stay, but the play ends with Nora leaving the house, her husband, and her children.

ANALYSIS

Here, Mrs. Linde radically disrupts the course of events in the play. While it would have been easier for her to ask Krogstad to get his letter back, thereby ensuring that life between the Helmers went on as normal, Mrs. Linde’s belief in honesty triumphs over her promise to Nora. This finally benefits Nora, as Torvald’s behaviour when he reads the letter allows her to see the reality of her situation and that she no longer wants to remain in her marriage.

In this act it is clear that Torvald is thinking of Nora far more as a possession that he can display in order to impress other people than a real person with her own thoughts and feelings. To him, Nora was at the party merely to perform for the enjoyment of him and others, not to have a good time herself.

Nora’s bitterness toward Mrs. Linde because she did not get Krogstad to retrieve the letter shows that she has cut herself off even from her close friends in her obsession with the secret of the debt. All the hope and innocence seems to have drained out of her, and she has become a much more serious, grave person.

In his speech we see that Torvald’s love and desire for Nora is revealed to be cosmetic, rather than an appreciation for whom she truly is as a person. He talks about his sexual desire for her with no consideration of whether she is feeling the same way at the moment; indeed, when she tells him that she doesn’t want to be with him that night, he dismisses her feelings by saying she must be playing a game. In reminding her that he is her husband, Torvald is suggesting that their marriage means Nora does not have the right to refuse sex with him, a commonly held belief at the time.

Nora is preparing to kill herself, perhaps the ultimate symbol of self-sacrifice. Her whispering murmurs on the stage suggest that she is becoming mad.

Throughout this whole section of the play Torvald only thinks of himself and doesn’t pause to consider the way Nora has been and will be affected by Krogstad’s threats, or that Nora did what she did purely out of love for him.

Nora has evidently undergone a transformation both visually and in the way she speaks to Torvald. For the first time, she is addressing him as an equal and demanding that he treats her with respect by listening and not interrupting.

Finally, Nora conducts what can be considered an unofficial divorce ceremony. Although Torvald doesn’t want her to go, the fact that he agrees to give her his ring and not to write or try to help her shows that he finally respects her wishes and ability to make decisions for herself.

CHARACTERS, CHARACTERIZATION AND

ROLE

NORA HELMER

Nora Helmer is the protagonist or the main character or the heroine in the play. Still a young woman, she is married to Torvald Helmer and has three children. Nora is by far the most interesting character in the play. Her whole life is a construct of societal norms and the expectations of others.

CHARACTER TRAITS

Impulsive and a spendthrift

-In her first moments onstage, we see her give the porter an overly generous tip.

-She comes in with tonnes of Christmas presents, and shrugs at the idea of incurring debt.

-Unbeknown to Torvald, Nora borrowed money so that they could afford a year-long trip to Italy.

Loving

-Nora borrowed money so that they could afford a year-long trip to Italy. Doctors said that Torvald would die without it—but that he shouldn’t know how bad his condition was.

-Nora brings home lots of Christmas presents for everybody in her house.

-She plays hide-and-go-seek with her kids.

Independent and farsighted

-In the past, Nora was always a passive child-like possession who followed Torvald’s orders, but towards the end of the play, she is an independent adult and is able to dominate Torvald.

Wise and intelligent

-Nora uses wisdom and intelligence to confront an emergency. She forges her father’s signature in order to secure a loan from Krogstad so as to save her husband’s life.

-Nora realizes that her understanding of herself, her husband, her marriage, and even her society was all wrong. She decides that she can no longer be happy in her life and marriage, and resolves to leave Torvald and her home in order to find a sense of self and learn about the world, a newly empowered woman boldly escaping the oppressive clutches of her old life.

-Nora has been leading a double life. She has not been thoughtlessly spending their money. Rather, she has been saving to pay off a secret debt.

Childlike, immature, ignorant and whimsical

-She happily accepts the pet names “little lark”, “little squirrel”, and “Little Miss Extravagant” that her husband calls her without any opposition. In fact she seems to enjoy and even play into it.

-The maturity level Nora exhibits demonstrates that the relationship between Torvald and Nora is more like father and daughter than husband and wife.

Irresponsible and reckless

-Her first act on stage is paying the porter. Though his service only costs sixpence, she gives him a shilling. (Pg 1) The casual way in which she gives it to him is indicative of her irresponsibility. She hands him the shilling and before he can thank her, she decides in the middle of the transaction that she is not patient enough to wait for change.

-She forges loan documents to raise money for an expensive trip to Italy. Even if the documents were not forged, Nora did not have any means to repay the loan anyway.

-She has never spent serious time with her husband of nearly a decade, and is always dumping her children on the nurse rather than bonding with them herself.

Dishonest and deceitful

-She falsely blames Mrs. Linde for smuggling forbidden macaroons into the house.

-She has been eating macaroons, something she has been forbidden by her husband, despite her promises of total obedience to him.

– At the beginning of her marriage, she secretly borrowed money from Nils Krogstad and forged her father’s signature in order to finance a trip to Italy that was necessary to save Torvald’s life.

Unfeeling

– She blames Mrs. Linde for smuggling forbidden macaroons into the house. Though she is just trying to hide her indiscretions, she does not care whom she hurts in the process.

Materialistic

-She is always trying to make herself happy by buying things: dresses, toys, candy etc., rather than doing anything meaningful with her life.

– She is infatuated with the acquisition of possessions.

Decisive, resolute and independent-minded

– At the end of the play, it becomes clear to Nora that “[she] had been living all these years with a strange man, and [she] had born him three children.”(Pg 117-118) This realization forces her into the real world and she ceases to be a doll. At the end of the above statement, she adds “Oh, I cannot bear to think of it!”

-She tells her husband, “Our home has been nothing but a playroom. I have been your doll-wife, just as at home I was papa’s doll-child.”(Pg 118) In the end, Nora has a sort of spiritual awakening. She walks out into the night alone but, for perhaps the first time in her life, she’s on the path to becoming a fully realized, fully independent human being.

– She decides to desert her family to go on a quest of personal enlightenment.

-The act of concealing the ill-gotten loan signifies Nora’s independent streak. She is proud of the sacrifice she has made. Although she says nothing to Torvald, she brags about her actions with her old friend, Mrs. Linde.

– Nora is independent enough to negotiate the loan to make her family’s holiday possible, and over the course of the play, Nora emerges as a fully independent woman who rejects both the false union of her marriage and the burden of motherhood.

Manipulative

-At the end of the play, Nora seats Torvald at the table and explains her situation to him. She does not let him speak until she has finished what she wants to say.

– Other examples of manipulation are having a nanny take care of her children, having Mrs. Linde repair her dress, behaving seductively around Dr. Rank, influencing Torvald to give her money, and most importantly convincing Krogstad to overlook the similarity between her penmanship and her “father’s.”

Selfish

– She does not want to forgive Torvald. She would rather start another life than try to fix her existing one.

A dreamer

– Until she comes to the realization that her life is a sham, she spends her whole life in a dream world in which she does not take anything seriously.

– In her dream world, Nora takes a back seat approach to life and becomes like an object, reacting to other’s expectations rather than advancing herself.

Trusting and naïve

-She trusted that Krogstad would not blackmail her and it therefore comes as a rude shock when he does so.

-Until she comes to the realization that her life is a waste, she spends her whole life in a dream world of naivety. In this dream world, Nora does not take life seriously, an attitude that led to many of the plot’s complications.

-She believes that Torvald loves her enough to take all blame upon himself, but she is mistaken. When she realizes that he is more concerned with appearances and respectability than with her happiness, she decides to leave him and find her own way in life.

-She naively thought that Torvald would selflessly give up everything for her. When he fails to do this, she accepts the fact that their marriage has been an illusion. Their false devotion has been merely play acting. She has been his “child-wife” and his “doll.”

Determined

-Whenever Nora would get money from Torvald, she would reserve half of it to repay the debt, determined to clear it all one fine day.

-She was so determined to save her husband that she committed fraud to do so.

Hardworking

-She has been secretly working odd jobs to pay back the debt.

Courageous, bold, daring and tenacious

-To save her husband from poor health, she committed fraud. She valued love over the law.

-She courageously confronts Torvald about the demeaning way he treats her at the end of the play.

-She slammed the door on her husband as she left.

-Although she has been forbidden from eating macaroons by her husband, she still does it anyway despite her promises of total obedience to him.

Calculating

-She is blackmailed by Krogstad, so she begs Torvald to let Krogstad keep his job.

-She flirts with Dr. Rank in the hope of borrowing money from him.

– She gets Christine to go and talk to Krogstad on her behalf regarding the withdrawal of the letter.

-She dances the tarantella to distract Torvald from the mail.

Principled and firm

-She decides against borrowing from Rank when he reveals that he is dying and is in love with her.

-She rejects Torvald’s drunken advances after the party.

Secretive

-She has never told Torvald where the money for their trip to Italy came from, as his pride would suffer.

-She also hides her thoughts and actions from her husband even when there is no real benefit in doing so.

Suicidal

-She contemplates committing suicide in order to eliminate the dishonour she has brought upon her family.

Rebellious

-She eats macaroons even though she has been forbidden by her husband.

ROLE

Advancement of the plot – Nora is the central character in the play around whom the play circulates. As a result of her borrowing money from Krogstad, the conflict in the play is born as Krogstad tries to blackmail her with that secret, in order to make Nora’s husband, Torvald, to allow him keep his former job at the bank. When Torvald refuses, Krogstad sends him a letter to inform him about Nora’s indebtedness. Torvald’s reaction to this letter not only betrays his true nature, but also leads to Nora leaving him and her children to seek independence and freedom.
Development of themes – Nora helps in developing themes such as love and marriage, women and feminity, lies and deceit, money and materialism, the sacrificial role of women, parental and filial obligations, the unreliability of appearances, gender roles, individual vs. society, growth and development and betrayal
Revealing character traits of other characters – Nora helps in revealing the character traits of other characters. For example, she helps in revealing Torvald as selfish and egoistic, naïve, strict, loving, hypocritical and hardworking. Through her, we learn that Krogstad was on one hand, vicious and ruthless, but on the other hand, merciful and forgiving. Dr Rank’s friendly but immoral nature is revealed through Nora. Mrs Linde traitorous nature is also revealed through her.
Enhancing style – Nora helps in enhancing the style of symbolism (through the symbols like the Christmas tree, tarantella, the dolls, macaroons), dramatic irony, foreshadow, etc.

TORVALD HELMER

Torvald Helmer is a lawyer who at the start of the play has recently been promoted to Bank Manager. He is married to Nora Helmer, with whom he has three children. He does not seem particularly fond of his children; even once saying that their presence makes the house “will only be bearable for a mother now!”(Pg 30) His best friend is Dr. Rank, who visits him every day.

Loving and affectionate

-He loves and is very affectionate towards Nora. That is why he showers her with endearments like “My little skylark”, “My little squirrel”, “My little singing bird,” “My pretty little pet,” “My little sweet-tooth,” and “My poor little Nora.” and “Little Miss Extravagant.” With every term of endearment, the word “little” is always included to show affection.

– His despair as Nora exits at the very end of the play suggests that, despite his patronizing and unjust treatment of her, Torvald really does love Nora.

Generous

-He treats Nora generously, giving her extra money when she asks for it.

Proud

-Typical of many contemporary heads-of-the-family, he is a proud specimen of a middle-class husband.

Morally upright

– He sees Krogstad as irredeemably morally tainted, and hence decides to give his job to Mrs Linde.

– He is keenly concerned with his place and status in society and wouldn’t allow anybody to threaten his reputation, including his own wife.

Selfish

-He considers Nora merely as an ornamental sex object instead of an equal partner in their marriage and the mother of his children.

-He maintains amorous fantasies toward his wife: he dresses her as a Capri fisher girl and encourages her to dance in order to arouse his desires.

-At the end of the play, Nora imagines that Torvald will defend her honour and not allow Krogstad to blackmail the Helmers. Nora imagines that Torvald would sacrifice his own reputation and future to save her, but Torvald tells her that he would not make the sacrifice, shattering Nora’s dream world. At this point it becomes clear to her that she had been living all these years with a strange man, and she had born him three children.

-He planned to cope with the scandal resulting from blackmail by stripping Nora of her spousal and motherly duties, but would keep her in the house for appearance sake.

-He is overly concerned with his place and status in society, and he allows his emotions to be swayed heavily by the prospect of society’s respect and the fear of society’s scorn.

Hardworking and diligent

-He spends a great deal of his time at home in his study working, avoiding general visitors and interacting very little with his children. In fact, he sees himself primarily as responsible for the financial welfare of his family and as a guardian for his wife.

Dictatorial, authoritarian and patronizing

-He restrains Nora with rules, much as a father would have to inhibit a child, forbidding her from eating macaroons and other temporal pleasures.

Manipulative

-He insists on Nora wearing the fish girl costume for the tarantella. The costume and dance are part of Torvald’s fantasy of gazing upon Nora from across the room at a party and pretending that she is something exotic. Torvald made Nora take on a foreign identity; he used her as a doll.

Unforgiving

-When he finds out about the debt, he fails to forgive her until he is sure that his reputation is safe.

Heartless and unfeeling

-At the end of the play, Torvald seems untroubled and even a little relieved at the thought of Dr. Rank’s death.

-When he finds out about Nora’s secret debt, he instantly turns on her until he confirms that his reputation is safe.

-His heartless and unfeeling nature makes Nora not to tell him the truth about her loan, and Dr Rank not to tell him about his imminent death.

Conservative and traditional

-Torvald’s focus on status and being treated as superior by people like Nils Krogstad points at his obsession with reputation and appearances.

-He has straightforward and traditional beliefs about marriage and society.

-When Nora tells him she is leaving him, Torvald at first reacts by calling her mad and saying she is acting like a stupid child.

-He is unable to cope with the disagreeable truths of life.

–However, he can be said to be flexible because when he realizes how resolute Nora is in her decision, he offers to change and desperately searches for a way to make her change her mind.

Shallow and vain

-He is incapable of understanding his wife or of properly returning her love.

Hypocritical and self-righteous

-Though he regards her as his wife, he never considers her an equal partner in the relationship.

-Many times throughout the play, Torvald criticizes the morality of other characters. He trashes the reputation of Krogstad, one of his lesser employees. He speculates that Krogstad’s corruption probably started in the home. Torvald believes that if the mother of a household is dishonest, then surely the children will become morally infected. He also complains about Nora’s late father. When Torvald learns that Nora has committed forgery, he blames her crime on her father’s weak morals.

-In the beginning of Act Three, after dancing and having a merry time at a holiday party, Torvald tells Nora how much he cares for her. He claims to be absolutely devoted to her. He even wishes that some calamity would befall them so that he could demonstrate his steadfast, heroic nature.

Of course, a moment later, such an opportunity arises. Torvald finds the letter revealing how Nora has brought scandal and blackmail into his household. Nora is in trouble, but Torvald fails to come to her rescue as he had promised.

Naïve

-Throughout the play, Torvald is oblivious to his wife’s craftiness. When he discovers the truth at the end, he is outraged.

ROLE

Advancement of the plot – Torvald is also a major character in the play who plays an instrumental role in the development of the plot. It is as a result of his illness and the subsequent one-year stay in Italy that caused Nora to get into a debt trap with Krogstad. When he refuses to reinstate Krogstad to his former job at the bank, he intensifies the conflict because this makes Krogstad to send him a letter exposing his wife’s secret. The fact that he at first refuses to forgive her leads to Nora’s sudden discovery that he was a selfish, egoistic man. She has no alternative but leave him and her children to seek independence and freedom.
Development of themes – Torvald also helps in developing themes such as love and marriage, pride, honour, respect and reputation, money and materialism, parental and filial obligations, the unreliability of appearances, gender roles, individual vs. society, and betrayal.
Revealing character traits of other characters – Torvald helps in revealing the character traits of other characters. For example, he helps in revealing Nora as impulsive and a spendthrift, childlike and immature, irresponsible and reckless, dishonest and deceitful, manipulative, calculating and traitorous. Through him also, we learn that Krogstad was unscrupulous, vicious and ruthless but merciful and forgiving.
Enhancing style – Torvald helps in enhancing the style of imagery through his pet names for Nora such as “My little skylark”, “My little squirrel”, “My little singing bird,” “My pretty little pet,” “My little sweet-tooth,” and “My poor little Nora.” He also enhances the style of symbolism like his insistence that Nora should wear the dancing costume, similes like when he says he will protect her “like a hunted dove,” dramatic irony, verbal irony, hyperbole, etc.

MRS CHRISTINE LINDE

Mrs. Linde is an old schoolfriend of Nora’s. She is a woman whose marriage was loveless, and based on a need for financial security, and who doesn’t have any children. She and Krogstad had been in love before, but he was too poor to support her family. She arrives in town in search of a job in order to earn money and survive independently.

CHARACTER TRAITS

Honest and Truthful

-She tells Krogstad that Nora must eventually conclude, through her own sufferings, that the only way of life which can survive crises is one based on truthful relationships.

-She believes very deeply in honesty and stops Krogstad from taking the letter he wrote to Torvald back, thereby ensuring that Torvald finds out about Nora’s secret.

-She insists that, “Helmer must know everything. This unhappy secret must come out!”(Pg 90) Even though she has the power to change Krogstad’s mind, she uses her influence to make certain that Nora’s secret is discovered.

Hardworking

-Towards the end of the play, she explains to Krogstad that she finds joy and meaning in work.

-She worked hard to support her helpless mother and two younger brothers since the death of her husband.

Independent-minded

-She arrives in town in search of a job in order to earn money and survive independently. In this way, she is a fairly modern woman.

Traitorous

-She stops Krogstad from taking the letter he wrote to Torvald back, thereby ensuring that Torvald finds out about Nora’s secret, which seems like betrayal to her friend Nora.

Conservative and traditional

-She tells both Krogstad and Nora that she is miserable without other people to take care of, thereby fitting into the traditional role of women as caretakers and nurturers. It is this conviction that causes her to marry Krogstad towards the end of the play.

Selfish and materialistic

-She ended up marrying another man in order to have enough money to support her dying mother and young brothers. Apparently, Krogstad was too poor at this time to marry her. This left Krogstad lost and embittered, unhappy in his own marriage, and is presented as the reason behind his moral corruption.

Resilient

-She has lived an independent life as a single working woman. She has struggled financially and now that she has no one to look after, she feels empty.

ROLE

Plot development– she lets the audience know the inner thoughts of the protagonist. She has a major effect on events that happen in the play.
Reveals character traits of Nora and Krogstad – Mrs. Linde functions as a convenient device for exposition. She enters Act One as an almost forgotten friend, a lonely widow seeking a job from Nora’s husband. However, Nora does not spend much time listening to Mrs. Linde’s troubles. Rather selfishly, Nora discusses how excited she is about Torvald Helmer’s recent success. Through Mrs Linde, Nora launches into a dramatic explanation of all her secret activities (obtaining a loan, saving Torvald’s life, paying off her debt). Mrs Linde therefore functions as the primary means by which the audience learns of Nora’s secret and her character traits. She is the first character to see that Nora is not a child.
Develops themes– she introduces the theme of deception. Through Mrs. Linde, Nora reveals that she has lied to save Helmer’s life and therefore deceived him with her cleverness.
Enhances style– she foreshadows how Nora will confront a bitter future after learning that her marriage is based on deception by recounting how she herself sacrificed her rights to love and self-determination by marrying for financial security.

DR RANK

Dr. Rank is a medical doctor who is best friends with Torvald and Nora, who he visits every day. He suffers from spinal tuberculosis; a condition he believes was caused by his father’s vices, which included having extramarital affairs and consuming too much luxurious food and drink. He is unmarried and lonely, and over the course of the play it is revealed that he is in love with Nora.

CHARACTER TRAITS

Friendly and loyal

-Nora explains how she always feels at ease around Dr. Rank because he does not have any expectations or demands of her.

-He visits the Helmers every day.

Immoral

– It is revealed that he is in love with Nora, his best friend’s wife.

Courageous

-He is unconcerned with what others think of him.

-He has accepted his fate and his impending death.

Cynical

-He rejoices when he finds out that his illness is terminal, and insists that neither Torvald nor Nora visits him in his dying days. As he predicted, he is not particularly missed by the other characters.

Trusting

– He trusts Nora completely. He refrains from telling Torvald of his imminent death because it is too “ugly” an idea for him to tolerate, but he does tell Nora, an indication of the bond between them. He talks with her about his coming death in a code that excludes Torvald and protects him from the harsh reality.

Hypocritical

-Dr Rank is not as straightforward as he appears. His real motive for visiting the Helmers is that he is in love with Nora.

Realistic

– On the subject of the costume party, Dr. Rank suggested that Nora should go as herself and that Torvald should be invisible. Under the surface, Rank is suggesting that Nora should not be a doll. With an invisible chaperon, Nora would not be dominated by a figure placing an identity over her.

ROLE

-To provide amusement for Nora as a change from the tiresome rules of Torvald, just as she used to seek the conversation of the maids as a refreshing change from the strictness of her father.

– Dr. Rank adds to the somber mood of the play; he is not essentially useful to the conflict, climax, or resolution.

NILS KROGSTAD

Nils Krogstad Krogstad is an employee at the bank at which Torvald is made manager. He leant Nora the money to take Torvald to Italy to recuperate. He is, at least at the beginning, the main antagonist: Everything is going well for the Helmers until Krogstad enters the story. Known to the other characters as unscrupulous and dishonest, he blackmails Nora, who borrowed money from him with a forged signature, after learning that he is being fired from his job at the bank. In the past, he too committed the crime of forgery, an act that he did not go to prison for but that nonetheless ruined his reputation and made it extremely difficult to find a respectable job.

CHARACTER TRAITS

Morally Corrupt

-In the past, he too committed the crime of forgery, an act that ruined his reputation, though he did not go to prison. But it made it extremely difficult for him to find a respectable job.

-He was once in love with Kristine Linde, who ended up marrying another man in order to have enough money to support her dying mother and young brothers. This left Krogstad lost and embittered, unhappy in his own marriage, and is presented as the reason behind his moral corruption.

-Torvald, who sees Krogstad as irredeemably morally tainted, decides to give his job to Mrs Linde.

Merciless, vicious and callous

-At first he treats Nora without mercy when demanding his money. He argues that no mercy has been shown to him in life.

Unscrupulous and dishonest

-He blackmails Nora with the threat of exposing her indebtedness unless she talked her husband Torvald into giving him back his job at the bank.

-Unless Nora persuades Torvald to keep Krogstad in his job (he later extends this to a promotion), he will tell Torvald about her loan and her forgery of her father’s signature.

Forgiving and remorseful

-At first he treats Nora without mercy on the basis that no mercy has been shown to him in life; however, after Mrs. Linde and he decide to marry, he becomes happier and rescinds his threats to Nora, saying he regrets his behaviour.

-He removes his threats to the Helmers and sends Nora’s bond back to her, relinquishing his power over her.

-After engaging in a conversation with his lost love, the widow Mrs. Linde, they reconcile, and once again their romance is reignited, Krogstad no longer wants to deal with blackmail and extortion. He is a changed man!

-Although Mrs. Linde suggests that he should leave the first letter in the mailbox so that Nora and Torvald can finally have an honest discussion about things, he later drops off a second letter explaining that their secret is safe and that the IOU is theirs to dispose of.

-He has been trying to remake his life after having made earlier mistakes.

– Although he has been labeled as corrupt and “morally sick,” Krogstad has been trying to lead a legitimate life. He complains, “For the last year and a half I have not had a hand in anything dishonourable, amid all that time I have been struggling in most restricted circumstances. I was content to work my way up, step by step.”(Pg 75) Then he angrily explains to Nora, “And be sure you remember that it is your husband himself who has forced me into such ways as this again. I will never forgive him for that.”(Pg 76)

– He is one of several examples in the play of a person being forced into morally questionable action as a result of the rigid and unmerciful forces of society.

Loving and responsible

-As soon as Mrs Linde tells him that she has always loved him and asks him to resume their relationship, he reveals himself as a more loving, joyful and merciful character.

-In matters concerning his children, we find the true measure of the man. Nils Krogstad is a good father and is not afraid of doing whatever it takes to make sure his family is secure.

– Despite the financial strain that he found himself under as a single parent Nils Krogstad still brought up his two boys as best he could.

– Although at times Krogstad is vicious, his motivation is for his motherless children, thus casting a slightly sympathetic light on his otherwise cruel character.

Hardworking and resilient

– Here is a man who is a single parent of two boys who is totally unsupported by the society he exists in. He has been dealt many blows by life, first by Mrs. Linde who rejected him on financial grounds many years ago, and then by the death of his wife. He has to work multiple jobs to support himself and his family; by day he is a lowly bank clerk and by night he is a moneylender and he even finds time to write for a paper.

Frank and sincere

-When Mrs. Linde proposes they resume their old relationship, Krogstad remains truthful and makes sure she is aware of his past deeds as well as what people think of him. He even makes sure she knows about his current dealings with the Helmers.

ROLE

He advances the plot – Krogstad initiates the conflict by attempting to blackmail Nora Helmer. He serves as a catalyst. Basically, he initiates the action of the play. He sparks the flames of conflict, and with each unpleasant visit to the Helmer residence, Nora’s troubles increase. In fact, she even contemplates suicide as a means of escaping his torments.
Develops themes – Krogstad helps to develop the themes of love and marriage, pride, honour, respect and reputation, money and materialism, parental obligations and individual vs. society.
Reveals character traits of other characters– Through him we are able to know that Nora is secretive and deceitful, and Mrs Linde is traitorous and materialistic for leaving him when he was poor.

NURSE

She is a nurse to both Nora and Nora’s children. Her name is Anne Marie. The nursemaid is an example of a woman in bad circumstances forced to do anything in order to survive.

CHARACTER TRAITS

Kind

-She was forced to give up her own child, who it is suggested was born out of wedlock.

Reliable and responsible

-When Nora first thinks of leaving, she considers the fact that her children will be raised by the nursemaid and, remembering what a good mother the nursemaid had been to her, decides that she would also raise Nora’s children well.

Immoral

-She gave birth to a baby out of wedlock,

Irresponsible

-She had to give up her own child in order to take up her position as the nursemaid at the Helmers.

Self-sacrificial

-She had to give up her own child in order to take up her position as the nursemaid at the Helmers. Nora finally leaves her children in her care, believing that they will be better off than they would be with her.

ROLE

Developing themes – She helps to develop the theme of the sacrificial role of women by giving away her child to strangers so that she could concentrate on her job.

IVAR, BOB, AND EMMY

These are Torvald and Nora’s young children. Raised primarily by Anne, the Nurse (and Nora’s old nurse), the children spend little time with their mother or father. The time they do spend with Nora consists of Nora playing with them as if she were just another playmate. The children speak no individualized lines in the play; they are “Three Children.” Their dialogue is facilitated through Nora’s mouth, and they are often cut entirely in performance.

CHARACTER TRAITS

Playful

-They asked their mother to play child games with them and they played hide-and-seek.

Insistent

When their mother showed reluctance to participate in the children’s game, they insisted until she gave in.

Loving

-They loved their mother dearly and would have wanted to spend more time with her and to continue playing children’s games with them.

ROLE

1. To bring out the character of Nora as a loving mother. She showers them with Christmas gifts and even plays children games with them. When Nora later refuses to spend time with them because she fears she may morally corrupt them, she acts on her belief that the quality of parenting strongly influences a child’s development.

HELEN

She is a housemaid employed by the Helmers.

CHARACTER TRAITS

Humble

-She answers Nora with a lot of humility.

Hardworking

-She does her work diligently.

PORTER

This is the porter who brings the Christmas tree to the Helmers house at the very beginning of the play.

CHARACTER TRAITS

Obedient

-He obediently delivers the Christmas tree to the Helmers house.

Grateful

-He thanks Nora for the tip that she gives him. (Pg 1)

Honest

-He honestly states his charges without exaggerating and was already giving her back the change when she told him to keep it. (Pg 1)

NORA’S FATHER

Though Nora’s father is dead before the action of the play begins, the characters refer to him throughout the play. Though she clearly loves and admires her father, Nora also comes to blame him for contributing to her subservient position in life.

CHARACTER TRAITS

Manipulative

-He manipulated Nora to do according to his wishes and whims. She complains that her father and her husband both treated her like a doll.

Immoral

-Torvald criticizes him as having been a morally crooked man who engaged in corrupt deals.

Insensitive

-The way he treated Nora was too insensitive for a father to treat his daughter.

THEMES

MARRIAGE AND LOVE

At the beginning of the play, Nora and Torvald appear to be very happily married, even to themselves. Nora talks joyfully about her love for Torvald, and Torvald refers to Nora using affectionate pet names. At first the Helmers seem happy, but over the course of the play, the imbalance between them becomes more and more apparent.

Their loving marriage stands in contrast with the lives of the other characters. For example, the marriage of Krogstad and Mrs. Linde was based on necessity rather than love, and were unhappy. Dr. Rank, on the other hand, was never married, and it is later revealed that he has silently loved Nora for years.

Yet although Nora and Torvald’s marriage is based on love (as opposed to necessity, as was the case with Krogstad and Mrs. Linde), it is still governed by the strict rules of society that dictated the roles of husband and wife. It is clear that Nora is expected to obey Torvald and allow him to make decisions for her.

At first it seems that Nora and Torvald both enjoy playing the roles of husband and wife in a way that is considered respectable by society. However, Nora soon reveals to Mrs. Linde that she secretly borrowed the money from Krogstad behind Torvald’s back, and therefore has already broken both the law and the rules of marriage at the time. This creates a dilemma: Nora broke the rules of marriage, yet did so in order to save her husband’s life – a true act of love.

By the end, the marriage breaks apart due to a complete lack of understanding. Nora Helmer, the “doll” wife, realizes after eight years of marriage that she has never been a partner in her marriage. At the play’s conclusion, she leaves her husband in order to establish an identity for herself that is separate from her identity as a wife and mother.

The main message of A Doll’s House seems to be that a true marriage is a joining of equals. The play centres on the dissolution of a marriage that doesn’t meet these standards.

There is a lot of talk about love in A Doll’s House. Throughout the play we hear of and see many different forms of love: familial, maternal, paternal, and fraternal. Romantic love even blossoms for two of the secondary characters, namely Krostad and Mrs Linde. However, for the main characters, the Helmers, true romantic love is elusive. They finally discovered that true love never existed between them.

WOMEN AND FEMININITY

Nora has often been painted as one of modern feminist heroines. Over the course of the play, she breaks away from the domination of her dictatorial husband, Torvald. Also throughout this play, there is constant talk of women, their traditional roles, and the price they pay when they break with tradition.

When A Dollʼs House was written in 1879, a wife was not legally permitted to borrow money without her husbandʼs permission. On her wedding day, a woman transferred from living under the authority of her father to under that of her husband.

Poverty had already forced women into the workplace early in the nineteenth century, and the Norwegian government passed laws protecting and governing women’s employment. By the middle of the 19th century, Norwegian women were permitted inheritance rights and the right to an education. But many of the rights provided to women favoured the lower economic classes. Employment opportunities for women were limited to low paying domestic jobs, teaching, or clerical work. Middle class women, such as Nora, noticed few of these new advantages. It was the institution of marriage itself that restricted the freedom of middle class women. Universal women rights were eventually achieved in 1913, making Norway the first country in Europe to have equal voting rights for men and women.

PRIDE, HONOUR, RESPECT AND REPUTATION

The men characters in A Doll’s House are obsessed with their reputation. Some have good names in their communities and will do anything to protect it; others have lost their good names and will do anything to get them back.

Honour

Honour is extremely important to Torvald; it is what motivates his behaviour. Early in the play, his value for honour is the reason he gives for sacking Krogstad, claiming that because he once displayed a lack of honour, it means that Krogstad is forever dishonoured. When he learns of his wife’s mistake, Torvald’s first and foremost concern is for his honour. He cannot appreciate the sacrifice that Nora has made for him; he is only concerned with how society will react to his family’s shame. For Torvald, honour is more important than family and far more important than love; he simply cannot imagine anyone placing love before honour. This issue brings out the glaring difference between Nora and Torvald.

Pride

Like honour, pride is another quality that Torvald upholds. He is proud of Nora in the same way one is proud of an expensive or rare item or possession. When her scandal threatens to be exposed, Torvald is very fearful of losing his public pride. Instead of accepting Nora with her misperfections, Torvald instead rejects her when she is most in need of his support. His pride in himself and in his possessions blinds him to Nora’s worth and value. Nora is left with no choice but to leave him. Only when she has made the decision to leave Torvald does she begin to develop pride in herself.

LIES AND DECEIT

The tension that runs throughout A Doll’s House comes from Nora’s fear of her secret being discovered. Her great terror being exposed leads her to tell a lie after a lie. When her web of lies finally reaches a climax, her marriage proves too weak to bear the strain.

At the beginning of the play, Nora appears to be a dutifully obedient and honest wife, however it is quickly revealed that she is hiding a serious secret from her husband—the fact that she borrowed money from Krogstad to finance a trip to Italy that she claims saved Torvald’s life. This confirms that all her statements about never disobeying or hiding anything from him were nothing but deceitful. When she reveals her dishonesty to Mrs. Linde, Mrs. Linde insists that she ought to confess to Torvald immediately, insisting that a marriage cannot succeed when husband and wife are not completely honest with each other.

But Nora is not alone in telling lies and being deceitful. Krogstad is also revealed to have committed a forgery. The fruits of their acts of deception are devastating: Krogstad’s reputation is ruined, and Nora is forced to leave her husband and family at the end of the play.

It should however be noted that the motivation behind Nora’s dishonesty was love – she lied in order to save her husband’s life. Furthermore, she wouldn’t have been deceitful if it weren’t for societal law dictating that women were not allowed to handle financial matters independently. Therefore Nora’s deceit was not the result of a personal flaw, but rather an attempt to commit a noble act of saving her husband’s life that went awry.

Dr. Rank also comes out as deceitful and dishonest. He has been deceiving both Nora and Torvald for years about the depth of his feelings for Nora. Only when she attempts to seek his financial help does Nora finally see beneath the surface to the doctor’s real feelings. He has been lusting for his best friend’s wife all those years. Nora is so shocked to discover this that she automatically decides not to ask Dr. Rank for financial assistance.

Torvald, who has been deceived throughout most of the play, is finally revealed in the final act to be the one most guilty of deception. He has deceived Nora into believing that he loved and cherished her, while all the while he had regarded her as little more than his property.

MONEY AND MATERIALISM

Throughout the entire play everyone is talking about money, as if it was a god. As the entire issue starts over a debt, the play revolves around money and who has it as well as who does not have it. It is a prevailing theme due to that.

In the very first scene, Nora gives the porter one shilling, telling him to “keep the change”, thus indicating her relaxed attitude to money and spending. The next scene with Torvald almost entirely revolves around the subjects of money, spending and borrowing, with Nora portrayed as a spendthrift. Torvald has very strong views on borrowing and debt. He says to her, “That is like a woman! But seriously, Nora, you know what I think about that. No debt, no borrowing. There can be no freedom or beauty about a home that depends on borrowing and debt.”(Pg 3)

A need for money affects all the major characters in A Doll’s House. In the beginning of the play it is revealed that Torvald was recently promoted and will receive “a big salary.”(Pg2) However, he still criticizes Nora for overspending, arguing that they need to be cautious financially. Mrs. Linde is in desperate need of a job following the death of her husband. Krogstad’s replacement at the bank leaves him threatening to reveal Nora’s secret in order to get his job back because he fears he will lose his source of income. Indeed, the bank works as a symbol for the pervasive presence of money in the characters’ lives.

Throughout the play A Doll’s House, the characters spend a good deal of time talking about their finances. Some are said to be doing quite well financially, and some have the promise of their finances improving in the future. Others are struggling to make ends meet. Either way, each character’s financial status seems to be a defining feature.

In the play, money symbolizes the power that the characters have over one another. In the first Act, Torvald’s ability to dictate how much Nora spends on Christmas presents shows his power over her. On the other hand, the debt that Nora owes Krogstad allows him to have power over her and Torvald. Both Nora and Mrs. Linde cannot earn large incomes because they are women; their inability to access significant amounts of money shows the power that men have over the women in this society.

It is also clear that, while earning money leads to power, it can also be dangerous. For instance, even if money actualized Nora and her family’s trip to Italy, the debt she owed Krogstad soon became a source of terror, dread, and shame. The thrill of obtaining money soon became a nightmare for her.

Krogstad is a moneylender, and money (or lack of it) has had a major effect on his life. We learn that Mrs Linde ended her relationship with him many years ago because of his lack of financial security, choosing to marry a richer man instead. Throughout his life Krogstad has been poor, struggling to support his family, and it is this dependency on financial income that leads him to blackmail Nora in an attempt to keep his job at the bank. Mrs Lindeʼs life has also been directly affected by money, or lack of it. Her late husbandʼs business collapsed, leaving her with nothing to live on, and since then she has had to work hard to survive.

Dr Rank is the only main character who appears to be comfortable financially, having inherited money from his late father. However, although he is financially comfortable he is terminally ill, referring to his body as being “bankrupt.”

Torvald in particular focuses on money and material goods rather than people. His sense of manhood depends on his financial independence. He was an unsuccessful lawyer because he refused to take “unsavory cases.” As a result, he switched jobs to the bank, where he will primarily be dealing with money.

THE SACRIFICIAL ROLE OF WOMEN

In A Doll’s House, Ibsen paints a bleak picture of the sacrificial role held by women of all economic classes in his society.

In order to support her mother and two brothers, Mrs. Linde found it necessary to abandon Krogstad, her true but poor lover, and marry a richer man.

The nanny had to abandon her own child to support herself by working as Nora’s and later as Nora’s children’s caretaker. As she tells Nora, the nanny considers herself lucky to have found the job, since she was “a poor girl who has got into trouble…” (Pg 50)

Though Nora is economically advantaged in comparison to the play’s other female characters, she nevertheless leads a difficult life because society dictates that Torvald be the marriage’s dominant partner. Torvald issues rules and looks down on Nora, and Nora must hide her debt from him because she knows Torvald would never accept the idea that his wife had helped save his life. Furthermore, she must work in secret to pay off her loan because it is illegal for a woman to obtain a loan without her husband’s permission.

Nora’s abandonment of her children can also be interpreted as an act of self-sacrifice. Despite Nora’s great love for her children, as seen in her interaction with them and her great fear of corrupting them, she chooses to leave them. Nora truly believes that the nanny will be a better mother and that leaving her children is in their best interest.

All the three women in the play have made some kind of personal sacrifice in their lives in order to fulfill the roles which society expects of them. Nora, besides risking her dignity by borrowing money on behalf of her family, she also has sacrificed all her own opinions, thoughts and ideas and adopted Torvaldʼs views as her own. Besides that, she has been saving every bit of money she had and working odd hours of the night to repay Krogstad. And at the end of the play she sacrifices her home, family and children for the sake of her own self-discovery.

Mrs Linde, after her husbandʼs death, continued to make personal sacrifices for the sake of her family, taking on any work she could to support them financially.

Anne-Marie, on the other hand, sacrificed motherhood for a respectable job, which was all too common for young unmarried mothers in the 19th century.

PARENTAL AND FILIAL OBLIGATIONS

There is a strong emphasis throughout the play on the importance of parental and filial responsibility, and of the effect that the actions of parents have upon their children.

Parental obligations

Nora, Torvald, and Dr. Rank believe that a parent is obligated to be honest and morally-upright, because a parent’s immorality is passed on to his or her children like a disease.

For instance, Dr. Rank has a disease that is the result of his father’s wickedness. Dr. Rank implies that his father’s immorality, which included affairs with many women, led him to contract a venereal disease that he passed on to his son, causing Dr. Rank to suffer for his father’s misdeeds. He talks about the unfairness of this, of the sins of the father being passed on to the son.

Torvald, on the other hand, talks about a parentʼs immorality being passed on to the children like a disease. He voices the idea that one’s parents determine one’s moral character when he tells Nora, “Almost everyone who has gone to the bad early has had a deceitful mother” (Pg 30) He speaks about Krogstad poisoning his own children with lies and immorality. He also refuses to allow Nora to interact with their children after he learns of her deceit; for fear that she will corrupt them.

Nora is referred to as being like her father, having inherited a lot of his qualities. It is also important to note that she never had a mother, with Anne-Marie fulfilling the maternal role in her life.

Anne-Marie was forced to give away her own child to take on the role of Noraʼs maid; in contrast Nora chooses to leave her own children at the end of the play.

Filial obligations

Filial means the duties, feelings or relationships which exist between a son or daughter and his or her parents.

The play suggests that children too have an obligation to protect their parents. Nora recognized this obligation, but she ignored it, choosing to be with, and sacrifice herself for, her sick husband instead of her sick father.

Mrs. Linde, on the other hand, abandoned her hopes of being with Krogstad and undertook years of labour in order to tend to her sick mother. Mrs Linde has fulfilled her filial responsibility by dedicating her life to care for her mother, at the expense of her own personal happiness. Her motherʼs illness has directly affected the life she has led and the personal decisions she has made.

Ibsen does not however pass judgment on either woman’s decision, but uses the idea of a child’s debt to her parent to demonstrate that familial obligation is not one way – it is reciprocal.

THE UNRELIABILITY OF APPEARANCES

Over the course of A Doll’s House, appearances prove to be quite misleading and hide the true reality of the play’s characters and situations. Our first impressions of Nora, Torvald, and Krogstad are all later proved quite wrong.

Nora, at first, seems a silly, childish woman, but as the play progresses, we see that she is intelligent, motivated, and, by the end of the play, a strong-willed, independent thinker.

Torvald, on the other hand, though he appears as the strong, benevolent husband, reveals himself to be cowardly, petty, and selfish when he fears that Krogstad may expose him to scandal.

Krogstad, who initially appears to be a vicious, ruthless blackmailer, later reveals himself to be a much more sympathetic and merciful character. He also turns out as an earnest lover. Indeed, the play’s climax is largely a matter of resolving identity confusion – we see Krogstad as a loving merciful man, Nora as an intelligent, brave woman, and Torvald as a helpless, sad man.

Situations too are misinterpreted both by the audience and by the characters. The seeming hatred between Mrs. Linde and Krogstad turns out to be love. Nora’s creditor turns out to be Krogstad and not Dr. Rank, as the audience and Mrs. Linde had thought. Dr. Rank confesses that he is not just a friend to Nora but instead he is in love with her, to Nora’s and the audience’s surprise. The seemingly ruthless Krogstad repents and returns Nora’s contract to her, while the seemingly kindhearted Mrs. Linde fails to help Nora, leading to Torvald’s discovery of Nora’s secret.

GENDER ROLES

A Doll’s House exposes the restricted roles of women during the time of its writing and the problems that arise from a drastic imbalance of power between men and women.

Throughout the play, Nora is treated like a child by the other characters. Torvald calls her his “pet” and his “property,” and suggests that she is not smart or responsible enough to be trusted with money. Neither Krogstad nor Dr. Rank take her seriously, and even Mrs. Linde calls her a “child.” Nora seems unperturbed by the views of others about her; even calling herself “little Nora” and promising that she would never dream of disobeying her husband.

However, there are clues that she is not entirely happy with the limited position she has as a woman. For example, when revealing the secret of how she borrowed money to finance the trip to Italy, she refers to it as her “pride” and says it was fun to be in control of money, explaining that it was “like being a man.” (Pg 21) Nora seems to wish to enjoy the privileges and power enjoyed by males in her society. She seems to understand the confinement she faces simply by virtue of her sex.

Nora’s dissatisfaction with her status as a woman intensifies over the course of the play. In the final scene she tells Torvald that she is not being treated as an independent person with a mind of her own. According to her, the bitter solution to this issue is to leave married life behind, despite Torvald’s begging that he will change. Nora’s problems arise because as a woman she cannot conduct business without the authority of either her father or her husband. When her father is dying, she must forge his signature to secure a loan to save her husband’s life. That she is a responsible person is demonstrated when she repays the loan at great personal sacrifice.

The men in this play have a very conservative view of the roles of women, especially in marriage and motherhood. Torvald, in particular, believes that it is the sacred duty of a woman to be a good wife and mother. Moreover, he tells Nora that women are responsible for the morality of their children. In essence, he sees women as childlike, helpless creatures detached from reality on the one hand, but on the other hand as influential moral forces responsible for the purity of the world through their influence in the home.

The men of A Doll’s House are in many ways just as trapped by traditional gender roles as the women. The men must be providers. They must bear the burden of supporting the entire household. They must be the undoubted kings of their respective castles. Besides providing for their families, the men are obsessed by a desire to achieve higher status. Respectability is of great concern to both Torvald and Krogstad. When Nora’s borrowing is revealed, Torvald’s first thoughts are for his reputation. On the other hand, Krogstad is obsessed with achieving success now that he has changed his character. He intends to one day take over Torvald’s job and run the bank.

By the end of the play, these traditional ideas are truly put to the test.

INDIVIDUAL VS. SOCIETY

Nora, a dutiful mother and wife, spends most of the play putting others before herself. She thinks little about herself to the extent of engaging in an act of forgery and taking a debt for the sake of her husband’s health. She doesn’t stop to worry about how these actions might impact the lives of her husband and children. Even when she plans to kill herself near the end of the play, it is not to hide her shame but rather because she thinks that if she is alive then Torvald will ruin himself in trying to protect her.

Similarly, Mrs. Linde admits that, without a husband or any family members to care for, she feels that her life is pointless. Therefore both women find a sense of meaning in their lives through serving others and performing the caring, obedient role that society requires of them.

However, Nora later learns that prioritizing her duty as a wife and mother cannot lead to real happiness. She realizes that while she thought she was sacrificing herself to protect her love, in fact no such love existed. It becomes clear that Torvald would never have sacrificed his reputation to protect her. She therefore decides to leave him in order to develop a sense of her own identity. The play ends with Nora choosing to put herself as an individual before society’s expectations of her.

Some characters, however, are more concerned about themselves as individuals rather than the society. A good example is Krogstad. Throughout most of the play, it seems that he cares more about his reputation than anything else. Punished by society for his act of forgery, he is desperate to reclaim respectability in the eyes of others. However, he realizes that he will only achieve happiness through truly reforming himself and regaining the personal integrity that he lost, rather than the outward respectability.

In a similar way to Nora, Krogstad learns that society’s view of him is meaningless if he doesn’t respect himself as an individual.

BETRAYAL

Betrayal is a theme of this play in several ways. Nora has betrayed her husband’s trust in several instances. She has lied about borrowing money, and to repay the money she must lie about how she spends her household accounts and she must lie about taking odd jobs to earn extra money. She also chooses to lie about eating macaroons which her husband has forbidden her.

Torvald betrays Nora when he rejects her pleas for understanding. Torvald’s betrayal of her love is clearly shown when he doesn’t want to understand that Nora took the loan because of his own welfare. To him, she threatened his otherwise good reputation in the eyes of the society, which was an unforgivable sin to him. This was the reality that Nora requires to finally awaken from; her previous view about her husband and their marriage was just but a sham.

Mrs Linde also betrays Krogstad when she opts to marry a richer man because Krogstad was too poor to help her sustain her sick mother and needy siblings.

GROWTH AND DEVELOPMENT

In Act I, Nora is portrayed as nothing more than a “doll,” a child who has exchanged a father for a husband without changing or maturing in any way. But as the play progressed, she realized that she had no identity separate from that of her husband. Torvald owned her just as he owned their home or any other possession. She was finally forced to face the reality of the life she was living. She realized in the final act that if she had to develop an identity as an adult, she must leave her husband’s home. When Nora finally gave up her dream for a miracle and, instead, accepted the reality of her husband’s self-centredness, she finally took her first steps toward maturity. She realized the inequity of her situation; she also recognized her own self worth. Her decision to leave is a daring one that indicates the seriousness of Nora’s desire to find and create her own identity.

THE HOME

The fact that the play is called A Doll’s House means that home might be a prevalent theme. Early on in the text, the home is seen as a thing of joy, a place of comfort and shelter. The idea of home is enmeshed with the idea of the happy family, which the Helmers seem to be.

Toward the play’s conclusion, however, the imbalance of power in the family becomes an issue. Now the seemingly happy home is revealed as having been an illusion – a doll’s house – that hid the gulf between the Helmers. The Helmers’ home is really more of a prison than a shelter.

The title, A Doll’s House, implies that everything is a façade, an illusion. Just like a doll that has a plastered smile on its face, the doll’s house hides the problems in the marriage.

STYLISTIC/LITERARY DEVICES

Stylistic or literary devices are techniques (ways to do things, styles, or forms) that authors use to get the attention of the reader which include playing with words, creating imagery, comparing and contrasting, or using metaphors, just to name a few. In A Doll’s House, the author has used a variety of stylistic devices, as discussed below.

SYMBOLISM

Symbols are objects, characters, figures, or colours used to represent abstract ideas or concepts. The following are the symbols used in the play:

Christmas and New Year Days

The action of the play is set at Christmas and New Year season. Christmas and New Year holidays are both associated with rebirth and renewal and several of the characters go through a kind of rebirth over the course of the play.

Both Nora and Torvald have a spiritual awakening, which could be seen as a “rebirth.” Nora’s trials and tribulations wake her up to the sorry state of her marriage. When the “wonderful thing” fails to happen, she realizes she will never be a fully realized person until she breaks away from her husband. And when she slams the door behind her, she is in a way reborn.

Nora is not alone in her spiritual awakening, however. Torvald’s last line, “The most wonderful thing of all?”(Pg 120) seems to indicate that he has also realized the complete inadequacy of his existence. By the end of the play, both Helmers have been reborn.

Krogstad and Christine are reborn as well. When these “two shipwrecked people…join forces,” (Pg 88) they each get a fresh start in life. Both of them view their renewed love affair as a chance for salvation. Krogstad hopes that it will help increase his standing with the community, and that Christine’s influence will make him a better person. Christine, on the other hand, is overjoyed that she will have someone to care for. She once again has purpose in her life.

Nora and Torvald both look forward to New Year’s as the start of a new, happier phase in their lives, a new beginning with no debts. In the New Year, Torvald will start his new job, and he anticipates with excitement the extra money and admiration the job will bring him. Nora also looks forward to Torvald’s new job, because she will finally be able to repay her secret debt to Krogstad. By the end of the play, however, the nature of the new start that New Year’s represents for Torvald and Nora has changed dramatically. They both must become new people and face radically changed ways of living. Hence, the New Year comes to mark the beginning of a truly new and different period in both their lives and their personalities.

In the end of the play, it resembles new beginnings as almost all the characters are starting new lives, Nora and Torvald separately, while Christine and Krogstad together.

Christmas Tree

The Christmas tree symbolizes Nora’s role in her household. She is only a decoration to be looked at. Her function in the household is pretty much the same as the tree. She is merely decorative and ornamental. She dresses up the tree just as Torvald dresses up her for the party. It’s interesting that she tells the maid not to let the children see the tree until it’s decorated.

The Christmas tree, therefore, a festive object meant to serve a decorative purpose, symbolizes Nora’s position in her household as a plaything who is pleasing to look at and adds charm to the home.

It also symbolizes family happiness and unity, as well as the joy Nora takes in making her home pleasant and attractive.

At the beginning of Act Two, the Christmas tree has been stripped of its ornaments and is only left with burnt-down candle-ends on its disheveled branches. Nora is alone in the room, walking about uneasily. Basically, Nora is a mess and so is the tree. She’s gotten the bad news from Krogstad, and as a result her mind is just as disheveled as the poor tree.

The tree seems to mimic Nora’s psychological state. It can be interpreted as symbolic of Nora’s disintegrating web of lies. The pretty decorations that Nora used to cover up her deceit are falling away. Soon the bare, ugly truth will emerge. This represents the end of Nora’s innocence and foreshadows the Helmer family’s eventual disintegration.

Macaroons

Torvald has banned Nora from eating macaroons. Although Nora claims that she never disobeys Torvald, this is proved false in the very opening of the play when Nora eats macaroons while she was alone in the living room. The macaroons are symbolic of Nora’s disobedience and deceit. She lies to Dr. Rank about having been given some by Mrs. Linde, and after giving her performance of the tarantella asks that macaroons be served at dinner, which indicates a close relationship between the macaroons and her inner passions, both of which she must hide within her marriage.

The tarantella

Tarantella takes its name from a spider, a Tarantula, which, according to the Italian legends, bites its victim to quick death. The only way to get rid of its poison is to dance so as to let the poison come out of the body with the sweat. Similarly, the wild dance of Nora is a symbolic expression of her tragic inner condition and, at the same time, a therapeutic instrument that gives her courage to face up the suicide that she plans to carry out. Nora dances the Tarantella at a time when she had accelerated anxiety, on the border of madness. So through the dance, her body was trying to express what couldn’t be said in words.

Like the macaroons, the tarantella symbolizes a side of Nora that she cannot normally show. It is a fiery, passionate dance that allows her to drop the mask of the perfect Victorian wife and express her desperate and tragic interior condition and her inner feelings. It is a dance of recovering from the madness of her fate; Tarantella has the power to heal Nora.

After the dance, in fact, she reemerges matured and able to look death in the eyes.

It is important to note that the rehearsal of Tarantella is the first moment in which Nora doesn’t obey what Torvald commands. Her repressed feelings are not allowed to come out in her marriage, the only way she can express them is through a performance. And her performance is wild and hysteric. Through the dance Nora liberates herself from her sexual doll’s role, which is a transformation from an old existence to a new one.

The Doll’s House

The title of the play A Doll’s House is also symbolic. It represents something impermanent or short-lived.

There are a few mentions of dolls early on in the play; for example, when Nora shows Torvald the dolls she bought for her daughter, and says that the fact that they are cheap doesn’t matter because she will probably break them soon anyway. This probably suggests that Nora is raising her daughter for a life similar to her own. It also foreshadows Nora breaking up her family life by leaving Torvald.

When Nora plays with her children she also refers to them as her “little darlings.” (Pg 42) However, it is not until the end of the play that the metaphor becomes explicitly clear. Nora tells Torvald that both he and her father treated her like a doll, and cites this as one of the reasons why she has become dissatisfied and disillusioned with her life with him.

The dance costume

At the end of the play, Nora decides to leave Torvald. The next thing Nora does is to change out of her fancy dance dress. Torvald bought this dress for Nora to wear at a costume party because he wanted her to appear as a “Neapolitan fish girl”. As one would put clothes on a doll, Torvald dresses Nora. When she sheds this dress, she is symbolically shedding her past life with Torvald and her doll-like existence.

Dr. Rank

Dr. Rank is a symbol of moral corruption within society. He has been lusting for Nora secretly. His illness is symbolic of the moral illness of the society as represented by himself, Krogstad and, by extension, Torvald.

Mrs Linde

Mrs Linde is a symbol of a modern, independent woman. She arrives in town in search of a job in order to earn money and survive independently. She perhaps also symbolizes hollowness in the matriarchal role.

Torvald Helmer

Torvald Helmer is a symbol of a male dominated, authoritative, and autocratic society.

The slamming of the door

The slamming of the door symbolizes the finality of the relationship between Torvald and Nora Helmer.

USE OF FIGURATIVE LANGUAGE

IMAGERY

Metaphors

A metaphor is a comparison without using the terms ‘like’ or ‘as.’ Henrik Ibsen uses quite a number of metaphors in A Doll’s House. These include the following:

Torvald’s pet names for Nora

-He calls her “featherhead,” “songbird,” “squirrel,” “hunted

dove…saved from hawk’s claws,” and “skylark.” When she leaves him, he calls her a “heedless child.” All these metaphors are, on one hand, aimed at reflecting Nora’s apparently innocent, carefree nature, and on the other hand, they suggest that her husband does not think of her as a proper adult because she is a woman.

-Another metaphor is where Torvald says, “…how much it costs a man to keep such a little bird as you.” Here, Torvald is comparing Nora to a bird by saying that people would not expect her to spend as much money as she does. The “bird” reference means that birds are typically low maintenance, but Nora is not.

The doll

-In Act 3, Nora tells Torvald that both her father and Torvald have treated her like a doll-child, with no opinions of her own, and have only played with her. Both men, she says, have committed “a great sin” against her in discouraging her from growing up. Torvald’s pet names for her are prefaced by “little,” showing that he sees her as a child.

Big black hat

-In Act 3. Dr. Rank has a coded conversation with Nora (designed to protect Torvald from unpleasant truths) in which he says he will attend the next fancy dress ball wearing “a big black hat” that will make him invisible. This is a way of saying that he will be dead.

Other metaphors

-Krogstad is labeled “morally diseased” because of the incriminating forged bond and the forged documents that tarnished his reputation.

-Nora and Torvald crumbling marriage and home are referred to as a “doll’s house” to mean their impermanency.

– Krogstad uses this metaphor, “I am a shipwrecked man clinging to a bit of wreckage” (Pg 87) to describe how he felt when Mrs. Linde chose to marry her late husband instead of him. Mrs. Linde replies that she had her mother and younger brothers to take care of and she needed financial stability, which Krogstad could not offer her. In this metaphor, Krogstad might be suggesting that he is still in love with Mrs. Linde.

-Torvald refers to his wife as his “frightened little songbird” and promises her that his “big broad wings” would protect her.

-The title of the play A Doll’s House is an extended metaphor. It is comparison of a small toy with that of a perfect house. It compares Nora’s relationship with every man in her life to that of a young child playing with her, merely a pretty plaything.

Similes

A simile is a comparison by use of the terms ‘like’ or ‘as.’ Similes are used in different places in the story to compare certain necessary ideas.

Examples

-“It was like being a man.”(Pg 21) This simile was used by Nora to compare the role she played in sustaining the family during their one-year stay in Italy. It made her feel like a man supporting them for all that time.

-Torvald brags that he will protect Nora “like a hunted dove that [he has] saved from the talons of a hawk.” Here, he wants to emphasize his commitment in ensuring Nora of her safety.

IRONY

Irony is a figure of speech in which words are used in such a way that their intended meaning is different from the actual meaning of the words. It may also be a situation that ends up in quite a different way than what is generally anticipated. There are three types of irony evident in A Doll’s House, namely: verbal, situational and dramatic irony.

Dramatic irony

Dramatic irony occurs when the audience is more aware of what is happening than one, some or all the characters on stage.

The full significance of a character’s words or actions is clear to the audience or reader although unknown to the character. In other words, the audience’s or reader’s knowledge of events or individuals surpasses that of the characters.

Examples

-This happens in A Doll’s House near the opening of the play when Nora eats macaroons. When Torvald then asks Nora if she has been eating sweets, she lies and says she has not. Nora and the audience know this is a lie and so know more than Torvald, making this a situation of dramatic irony.

-Torvald tells Nora, “That is like a woman! But seriously, Nora, you know what I think about that. No debt, no borrowing. There can be no freedom or beauty about a home that depends on borrowing and debt.”(Pg 3) But nevertheless, she has borrowed money from Krogstad which she has been paying for a long time without his knowledge.

-The reader is aware that Nora borrowed money from Krogstad without her husband’s permission. Nora also forged her father’s name to gain the money. She says, “You don’t know all. I forged a name.” In the following conversation between Nora and Christine it is clearly stated that Torvald does not know of Nora’s actions

Mrs. Linde: And since then have you never told your secret to your husband?

Nora. Good heavens, no! (Pg 20)

-Another example of dramatic irony in A Doll’s House is when Nora wants to practise a dance called the Tarantella. When Torvald goes to look in the letter box Nora says, “Torvald please don’t. There is nothing in there.” (Pg 80) The reader knows there is a letter in the mailbox that has been dropped by Krogstad. The reader also knows that Nora has not forgotten the dance as she claimed, she was just pretending. The reader knows this when Torvald goes to check the mail and Nora begins to play the Tarantella. Nora then says, “I can’t dance tomorrow if I don’t practise with you.” (Pg 81) The reader knows that all Nora is trying to do is keep Torvald from reading the mail which contains a letter from Krogstad.

-Dramatic irony is evident throughout the text to indicate Nora’s exit from her marriage with Torvald. Some escalating events have happened in the three acts to give clues to the audience that she has already decided to leave.

Examples

i) Nora to Nurse regarding the children:

Nora: Yes, but, nurse, I shall not be able to be so much with them now as I was before.

Nurse: Oh well, young children easily get accustomed to anything.

Nora: Do you think so? Do you think they would forget their mother if she went away altogether? (Pg 50)

ii) Nora to Torvald:

Nora: “Torvald, you will be sorry for not letting me stay, even

for just half an hour.”

She knows that the letter is still in the mailbox and doesn’t want Torvald to find out about the contract.

iii) Nora to Mrs Linde:

Nora: “You all think I’m incapable of doing anything serious…or of ever having to face the brutality of life.”

Situational irony

Situational irony occurs when something entirely different happens from what the audience may be expecting, or the final outcome is opposite to what the audience is expecting.

Examples

-Situational Irony is present when Nora is discussing Krogstad’s forgery with her husband in Act 1. Minutes before this conversation, Krogstad approached Nora about her own forgery of her father’s signature.

-There is very little hint that Nora is going to leave Torvald until the end of the play. At the beginning of the play she acts as if she loves him very much. Not until she says “Or if anything else should happen to me – anything, for instance, that might prevent me from being here” does anyone think about Nora leaving Torvald. At the end of the play she calls Torvald a “stranger” and walks out.

-It is ironic that Torvald states that he awaits the moment when Nora will be in trouble so that he can rescue her. When in fact the truth comes out and Torvald has been given his opportunity to rescue Nora, all he is concerned with is his reputation. He yells at her. He insults her by calling her feather brain. He screams at her, telling her to go to her room. He is not interested in how he can rescue her. He is interested in how he can get out of this mess without ruining his good name.

-When Krogstad returns the IOU document, Torvald exclaims that he is saved and that he has forgiven Nora. When Nora asks if she is saved, Torvald exclaims that she is, of course. Only moments earlier, he was furious with her. Ironically, he did not even consider that she had borrowed the money to in fact save him.

-Situational irony is also evident earlier on in the play during Nora’s chat with Mrs. Linde, where she talks, or rather brags about her husband getting promoted as the manager of the bank. She says, “I feel so relieved to have heaps of money and not need to have any anxiety…” (Pg 11-12) Here, Nora visualizes a happy and blissful life with Helmer. However, there is irony in what she says because later on in the play, her marital relationship will be shattered and she will leave her husband and all the “money” that she had visualized, for an uncertain future away from Torvald after realizing that the world she was living in was equivalent to the world of a puppet, or rather, a doll.

Verbal irony

Verbal irony occurs when a speaker’s intention is the opposite of what he or she is saying.

Examples

-Verbal irony is present when Helmer says, “Is that my little skylark twittering out there?” (Pg 2) He is not really asking if Nora is a bird. He is not even saying that she is twittering like a bird. He is just asking if it is his wife, Nora, and if she is saying something. When Torvald Helmer says, “Is it my little squirrel bustling about?” (Pg 2) he does not think that Nora is a squirrel either.

-Nora has her share of verbal irony too. When she is sitting down talking to Mrs. Linde she says, “There now, it is burning up.” The place is not literally burning up. The house is not on fire. Nora is just stating that the temperature inside the house is hot.

– When Nora is chatting with Mrs. Linde, where she says “just fancy, my husband has been made manager of the Bank!”(Pg 11) where she talks, or rather brags about her husband getting promoted as the manager of the bank. She says, “I feel so relieved to have heaps of money and not need to have any anxiety…” (Pg 11-12) The reader is tempted to think that her life and that of her family is one smooth ride. But it emerges that she is deep in debt and even has to work extra hours at night in order to keep up with the payments.

FORESHADOW

Foreshadowing refers to clues that point to events that will happen later.

Examples

-Nora’s early rebellion of eating the macaroons against Torvald foreshadows her later rebellion

-The way Torvald always called Nora “My little skylark”, “My little squirrel”, “My little singing bird,” “My pretty little pet,” “My little sweet-tooth,” and “My poor little Nora.” was a foreshadow. She ends up saying something like “I’m just your little dove” in the later Acts when she decides to leave him. She acknowledges the fact there was never actually love between them; she was just his play toy, hence the name of the play, A Doll’s House.

-In the following conversation between Nora and Anne-Marie, there is use of foreshadow.

Nora: Yes, but, nurse, I shall not be able to be so much with them now as I was before.

Nurse: Oh well, young children easily get accustomed to anything.

Nora: Do you think so? Do you think they would forget their mother if she went away altogether? (Pg 50)

Nora eventually leaves her family, which was why she asked Anne-Marie how she possibly could have done it.

-Torvald’s stubbornness about denying Krogstad the banking job has complicated Nora’s attempt to continue hiding her little secret. We know there is going to be trouble later on. Nora’s secret is bound to come out. Ibsen has foreshadowed an ironic inevitability.

-Mrs. Linde plays the role of foreshadowing the future of Nora and a mirror to Nora’s character. She delves into the mistakes Nora will make and views her for what she truly is. She is the wise woman who has hindsight of what becomes of women who spend their money and borrow. She provides exposition to the play because she is the only one Nora can discuss her history with without consequences. Talking to Mrs. Linde provides an opportunity for the audience to understand Nora’s character.

-Mrs. Linde shares with Nora that her husband had died and that, due to her habits and his unstable business, she was now poor and struggling to make ends meet. She seems to be foreshadowing Nora’s impending fate.

CONTRAST/JUXTAPOSITION

Contrast or juxtaposition involves two characters or things being placed together with a contrasting effect.

Examples

-The father-daughter relationship between Nora and her father and that of Nora and Torvald is contrasted in the final Act. Nora makes this connection that life with her father was like life with Torvald. Nora’s father would force his beliefs on her and she would comply with them lest she upset him; she would bury her personal belief under Papa’s. According to Nora, Torvald was guilty of the same things. A good example was his insistence on her wearing the fish girl costume and his frustration over her inability to grasp the tarantella.

-Krogstad and Nora are also contrasted. The more we learn of Krogstad, the more we understand that he shares a great deal with Nora Helmer. First of all, both have committed the crime of forgery. Moreover, their motives were out of a desperate desire to save their loved ones. Also like Nora, Krogstad has contemplated ending his life to eliminate his troubles but was ultimately too scared to follow through.

– Dr. Rank’s treatment of Nora is contrasted sharply with that of Torvald. Rank always treats Nora like an adult. He listens to her and affords her a dignity, which is definitely missing in Torvald’s treatment.

– Mrs Linde’s relationship with Krogstad also provides a point of comparison with that of Nora and Torvald.

-Nora and Mrs Linde are also contrasted. Whereas Mrs. Linde took responsibility for her sick parent, Nora abandoned her father when he was ill. Mrs. Linde’s account of her life of poverty underscores the privileged nature of the life that Nora leads. Her sensible worldview contrasts sharply with Nora’s somewhat childlike outlook on life.

FOIL

A foil is a literary character who contrasts another character in order to highlight certain aspects of the other character.

Examples

-Mrs Linde’s life’s journey from independence to marriage is a foil to Nora’s journey in the opposite direction.

-Dr Rank is a foil to Torvald in that he treats Nora as an intelligent human being and she in return speaks more openly to him than she does to her husband.

-Mrs. Linde is the character that really makes Nora look bad in comparison and acts as a foil for Nora. In fact, you could argue that all the characters act as foils for Nora.

MOTIFS

Motifs are recurring structures, contrasts, or literary devices that can help to develop and inform the text’s major themes.

Examples

Nora’s definition of freedom

-Nora’s understanding of the meaning of freedom recurs in the course of the play. In the first act, she believes that she will be totally “free” as soon as she has repaid her debt, because she will have the opportunity to devote herself fully to her domestic responsibilities. After Krogstad blackmails her, however, she reconsiders her outlook regarding freedom and questions whether she is happy in Torvald’s house, subjected to his orders and commands. By the end of the play, Nora seeks a new kind of freedom. She wishes to be relieved of her familial obligations in order to pursue her own ambitions, beliefs, and identity.

Use of letters

-Many of the plot’s twists and turns depend upon the writing and reading of letters. Krogstad writes two letters: the first reveals Nora’s crime of forgery to Torvald; the second retracts his blackmail threat and returns Nora’s promissory note.

-The first letter, which Krogstad places in Torvald’s letterbox near the end of Act Two, represents the truth about Nora’s past and initiates the inevitable dissolution of her marriage. The second letter releases Nora from her obligation to Krogstad and represents her release from her obligation to Torvald.

-The two letters have exposed the truth about Torvald’s selfishness, and Nora can no longer participate in the illusion of a happy marriage.

-Dr. Rank communicates his imminent death through another form of a letter: a calling card marked with a black cross in Torvald’s letterbox. By leaving his calling card as a death notice, Dr. Rank politely attempts to keep Torvald from the “ugly” truth, as he had said earlier about his best friend, Torvald.

Other letters include Mrs. Linde’s note to Krogstad, which initiates her life-changing meeting with him, and Torvald’s letter of dismissal to Krogstad.

HYPERBOLE

Hyperbole refers to extreme exaggeration of statements or claims which makes someone or something sound bigger, better or more than they are.

Examples

i) Nora: Yes, that’s just it.

Helmer: Now you have destroyed all my happiness. You have ruined all my future. It is horrible to think of! I am in the power of an unscrupulous man; he can do what he likes with me, ask anything he likes of me, give me any order he pleases – I dare not refuse. And I must sink to such miserable depths because of a thoughtless woman! (Pg 104)

It is a hyperbole because although Nora may have caused a major accident of forging a signature and hiding it from Helmer, it is not obvious it will affect his future. Helmer is exaggerating that his happiness is destroyed because he feels betrayal and anger, just to show the seriousness of the shame that Nora has caused.

ii) Linde: But now I am quite alone in the world – my life is so dreadfully empty and I feel so forsaken.

This is a hyperbole because Mrs. Linde is exaggerating about her situation.

iii) NORA: I should like to tear it into a hundred thousand pieces.

It is a hyperbole because Nora cannot possibly be able to tear the letter into a thousand pieces.

REVISION QUESTIONS

CONTEXT QUESTIONS

Where is A Doll’s House set?

-In a middle-class living-room; nice but not extravagant furniture; piano, fire, winter’s day; Norway

Why does Nora tell the porter to hide the tree?

-So the kids don’t see it before it’s decorated

What does Nora sneak at the beginning of the play?

-Macaroons

What does Torvald call Nora?

-Squirrel, skylark

Why does Torvald scold Nora at the beginning of the play?

-for buying things and being careless with money

When is Torvald getting a new job? Why does he want Nora to wait to spend money?

-New Year; what if he gets hit in the head with a brick and therefore never gets the job?

How does Nora suggest they buy things before the promotion?

-borrowing from people (Helmer thinks she’s just a typical woman who doesn’t understand debt)

What does Nora want Torvald to get her for a present? Why doesn’t he want to give it to her?

-money; he calls her a spendthrift and says she’ll just waste all the money on unimportant things

What does Torvald think runs in Nora’s family?

-spending too much money (like her father)

Does Nora want to be like her dad? Does Torvald want this?

Yes, no

What is the first thing Nora lies about?

-she denies that she would ever eat macaroons because she promised not to…but she did

What did Nora do the Christmas before this one?

-locked herself away to make paper flowers to decorate the tree (the cat ripped up the flowers, though)

Who is Mrs. Linde?

-Christine; Nora’s childhood friend; they haven’t seen each other in 9 or 10 years

What does Nora think of Mrs. Linde’s appearance?

-she looks pale

What work has Nora had to do?

-sewing, embroidery, etc.

What secret does Nora tell Mrs. Linde first?

-when Torvald worked too hard in the first year of their marriage and got sick and had to go to Italy to recover, she borrowed money from her father in order to pay for it (at the time: her father was dying, she was pregnant, and she couldn’t go see him)

What is Christine’s backstory?

-she never loved her husband, she just married Mr. Linde because she had a sick mother and two little brothers to support; when he died, his business fell apart and she was left with nothing; she had been working ever since; recently, her mom died and her brothers are now old enough to take care of themselves.

What does Christine ask of Nora?

-if Torvald can give her a job at the bank he’ll be managing

Why does Christine think it’s nice of Nora to care so much about her troubles?

-because she says Nora has never had any of her own

What does Nora reveal to Mrs. Linde after being insulted because Christine said she didn’t really know trouble?

-she didn’t actually borrow money from her father; she borrowed it from someone else, but Torvald doesn’t know this; he didn’t even know how sick he was and that the Italy trip was to make him survive; she pretended that she just wanted to travel abroad and secretly borrowed the money to spare his pride

When does Nora plan on telling Torvald the truth?

-when she’s old and he’s not attracted to her any more

How has Nora been paying back the loan?

-lots of odd jobs secretly, scrimping on buying new clothes for herself (cheap clothes look good on her so Torvald doesn’t notice); last winter she copied letters late every night (perhaps she didn’t really spend all that time making tree decorations?)

What does Nora dream?

-a rich old man would come and give her all the money she needs

How does Mrs. Linde first claim to have known Krogstad?

-he was a lawyer in her area

What is Krogstad’s backstory?

-had an unhappy marriage, now has several children, got himself into bad business troubles

What does Mrs. Linde say when Dr. Rank tells her and Nora what a bad guy Krogstad is?

-people should try to help the diseased (Dr Rank says Krogstad is morally diseased)

What does Torvald say when Nora asks if Christine can have a job?

-yes; she’s come at a good time

What does Krogstad first ask Nora? Then what does he ask of her?

-if Mr. Helmer is giving Christine a job; if she will use her influence to get him a job (she says she doesn’t have much influence)

Why is Nora no longer afraid of Krogstad when he first comes to visit?

-come New Year she’ll be out from under his thumb

Why does Krogstad want to regain the community’s respect?

-for the sake of his sons

Why does Nora tell Krogstad not to tell Torvald about the money?

-that would only confirm what a bad person Torvald sees Krogstad as saying it would make everything unpleasant

What does Krogstad tell Nora that makes the whole thing worse?

-she forged her father’s signature; he can prove it because the signature is in Nora’s handwriting and she dated the document after her father died; basically he can prove that not only does she owe money but she broke the law (Nora says she didn’t have time to think about it because her husband was dying)

When does Nora start decorating the Christmas tree?

-after Krogstad leaves after blackmailing her

Why does Torvald think Krogstad was at the house?

-to get Nora to put in a good word for him

What does Nora ask Torvald to do for her?

-pick out her costume for the Stenborgs’ party because she’s helpless without his taste

What crime did Krogstad commit? Why does Torvald not like him?

-forgery; he never admitted his guilt, he got off through loopholes, he is corrupt and has lived a life of lies even around his family

How does the Christmas tree look at the start of Act 2?

-stripped and bedraggled

Who is Anne?

-nurse; she got pregnant by a bad guy and had to give up her child

What costume is Nora going to wear to the Stenborgs’ party?

-Neapolitan fisher girl

What dance does Torvald want Nora to do?

-tarantella

What does Christine offer Nora?

-to fix up her costume (if she can come by and see Nora in it)

What is Dr. Rank sick with? How did he get it?

-consumption of the spine; his promiscuous father

What can Nora not talk about with Torvald? Why?

-her old school friends; he gets jealous

Who does Christine guess lent Nora the money?

-Dr. Rank

What does Christine think of Nora and Dr. Rank’s relationship?

-they should stop talking to each other so much because he’s probably interested in Nora

Who does Nora think about borrowing money from to pay Krogstad?

-Dr. Rank

What does Nora tell Torvald she’ll do anything he asks of her for? What does Torvald say?

-if he’ll not fire Krogstad; he’s given his job to Mrs. Linde

What reasoning does Nora give as to why Torvald should not fire Krogstad?

-he writes small newspapers and he might write nasty articles about Torvald if he’s fired (like when her father got in legal trouble and was slandered by the papers); he should fire someone else and let Mrs. Linde and Krogstad both have jobs

What does Torvald think of Nora’s dad’s business?

-his dealings were shady

Why won’t Torvald give in to Nora’s request for a job for Krogstad?

-he already said no and told everyone at the bank he’s firing Krogstad, and he’s worried that if he changes his mind people will think his wife rules him; also, he knew Krogstad when they were kids and Krogstad insists on calling Torvald his first name which is really embarrassing given Krogstad’s corrupt past

What finally makes Torvald send a letter firing Krogstad?

-Nora calls him petty for worrying about his affiliations with Krogstad

Why does Torvald forgive Nora’s behavior in wanting a job for Krogstad?

-shows how much she loves him; but if trouble comes from the firing, he can handle it

What does Dr. Rank ask of Nora?

-to keep Torvald away from Dr. Rank’s sick room, because he is sensitive to such unpleasantness and Rank doesn’t want to upset him

How will Dr. Rank inform Nora of his death?

-business card with black cross in their mailbox when he starts to die (Nora says it’s morbid and depressing)

Who does Dr. Rank think will replace him as Nora’s friend after he dies?

-Mrs. Linde

How does Nora react when Dr. Rank tells her he loves her?

-she says it’s inappropriate; it makes her uncomfortable, he shouldn’t have said that

How does Nora respond when Rank tells her that she seems like she loves him more than Torvald?

-she says the people you have fun with aren’t the same as the ones you love; when she was little she loved her father the most, but had fun gossiping with the maids

What relationship does Nora liken her relationship with Torvald to?

-her relationship with her father

Why is Krogstad surprised Torvald would fire him at first?

-knowing what power he has over them

What does Krogstad offer?

-that the three of them can settle the matter, and no one else has to be involved

How does Krogstad plan to use the blackmail?

-to blackmail Torvald into giving him a promotion; in a year it’ll be him running the bank instead of Torvald

What does Nora threaten to do? What does Krogstad say?

-commit suicide; doesn’t believe she’ll do it, and it wouldn’t do any good because he would still be able to ruin her reputation which Torvald would never let happen

What does Nora expect Torvald to do when he finds out?

-take all the blame on himself; a wonderful, yet terrible thing

How does Mrs. Linde reassure Nora?

-says she will go and convince Krogstad to ask Torvald for the unopened letter back because they used to be close

What does Nora ask Torvald to help her with?

-practising the tarantella (she does it, and he says she has a lot of practising to do, which she agrees with and says he will have to help her every moment between then and the party)

What does Torvald guess when Nora says he will have no time to open letters since he’s helping her practise? What does he agree to do?

-that there’s a letter from Krogstad; wait till after the party to open it

Why is Nora kind of glad that Torvald’s going to find out?

-the wonderful thing, the miracle will happen – he’ll jump to her rescue and take all the blame

Where does Mrs. Linde wait for Krogstad? Why?

-at the Helmers’ house while they’re at the party; there’s no private entrance at the place where she’s staying

What is Krogstad and Mrs. Linde’s past?

-she broke up with him abruptly for the man she married because she needed his money to support her family

What does Krogstad say he has been like since Christine left him? What does she say?

-a man lost at sea on a wreck; she feels the same, and the shipwrecks should get together

Why does Mrs. Linde tell Krogstad she came to town? What does he say?

-for him: she needs someone to work for, to help, or she feels like she has no purpose in life; Krogstad calls her hysterical and says she’s just looking for a chance at self-sacrifice

What does Mrs. Linde say when Krogstad asks if she only got back together with him for Nora?

-she sold herself in the past and would never do it again

Why does Christine tell Krogstad not to get his letter back?

-all the lies in the Helmers house need to come to light

Did Nora want to leave the party so early?

-no

Why does Mrs. Linde say she is at their house?

-to see Nora’s costume

What did Torvald think of Nora’s performance at the party?

-a bit too realistic, she was a bit too much like a Neapolitan fisher girl for his taste (but the other guests loved it)

What does Torvald think Mrs. Linde should do instead of knitting?

-embroidery (more tasteful since knitting needles looks Chinese)

What does Torvald pretend when the Helmers go to parties?

-Nora’s not his wife, but his secret lover

What does Dr. Rank say he will be at the party next year?

-invisible

Why does Dr. Rank tell Torvald he had a right to drink a lot at the party?

-he did medical research all day and has found something for sure and was very productive (but really, he found out he’s definitely going to die)

What does Torvald notice about the mailbox? What is Nora’s response?

-someone tried to pick the lock; one of Nora’s hairpins is jammed into it; she blames it on the kids

What does Torvald find in the mailbox from Rank?

-two cards with black crosses on them

What does Torvald tell Nora after he says he’s glad to have her?

-he sometimes wishes she were in trouble so he could save her

What does Torvald do after reading Krogstad’s letter?

-screams at Nora, says she is just as disgusting as her father was; says his happiness is now destroyed because he will have to obey Krogstad

What does Torvald say in response to Nora’s threatening to commit suicide?

-it won’t do any good because Krogstad will still have power and suspect him as an accomplice (he doesn’t really care that she’d be dead?!)

What is Torvald’s solution?

-Nora can still live in the house to keep up appearances, but the relationship is over and she’s not allowed near the children because she’ll corrupt them

How does Krogstad fix the problem?

-sends another letter with the forgery and says he’s ashamed he tried to blackmail them

Why does Torvald say his love for Nora is even deeper now?

-after having forgiven her from the bottom of his heart; his possession of her has grown even greater; she shouldn’t worry because hewill continue to guide her through life as if she were a child

What does Nora complain about after Torvald forgives her?

-she says they have never had a serious conversation before now

Who does Nora accuse of treating her like a doll?

-Torvald and her father; they dressed her up and made her into what they wanted her to be

Why does Nora say she’s leaving Torvald and the kids?

-she has a duty to herself that she’s never fulfilled; she realizes she’s never been happy with Torvald; she will spend the night at Mrs. Linde’s

Why does Torvald admonish Nora for wanting to leave?

-he says she’s forsaking her sacred duties to her husband and children; then he says what Jesus would do

What does Nora need to learn by leaving?

-whether she’s just too ignorant to understand society, as Torvald says, or if society’s wrong

Why does Nora realize she never loved Torvald?

-she realizes he isn’t the man she thought he was when the “miracle” of him taking the blame from her didn’t happen

When does Nora say she would come back?

-if they had a true marriage instead of just living together

What does Helmer end with?

“The most wonderful thing of all”

ESSAY QUESTIONS

What is important about the title? Who is the “doll” Ibsen refers to?
Who is the more significant female character in terms of plot development, Nora or Christine? Explain your answer.
Do you think Christine’s decision not to prevent Krogstad from revealing the truth to Torvald is a betrayal of Nora? Does this act ultimately hurt or benefit Nora?
How does Henrik Ibsen reveal character in A Doll’s House? Is Nora a sympathetic character? Did your opinion of Nora change from the beginning of the play to its conclusion?
Does the play end the way you expected? Do you think this was a happy ending?
A Doll’s House is generally considered a feminist work. Do you agree with this characterization? Why or why not?
What does the Tarantella dance symbolize in A Doll’s house?

Teachers' Resources

Performing Arts Grade 7 CBC Free Schemes of Work

October 5, 2025 Maverick John Leave a comment

PERFORMING ARTS

GRADE 7 SCHEMES OF WORK

NAME OF THE TEACHER SCHOOL YEAR III TERM

Ref used:

Grade 7 Performing arts curriculum Design

MTP Grade 7 Performing Arts Teachers Guide

MTP Grade 7 Performing Arts Learners Book

Week	LSN	STRAND	Sub-strand	Specific Learning Outcomes	Key Inquiry Question(s)	Learning Experiences	Learning Resources	Assessment Methods	Ref
1	1	PERFORMING	Verse – Persona’s point of view	By the end of the lesson the learner should be able to: Knowledge a) discuss how a persona’s point of view expresses meaning in a verse Skill b) rehearse the verse to internalize the persona. Attitude c) Develop curiosity in defining the persona point of view.	1. How do we use verse performance to express issues in society? 2. How do we make the presentation of a verse interesting and memorable?	The learner is guided to: • read the verse script to brainstorm, with other learners, the meaning conveyed by the persona’s view • rehearse the verse to internalize the persona’s message	Exercise books ICT devices Observation checklists Portfolio Anecdotal record *MTP Performing Arts T.G pg. 129-131* *MTP Performing Arts P.b pg. 130*	Oral presentations Written tests Question and answer Assignments Participatory assessment Peer assessment
	2		Verse – using voice techniques in a performance	By the end of the lesson the learner should be able to: Knowledge a) state the persona’s point of view. Skill b) perform a verse using voice techniques to convey the intended message. Attitude c) Develop curiosity in defining the persona point of view.	1. How do we use verse performance to express issues in society? 2. How do we make the presentation of a verse interesting and memorable?	The learner is guided to: • perform a verse before an audience while employing the use of voice, body and movement to deliver the message	Exercise books ICT devices Observation checklists Portfolio Anecdotal record *MTP Performing Arts T.G pg. 131-132* *MTP Performing Arts P.b pg. 130-131*	Oral presentations Written tests Question and answer Assignments Participatory assessment Peer assessment
	3		Verse – Using body and space appropriately in a performance	By the end of the lesson the learner should be able to: Knowledge a) state ways in which verse performance express issues in the society. Skill b) use body and space appropriately in performing a verse to convey the intended message. Attitude c) Develop curiosity in defining the persona point of view.	1. How do we use verse performance to express issues in society? 2. How do we make the presentation of a verse interesting and memorable?	The learner is guided to: • perform a verse before an audience while employing the use of voice, body and movement to deliver the message	Exercise books ICT devices Observation checklists Portfolio Anecdotal record *MTP Performing Arts T.G pg. 132-133* *MTP Performing Arts P.b pg. 131-133*	Oral presentations Written tests Question and answer Assignments Participatory assessment Peer assessment
2	1		Verse – appreciating verse performance	By the end of the lesson the learner should be able to: Knowledge a) list ways we can make presentation of a verse interesting and memorable. Skill b) Watch a video clip on verse performance features. Attitude c) appreciate the use of poetic language, voice, paralinguistic features and stage movements as complementary elements in verse performance.	1. How do we use verse performance to express issues in society? 2. How do we make the presentation of a verse interesting and memorable?	The learner is guided to: • watch live or recorded verse performances to identify performance features • reflect on individual performance based on appraisal from other learners.	Exercise books ICT devices Observation checklists Portfolio Anecdotal record *MTP Performing Arts T.G pg. 133-134* *MTP Performing Arts P.b pg. 134*	Oral presentations Written tests Question and answer Assignments Participatory assessment Peer assessment
	2		Skit –describing a scenario for a skit	By the end of the lesson the learner should be able to: Knowledge a) describe a scenario on a selected theme on a pertinent and contemporary issue Skill b) audition and cast appropriately for the skit. Attitude c) appreciate the use of skit in addressing pertinent issues in society.	1. How do we present a skit on stage? 2. How do we role-play a character on stage? 3. What role do props and costumes play to enhance the performance of a skit?	The learner is guided to: • research and create a scenario, in a group, based on a contemporary issue such gender education, animal welfare education • watch a live or recorded performance and is guided to execute plot • examine a given scenario of a skit, identify and take up a role, in a group	Exercise books ICT devices Observation checklists Portfolio Anecdotal records Internet connectivity *MTP Performing Arts T.G pg. 134-138* *MTP Performing Arts P.b pg. 133-135*	Oral presentations Written tests Work reports
	3		Skit – interpreting milestones and performing in a skit	By the end of the lesson the learner should be able to: Knowledge a) define milestones in a skit. Skill b) interpret milestones and perform a devised skit, in groups. Attitude c) appreciate the use of skit in addressing pertinent issues in society.	1. How do we present a skit on stage? 2. How do we role-play a character on stage? 3. What role do props and costumes play to enhance the performance of a skit?	The learner is guided to: • perform a skit on a pertinent issue in the society before an audience in the school and the community	Exercise books ICT devices Observation checklists Portfolio Anecdotal records Internet connectivity *MTP Performing Arts T.G pg. 134-138* *MTP Performing Arts P.b pg. 133-135*	Oral presentations Written tests Work reports
3	1		Skit – manipulating voice, body and space to deliver a skit	By the end of the lesson the learner should be able to: Knowledge a) identify the role of props and costumes. Skill b) manipulate voice, body and space to effectively deliver a message using a skit Attitude c) appreciate the use of skit in addressing pertinent issues in society.	1. How do we present a skit on stage? 2. How do we role-play a character on stage? 3. What role do props and costumes play to enhance the performance of a skit?	The learner is guided to: • rehearse the skit focusing on character development through use of voice, body and space	Exercise books ICT devices Observation checklists Portfolio Anecdotal records Internet connectivity *MTP Performing Arts T.G pg. 140-141* *MTP Performing Arts P.b pg. 137-138*	Oral presentations Written tests Work reports
	2		Skit –use of props, costumes and make up in a skit	By the end of the lesson the learner should be able to: Knowledge a) identify the role of props and costumes in a play. Skill b) use props, costume and makeup to enhance performance of a skit on a given theme. Attitude c) appreciate the use of skit in addressing pertinent issues in society. Project a) stage a five-minute skit in class/school	1. How do we present a skit on stage? 2. How do we role-play a character on stage? 3. What role do props and costumes play to enhance the performance of a skit?	The learner is guided to: • use appropriate costumes and props in performance *Project Task:* • script a skit based on a pertinent and contemporary issue in society • cast for the skit • rehearse the skit • design and collect costume and props for the skit • stage a full presentation of the skit in class • get feedback from the class.	Exercise books ICT devices Observation checklists Portfolio Anecdotal records Internet connectivity *MTP Performing Arts T.G pg. 141-144* *MTP Performing Arts P.b pg. 138-140*	Oral presentations Written tests Work reports
	3	CRITICAL APPRECIATION	Kenyan Folk Music – medium of performance in Kenyan folk music	By the end of the lesson the learner should be able to: Knowledge a) identify the medium of performance in a Kenyan folk music performance. Skill b) listen to Kenyan folk music and identify its components Attitude c) appreciate different cultures through analyzing folk music from diverse Kenyan communities.	1. What values can one acquire from folk music in Kenya?2. How are different styles of performance applied in Kenyan folk music?3. How do different performance media enrich folk music?	The learner is guided to: • watch videos or live performances of Kenyan folk music and is guided to identify the medium of performance (vocal and/or vocal and instrumental) • watch videos or live performances and identify the components in the folk music (performers, songs, instruments/voice, costumes and props, dance movements)	Exercise books ICT devices Observation checklists Portfolio Anecdotal records Internet connectivity *MTP Performing Arts T.G pg. 145-150* *MTP Performing Arts P.b pg. 141-144*	Oral presentations Written tests Work reports
4	1		Messages and values in Kenyan folk music performance	By the end of the lesson the learner should be able to: Knowledge a) discuss the messages and values in Kenyan folk music performances b) describe the style of performance in Kenyan folk music using appropriate terminology. Skill c) express personal feelings towards Kenyan folk music experienced from performances Attitude d) appreciate different cultures through analyzing folk music from diverse Kenyan communities.	1. What values can one acquire from folk music in Kenya?2. How are different styles of performance applied in Kenyan folk music?3. How do different performance media enrich folk music?	The learner is guided to: • individually and in groups discuss messages and values portrayed in Kenyan folk music • discuss the styles of traditional performance ( solo, solo-response and choral) • discuss personal feelings, mood and attitudes experienced from folk performances watched or listened to • analyze Kenyan folk music performances within the community and on mass media.	Exercise books ICT devices Observation checklists Portfolio Anecdotal records Internet connectivity *MTP Performing Arts T.G pg. 151-153* *MTP Performing Arts P.b pg. 145-147*	Oral presentations Written tests Work reports
	2		Kenyan Folk dance – components of Kenyan folk dance performance	By the end of the lesson the learner should be able to: Knowledge a) analyze the components of a Kenyan folk dance performance Skill b) examine the messages and values in a Kenyan folk dance performance. Attitude c) appreciate analyzing folk dance from diverse Kenyan communities.	1. What constitutes a folk dance? 2. How can a dance be used as a medium of communication? 3. What should one consider in analyzing a Kenyan folk dance?	The learner is guided to: • watch live or recorded performances of Kenyan folk dances for general appreciation • listen to or watch performances, pick out and discuss distinct components of a Kenyan folk dance; song, ornamentation, instrumental accompaniment, costumes and décor, props and artifacts, dance steps, formations and patterns, transitions, audience, division of roles	Exercise books ICT devices Observation checklists Portfolio Anecdotal records Internet connectivity *MTP Performing Arts T.G pg. 154-156* *MTP Performing Arts P.b pg. 149-150*	Oral presentations Written tests Work reports
	3		Analyzing folk dance from diverse Kenyan communities	By the end of the lesson the learner should be able to: Knowledge a) Outline the main events in a Kenyan folk dance performance. Skill b) critique a Kenyan folk dance using a given criterion. Attitude c) appreciate analyzing folk dance from diverse Kenyan communities.	1. What constitutes a folk dance? 2. How can a dance be used as a medium of communication? 3. What should one consider in analyzing a Kenyan folk dance?	The learner is guided to: • watch live or recorded dance performances and discuss messages and values in a Kenyan folk dance • write a summary of events (plot) in a Kenyan folk dance performance individually and in groups • critique recorded or live performances of Kenyan folk dances from various communities using a given criteria	Exercise books ICT devices Observation checklists Portfolio Anecdotal records Internet connectivity *MTP Performing Arts T.G pg. 157-159* *MTP Performing Arts P.b pg. 150-151*	Oral presentations Written tests Work reports
5	1		Narrative – themes addressed in the script and performance of a narrative.	By the end of the lesson the learner should be able to: Knowledge & skill a) identify and illustrate the theme addressed in the script and performance of a narrative b) identify the various narration techniques used in the narration process and state their effectiveness in the narration process Attitude c) appreciate the narrative as a tool of addressing contemporary issues in society.	1. How can pertinent issues in society be addressed in narratives?2. How does character development in narratives mirror the society?	The learner is guided to: • stage live performance of a narrative on guided themes • evaluate others’ performance in groups or pairs • identify and discuss how pertinent issues in society are highlighted through performance of narratives	Exercise books ICT devices Observation checklists Portfolio Anecdotal records Internet connectivity *MTP Performing Arts T.G pg. 159-163* *MTP Performing Arts P.b pg. 151-154*	Oral presentations Written tests Work reports
	2		Use of body, voice and space aids to effectively communicate the intended message	By the end of the lesson the learner should be able to: Knowledge a) explain how the use of body, voice and space aids to effectively communicate the intended message. Skill b) demonstrate the use of body, voice and space. Attitude c) appreciate the narrative as a tool of addressing contemporary issues in society.	1. How can pertinent issues in society be addressed in narratives?2. How does character development in narratives mirror the society?	The learner is guided to: • watch live performances of narrative to identify and discuss the various performance elements; body, space, voice, message, storyline and narration style • discuss the role of the audience in the narration process	Exercise books ICT devices Observation checklists Portfolio Anecdotal records Internet connectivity *MTP Performing Arts T.G pg. 163-164* *MTP Performing Arts P.b pg. 154-156*	Oral presentations Written tests Work reports
	3		Use of costumes and props to enhance communication in a narrative	By the end of the lesson the learner should be able to: Knowledge a) explain the role of the costumes and props in a narrative Skill b) Demonstrate the use of costume and props enhances communication in narrative Attitude c) appreciate the narrative as a tool of addressing contemporary issues in society.	3. What attributes define an effective narrator?4. What is the role of costume and decor in the narration process?	The learner is guided to: • visit a performing gallery nearby and participate in watching and critiquing the narrative performances • discuss how props and costume enhance communication in the narration process in groups or pairs	Exercise books ICT devices Observation checklists Portfolio Anecdotal records Internet connectivity *MTP Performing Arts T.G pg. 165-167* *MTP Performing Arts P.b pg. 156-158*	Oral presentations Written tests Work reports
6	1		Verse –criteria for evaluating a verse performance	By the end of the lesson the learner should be able to: Knowledge a) examine the criteria for evaluating a verse performance Skill b) watch a video clip on verse performance. Attitude c) appreciate the use of verse to communicate issues in society.	1. What does one look for when evaluating a verse?2. What is the role of verse in shaping the learner’s perspective in society?	The learner is guided to: • research on the criteria for evaluating a verse; body, voice, movement, theme, poetic language • watch live or recorded performances of verse while paying attention to key events in the verse • constructively evaluate verse performances with a view to suggesting improvements	Exercise books ICT devices Observation checklists Portfolio Anecdotal records Internet connectivity *MTP Performing Arts T.G pg. 168-170* *MTP Performing Arts P.b pg. 158-159*	Oral presentations Written tests Work reports
	2		Examine main characters and events in verse	By the end of the lesson the learner should be able to: Knowledge a) examine main characters and events in a verse. Skill b) comment on the significance of the values promoted in verse Attitude c) appreciate the use of verse to communicate issues in society.	1. What does one look for when evaluating a verse?2. What is the role of verse in shaping the learner’s perspective in society?	The learner is guided to: • analyze, in groups, traits of characters presented in the verse and relate them to own experiences	Exercise books ICT devices Observation checklists Portfolio Anecdotal records Internet connectivity *MTP Performing Arts T.G pg. 171* *MTP Performing Arts P.b pg. 159-160*	Oral presentations Written tests Work reports
	3		Use of body, voice and space in a verse	By the end of the lesson the learner should be able to: Knowledge a) state the meaning of a verse. Skill b) analyze the use of body, voice and space to effectively communicate topical concerns in verse. Attitude c) appreciate the use of verse to communicate issues in society.	1. What does one look for when evaluating a verse?2. What is the role of verse in shaping the learner’s perspective in society?	The learner is guided to: • present orally and in writing own or group appraisal of a performance • relate stage conflicts and resolutions in a verse to real life situations.	Exercise books ICT devices Observation checklists Portfolio Anecdotal records Internet connectivity *MTP Performing Arts T.G pg. 171* *MTP Performing Arts P.b pg. 160*	Oral presentations Written tests Work reports
7	1		Using verse to communicate issues in society	By the end of the lesson the learner should be able to: Knowledge a) analyze the use of body, voice and space to effectively communicate topical concerns in verse. Skill b) demonstrate the use of a verse to communicate issues in the society. Attitude c) appreciate the use of verse to communicate issues in society.	1. What does one look for when evaluating a verse?2. What is the role of verse in shaping the learner’s perspective in society?	The learner is guided to: • brainstorm with others how the various performance elements combine to aid delivery of the message	Exercise books ICT devices Observation checklists Portfolio Anecdotal records Internet connectivity *MTP Performing Arts T.G pg. 172* *MTP Performing Arts P.b pg. 161*	Oral presentations Written tests Work reports
	2		Skit – use of the plot to communicate the message in a skit	By the end of the lesson, the learner should be able to: Knowledge a) discuss how character development is achieved in a skit Skill b) examine how plot is used to communicate the intended message in a skit. Attitude c) appreciate the role of the skit in society	1. How is plot used in communicating a message in a skit? 2. How is a character developed in a skit?	The learner is guided to: • watch live or recorded performances of skits and discusses how the various elements aid in communicating the intended message (scenario, storyline, milestones, plot, conflict, characterization, language, improvisation, use of voice and body)	Exercise books ICT devices Observation checklists Portfolio Anecdotal records Internet connectivity *MTP Performing Arts T.G pg. 173-176* *MTP Performing Arts P.b pg. 162-164*	Oral presentations Written tests Work reports
	3		How themes in a skit can help address moral issues	By the end of the lesson, the learner should be able to: Knowledge a) evaluate how thematic concerns in a skit can help addresses moral issues Skill b) analyze how body, voice and space can effectively be used to communicate messages in a skit. Attitude c) appreciate the role of the skit in society	3. How do themes addressed in skits shape moral issues in society?	The learner is guided to: • watch live or recorded performances of skits and discusses how the various elements aid in communicating the intended message (scenario, storyline, milestones, plot, conflict, characterization, language, improvisation, use of voice and body)	Exercise books ICT devices Observation checklists Portfolio Anecdotal records Internet connectivity *MTP Performing Arts T.G pg. 177* *MTP Performing Arts P.b pg. 164-165*	Oral presentations Written tests Work reports
8	1		Role of costumes and make up in depicting characters in a skit	By the end of the lesson, the learner should be able to: Knowledge a) state the role of skit in addressing issues in the society. Skill b) examine the role of costume and make-up in depicting the intended characters in a skit. Attitude c) appreciate the role of the skit in society	4. In what ways can body, voice and space be used to communicate in a skit? 5. What is the role of costume and makeup in a skit?	The learner is guided to: • evaluate performances by others to appraise qualities of a good performance. (storyline, acting, language and style, costume and make-up, props, use of space)	Exercise books ICT devices Observation checklists Portfolio Anecdotal records Internet connectivity *MTP Performing Arts T.G pg. 178-179* *MTP Performing Arts P.b pg. 165-167*	Oral presentations Written tests Work reports
	2	PERFORMING ART IN SOCIETY	Performing arts in society – social and economic role of performing arts	By the end of the lesson, the learner should be able to: Knowledge a) discuss the social and economic role of Performing Arts in Society. Skill b) watch a video on songs, dances, narratives and skit to identify economic and social roles of P.A in society. Attitude d) appreciate the place of Performing Arts in society	1. Why do we need Performing Arts in society?	The learner is guided to: • watch live or recorded songs, dances, verses, narratives and skits to identify the social and economic roles of Performing Arts in society • in groups research in the community and in the digital space and reports on the role of Performing Arts in society	Exercise books ICT devices Observation checklists Portfolio Anecdotal records Internet connectivity *MTP Performing Arts T.G pg. 181-184* *MTP Performing Arts P.b pg. 168-169*	Oral presentations Written tests Work reports
	2		Performing arts platform	By the end of the lesson, the learner should be able to: Knowledge a) state the social and economic roles of PA in the society Skill b) utilize the Performing Arts platforms and contexts in furthering the role of Performing Arts in society. Attitude c) appreciate the place of Performing Arts in society	2. How can Performing Arts products be availed to the wider society?	The learner is guided to: • participate in festivals, celebrations, ceremonies and talent fairs within and without the school to exhibit or illustrate the place of Performing Arts in society • discuss how Performing Arts can be utilized to address societal issues such as: peace, integrity	Exercise books ICT devices Observation checklists Portfolio Anecdotal records Internet connectivity *MTP Performing Arts T.G pg. 184-185* *MTP Performing Arts P.b pg. 169-170*	Oral presentations Written tests Work reports
	3		Applying lessons learnt in performing arts	By the end of the lesson, the learner should be able to: Knowledge a) identify lesson leant from performing arts. Skill b) apply lessons learnt in Performing Arts to real life situations. Attitude c) appreciate the place of Performing Arts in society	3. How can lessons learnt in Performing Arts be applied in real life situations?	The learner is guided to: • discuss lessons learnt from Performing Arts and identify	Exercise books ICT devices Observation checklists Portfolio Anecdotal records Internet connectivity *MTP Performing Arts T.G pg. 185-186* *MTP Performing Arts P.b pg. 170-171*	Oral presentations Written tests Work reports
9	1-3	END YEAR ASSESSMENT/CLOSING.

Teachers' Resources

Free class 7 revision papers

October 5, 2025 Maverick John Leave a comment

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Form 2 Complete Termly Exams With Marking Schemes Free PDF Downloads

October 5, 2025 Maverick John Leave a comment

Form 2 Complete Termly Exams With Marking Schemes

KISWAHILI F2 QS.docx
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HISTORY F2 MS.docx
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MATH F1 QS.docx
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HISTORY F2 QS.docx

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MOKASA JOINT HISTORY AND GOVERNMENT PAPER 2 EXAMS FREE

October 5, 2025 Maverick John Leave a comment

311/2

HISTORY AND GOVERNMENT

PAPER 2

TIME: 2½ hours

FORM FOUR EXAMINATION

Kenya Certificate of Secondary Education

HISTORY AND GOVERNMENT

Paper 2

MOKASA II

QUESTIONS & MARKING SCHEME

Instructions to Candidates

Answer ALL questions Section A
Any THREE Section B
Any TWO Section C

SECTION A (25 marks)

The meaning of Anthropology. (1 mark)

The study of human beings, their origin, development, customs, beliefs and social

relationships / way of life.

Archaeological site in Tanzania. (2 marks)
- Olduvai Gorge
- Peninj
- Leatoli
- Garusi
- Isimila / Eyasi
- Nyabusora
- Apis Rock

Inventions that promoted the development of Agriculture plough in Egypt –

Shadoof – Bronze hoes – ox – drawn plough. (2 marks)

Negative impacts of Trans-Saharan trade. (2 marks)

(i) led to increased warfare in the region.

(ii) Intensified slave trade which created insecurity and misery.

(iii) Demand for Ivory led to destruction of wildlife.

Meaning of telecommunication. (1 mark)

Technology of receiving and sending messages by television, telegraph etc / electronic

media.

Two reasons why the industrial revolution intensified the scramble for colonies in

Africa. (2 marks)

Colonies were perceived as sources of raw materials.
Colonies were seen as market for industrial goods.
Colonies were seen as regions of possible investment.

Main reason for growth of ancient city of Meroe. (1 mark)

Abundant wood fuel for the smelting of iron.

Two officials who assisted the Shona Emperor to administer the Kingdom. (2 marks)

(i) The head cook

(ii) Gate keeper / Chancellor

(iii) Court steward / Chamberlain / Chancellor

Queen mother
Head drummer
Head of the army
Senior son in law
King’s sister
Treasurers
9 Principal wives of King

One Africans Country that was colonized by Italy. (1 mark)

Libya
Italian Somaliland

One reason why the Africans in Tanganyika were against the use of Akidas by the

German Colonial administrators. (1 mark)

Akidas were foreigners
Akidas took African vacancies in administering their country.
Akidas were harsh / brutal / whipped Africans

Two political benefits enjoyed by Assimilated Africans in Senegal. (2 marks)

(i) They were allowed to send representatives in French chamber of deputies.

(ii) They were allowed to vote during elections

Subjected to French judicial System

Main reason for the failure of the League of nations to pressure world peace. (1 mark)

(i) Rearmament of Germany / Lack of military wing.

One function of the United Nations International court of Justice. (1 mark)

(i) Settling disputes over international borders

(ii) Handling other international disputes

(ii) Handling cases of human rights violation and crimes against humanity.

Two social achievements of the commonwealth. (2 marks)

(i) Enabled members to share technological information to promote educational

research.

(ii) Improved co-operation among members

Cultural exchange
Engagement in games etc.

One function of ECOWAS council of ministers. (1 mark)

(i) The general management of the organization

(ii) Advising the authority of Head of State.

Giving direction to the subordinate organs.

Two political changes introduced by Mobutu SeseSeko which led to dictatorship in

Democratic Republic of Congo. (2 marks)

He banned all the political parties except the people’s revolutionary movement which led
He amended the constitution stripping parliament its powers.
Abolished federal system / centralized powers around himself.
Civil servant were appointed by centralized powers around himself.
Declared himself a life president.

Two qualifications for election to the council of states in Idia. (2 marks)

(i) Indian citizen

(ii) One must be above the age of 30 years

(iii) One must be registered as a voter

One must be a resident of the state in which one is contesting.

SECTION B (45 MARKS)

(a) Three limitations of electronic sources of information on history and government.

(3 marks)

Can only be used in areas with electricity.
They are expensive to obtain and use.
They require experts.
They are subject to bias of the producer.
Contain unrealistic / exaggerated information inaccurate as they depict what appeals to the public.

(5×1= 5 marks)

(b) Six reasons why man lived in groups during Stone Age period. (12 marks)

(i) To help each other in times of hardships.

(ii) To give moral support / encouragement to each other.

(iii) To share resources.

(iv) To share work / duties.

For companionship.
For security reasons.

(6×2 = 12 marks)

(a) Three disadvantages of the open-field system of farming in Britain. (3 marks)

(i) Land was not fully utilized as land was left fallow.

(ii) Cart tracks and paths wasted land.

(iii) There was wastage of labour / time duet to ploughing fallow fields and leaving

idle.

(iv) Discouraged livestock rearing due to spread of diseases on common grazing

grounds.

It was difficult to practice selective breeding of livestock.
It was not easy to get enough way for winter breeding and farmers slaughtered animals in autumn.
Discouraged the use of machines.

(b) Six remedies that should be put in place by third world countries to prevent food

shortage. (12 marks)

Land reclamation through irrigation and drainage of swamps.
Agricultural policies to be reformulated from concentration on cash crops to paying more attention to food production and encourage indigenous crops.
Provision of extension services e.g. advice research, information dissemination and training of Agricultural officers.
Infrastructural development be improved e.g. transport, storage and marketing.
Reforestation programmes by planting more checking soil erosion, protection of water catchment areas and intercropping.
Intensive agricultural research undertaken to develop drought resident crops, control pests.
Stoppage of civil string through peaceful conflict resolutions and democracy.
Subsidies / reduction of taxes on farm inputs to increase production.
Control the rate of population growth through family planning.

(6×2 = 12 marks)

(a) Three uses of canals. (3 marks)

(i) To shorten trade routes in transport.

(ii) Linking industrial centres to other parts.

(iii) Water from canals can be used to irrigate land.

(iv) Canals supply water to town.

Acts as safety measures for controlling floods.

(b) Six factors that promoted industrialization in South Africa. (12 marks)

(i) Availability of natural resources / minerals / raw materials which were processed

by her industries.

(ii) Industrial goods being of high quality can compete favourably for market

internally, continentally and the rest of the world.

(iii) High population in South Africa has contributed to both skilled and unskilled

labour.

(iv) Development of sources of energy HEP, coal.

Well developed network of roads, water, air and railway transport to transport raw materials and manufactured goods.
Availability of capital mainly from trade in minerals earns the country foreign exchange.
Political stability after the end of apartheid rule has encouraged trade and investment.
Government support through good polices of promoting industrialization in the country and encouraged local and foreign investors.

(6×2 = 12 marks)

21.(a) Three social factors that led to the scramble and partition of Africa. (3 marks)

(i) Missionary factor / the need to spread Christianity / protect missionaries.

(ii) Need to settle their surplus population of Europe.

(iii) Abolition of Slave trade.

(iv) To civilize Africans through western education and medicine.

(v) Role of influential individuals.

(vi) Racism and paternalism.

(b) Six characteristics of direct rule in Zimbabwe. (12 marks)

(i) There was a large number of European settlers hence influenced the system of

administration.

(ii) Many British settlers believed that the territory was pre-ordained to be white

settler colony.

(iii) The colony was administered by a commercial company.

(iv) It was headed by an administrator, followed by other Europeans.

(v) The method of administration was applied to the Africans.

(vi) Legislative council was started, giving Europeans political rights for self

government.

Europeans acquired large tracks of land hence compel Africans to provide labour.

(6×2 = 12 marks)

SECTION C (30 MARKS)

(a) Five factors that led to the growth of Asante Kindom. (5 marks)

(i) Agriculture – The land was fertile and well – watered hence suitable for mixed

farming.

(ii) Trade – Through Trans- Atlantic Trade, the empire acquired wealth.

(iii) Unity and stability – Golden stool and centralized political system under

Asaintehene, provided unity.

(iv) Efficient standing Army – It had a large Army strengthened by introduction of

guns and gun powder.

(v) Conquest – The united against oppression and created a new Kingdom around

Kumasi which was closely knit.

(vi) Able leadership – They had able rulers like Obiri, Osei Tutu and Opoku who

united the people.

b) Five functions of the Lukiiko among the Bunganda Kingdom. (10 marks)

(i) Made laws for the Kingdom.

(ii) Advised the Kabaka.

(iii) Represented the interests of the people.

(iv) Acted as the court of appeal / settled disputes.

(v) Directed the collection of taxes in the Kingdom and planned government

expenditure.

(vi) Checked the activities of government.

The Bataka were the minor chiefs in charge of clans who were answerable to the

Mugema.

(5×2 = 10 marks)

(a) Five roles played by Kwame Nkrumah in liberation of Africa. (5 marks)

(i) He formed a political party, the CPP which fought for independence in Gold

coast.

(ii) He wrote a newspaper, the Accra Evening News which articulated the demands of

people of Ghana

(iii) He held rallies mobilize Ghanaians towards the struggle for independence.

(iv) He organized industrial boycotts encouraging the people to boycott European

goods.

(v) His arrest and detention in 1950 turned international attention to the plight of

Ghanaians.

(vi) He attended international conferences highlighting the cause of the Gold Coast

independence.

He attended constitutional negotiations which red to Ghana’s Independence.

(6×2 = 12 marks)

(b) Five political challenges faced by Tanzania since independence. (10 marks)

(i) Army muting over delayed promotion of Africa in 1964.

(ii) Riots by students of university of Dar-e-salam who opposed forceful service in

National Youth Service.

(iii) The assassination of AbeidKarume 1972 caused tension.

(iv) Attack of Tanzania by Dictator Idi Amin put the country into costly and

unnecessary war.

(v) The country had large influx of refugees.

(vi) Resignation of AboudJumbe strained relations between Zanzibar and main land

Tanzania.

(vii) Re-introduction of multi-party democracy reignited ethnic differences and

regionalism.

(viii) Failure of ujamaa policy weakened public confidence in the government.

Personality differnces between Nyerere, Amin and Jomo Kenyatta undermined regional cooperation.
Collapse of EAC
Border closure.

(5×2 = 10 marks)

(a) Five roles of the state governments in USA. (5 marks)

(i) It maintains law and order using state police.

(ii) It makes state laws using the state legislative.

It provides social amenities to its citizens, e.g. education, health and public

works.

It generates revenue from a number of internal sources.
It administers Justice using the state law courts.
It provides administrative structure within the states such as counties, municipalities and townships.

(5×2 = 10 marks)

(b) Five factors that may limit the supremacy of parliament in Britain. (10 marks)

(i) Local authorities make and pass by – laws without consulting parliament.

(ii) Legislation made by parliament may be altered by a future parliament.

(iii) As parliament makes laws, it takes into consideration the moral values of the

society.

(iv) The Action of parliament are heavily influenced by public opinion.

(v) Before legislation is made in parliament, the interest of the affected institutions are taken into accounts.

International law is taken into account when parliament is making laws.

(5×2 = 10 marks)

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Form 1 History Exams and Marking Schemes Free

October 5, 2025 Maverick John Leave a comment

Form 1 History Exams and Marking Schemes Free

NAME………………………………………..ADM NO…………………….

CLASS………………………………………..DATE………………………….

TIME: 2HRS

HISTORY & GOVERNMENT

FORM ONE TERM 3

JOINT EXAMINATION

Instructions

Write your name and admission number in the spaces provided above
Write your name, ADM no, class and date of the examination in the spaces provided above.
This paper consists of 23 questions
Check if all the pages are printed
Answer all the questions on the spaces provided.
Answer all the questions in English.

Name four early visitors to the East Africa Coast upto 1500AD. (4mks)

List four items that were exported from East Africa Coast to outside World by the traders upto 1500 AD (4mks)

Which four items did Coastal people of East African Coast get from outside world and took up the interior. (4mks)

State six positive impacts of the Indian Ocean trade on the people of East Africa. (6mks)

……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………

Name the first Portuguese to sail and appear in East African Coast in 1498. (1mk)

………………………………………………………………………………………………………….

State 6 reasons for Portuguese success (6mks)

List the duties of the Portuguese captains along the Coast and Mozambique. (4mks)

Outline six reasons responsible for the decline of the Portuguese rule of the Coast. (6mks)

Give six negative results of Portuguese rule in East Coast of Africa. (6mks)

Give five reasons why Seyyid said moved his capital from Muscat to Zanzibar. (5mks)

State four effects of Oman rule. (4mks)

State five reasons for coming of Christian missionaries in Kenya (5mks)

Outline five factors that facilitated spread of Christianity in Kenya. (5mks)

State six problems experienced by Christian missionaries in Kenya. (6mks)

Identify six positive effects of missionary activities. (6mks)

Define the term citizenship. (2mks)

……………………………………………………………………………………………………………………………………………………………………………………………………………………

State two ways of becoming a citizen in Kenya. (2mks)

State three limitations to the right of life. (3mks)

Identify three economic responsibilities of a Kenyan citizen. (3mks)

……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………….

Identify four values of a good citizen. (4mks)

Name four factors promoting National unity. (4mks)

State five factors limiting National Unity (5mks)

Identify five methods of resolving conflicts (5mks)

________________________________________________________________________________________

JOINT EXAMINATION

FORM 1 HISTORY TERM 3

MARKING SCHEME

Name four early visitors to the East Africa Coast up to 1500AD. (4mks)

Greeks
Romans
Persians
Phoenicians
Chinese
Arabs

List four items that were exported from East Africa Coast to outside World by the traders up to 1500 AD (4mks)
Gold
Ivory
Iron
Skins
Rhinocerous horns
Tortoise shells
Slaves

Which four items did Coastal people of East African Coast get from outside world and took up the interior. (4mks)

Guns
Glass
Beads
Swords
Porcelain bowls
Daggers
Iron wire
Cloth of silk, spice

State six positive impacts of the Indian Ocean trade on the people of East Africa. (6mks)

Foundation and growth of city states
Arab settlement
New administrative systems
Social stratification
Contact with the interior
Spread of Islam
Arab and Persian architecture
Sharia law
Emergence of the Swahili

Name the first Portuguese to sail and appear in East African Coast in 1498. (1mk)

Vasco da gama

State 6 reasons for Portuguese success (6mks)

Superior weapons
Better naval power
Disunity of Coastal City states
Weakness of rival navies
Portuguese Alliance with the localities
Reinforcement from India
Lack of resistance from Towns

List the duties of the Portuguese captains along the Coast and Mozambique. (4mks)

Collect tribute from local rulers
Impose custom duties on imports and exports
To suppress resistance or opposition of their rule
Supervised the ruling families.

Outline six reasons responsible for the decline of the Portuguese rule of the Coast. (6mks)

Inadequate personnel
Portuguese faced constant hostility and rebellion from Coastal people
Decline in trade made them lose revenue for administration
Distance between Portugal and East Africa Coast.
Portuguese were attacked by tropical diseases
At home, Portugal suffered annexation by Spain
In 1588, the Coast was invaded by Zimba warriors from Mozambique
The capture of eventual siege of Fort Jesus in 1696 by Omani Arabs

Give six negative results of Portuguese rule in East Coast of Africa. (6mks)

Taxation –the coastal towns were exposed to heavy taxation
Decline of Coastal Trade
Decline of Coastal towns
Slavery and slave trade
Segregation of local people
Suffering- they misruled the cities leading to misery.

Give five reasons why Seyyid said moved his capital from Muscat to Zanzibar. (5mks)

Zanzibar was loyal to him
It was green and pleasant with good climate
It had clean water
Its position was convenient for trade with the mainland
It had good harbours to anchor ships
Its climate and fertile soil were good for cultivation of cloves.

State four effects of Oman rule. (4mks)

Growth of slave trade
Growth of towns
Local ,regional and international trade to the rest of the world.
Linked East Africa Coast to the rest of the world
Spread of Islam religion.
Growth of plantation agriculture
Missionaries came to East Africa Coast in attempt to stop slave trade.

State five reasons for coming of Christian missionaries. (5mks)

Spread of Christianity
Spread of western civilization
Abolition of slave trade
Introduction of legitimate trade
Formation of missionary societies
The presence of Islam
Exploration and adventure

Outline five factors that facilitated spread of Christianity in Kenya. (5mks)

Early missionaries enjoyed the support of Seyyid Said.
Missionaries were quick to study African languages
At first most missionaries used Kiswahili in their work
African convents were used to spread gospel
Education and health influenced Africans into conversion
Building of Kenya-Uganda railway
Discovery of quinine-enabled missionaries fight disesses.

State six problems experienced by Christian missionaries in Kenya. (6mks)

Tropical diseases
Poor means of transport and communication
Inadequate essential supplies
Hostility from some communities
Insecurities from some communities and also wild animals
Scarcity of personnel(few)
Inter denominational differences
Hostility from slave traders
Islam in areas where Islam was prevalent –it was hard

Identify six positive effects of missionary activities. (6mks)

Spread Christianity
Introduction of western education
Rehabilitation centres
Medical services
Improvement in agriculture
Development of transport system
Translation of the Bible
Emergence of a new social class
Exploration
Rise of independent churches and schools
Representative of Africans.

Define the term citizenship. (2mks)

It is the status of being recognized under the custom or law as being legal member of a sovereign state or nation

State two ways of becoming a citizen in Kenya. (2mks)

By birth
By registration

State three limitations to the right of life. (3mks)

When a person acts in self-defense or property
When security officer suppress a riot, rebellion
When security officer act to prevent the escape of a person who is lawfully detained
When a security officer act to prevent an individual from committing a crime
During war
When a person is sentenced to death

Identify three economic responsibilities of a Kenyan citizen. (3mks)

Paying tax
Participating in development activities
Engaging in income generation
Protecting the environment
Fighting corruption

Identify four values of a good citizen. (4mks)

Nationalism
Patriotism
Morality
Integrity
Thrift
ethics

Name four factors promoting National unity. (4mks)

The constitution
Education
Fair distribution of resources
Social economic interactions
National currency
National; philosophies
National symbols eg national flag,national anthem,coat of arms, public seal
Government institution
National days and events.

State five factors limiting national Unity (5mks)

Religious conflict
Ethnicity
Racial intolerance
Corruption
Divisible politics
Economic regulation
Cultural conflict

Identify five methods of resolving conflicts(5mks)

Arbitration
Diplomacy
Legislation
Use of elders
Religious action
Community policing
International agreement

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KCSE MATHS TOP STUDENT REVISION RESOURCE

October 5, 2025 Maverick John Leave a comment

MATHEMATICS 1

PART I

SECTION A:

Use logarithm tables to evaluate (4 mks)

0.0368 x 43.92

361.8

Solve for x by completing the square (3mks)

2x² – 5x + 1 = 0

Shs. 6000 is deposited at compound interest rate of 13%. The same amount is deposited at 15% simple interest. Find which amount is more and by how much after 2 years in the bank (3mks)

The cost of 3 plates and 4 cups is Shs. 380. 4 plates and 5 cups cost Shs. 110 more than this. Find the cost of each item. (3mks)

A glass of juice of 200 ml content is such that the ratio of undiluted juice to water is 1: 7 Find how many diluted glasses can be made from a container with 3 litres undiluted juice (3mks)

Find the value of θ within θ < θ < 360^Oif Cos (2 θ + 120) = γ3 (3mks)

A quantity P varies inversely as Q² Given that P = 4 When Q = 2. , write the equation joining P and Q

hence find P when Q = 4 a (3mks)

A rectangle measures 3.6 cm by 2.8 cm. Find the percentage error in calculating its perimeter. (3mks)

Evaluate: 11/6 x ¾ – 11/12 (3mks)

½ of 5/6

A metal rod, cylindrical in shape has a radius of 4 cm and length of 14 cm. It is melted down and recast into small cubes of 2 cm length. Find how many such cubes are obtained ( 3mks)

A regular octagon has sides of 8 cm. Calculate its area to 3 s.f. (4mks)

Find the values of x and y if ( 2 mks)

3 x 1 = 2

2 1 -1 y

An equation of a circle is given by x² + y² – 6x + 8y – 11 = 0 (3mks)

Find its centre and radius

In the figure given AB is parallel to DE. Find the value of x and y

A line pass through A (4,3) and B(8,13). Find (6 mks)

(i) Gradient of the line

(ii) The magnitude of AB

(iii) The equation of the perpendicular bisector of AB.

A train is moving towards a town with a velocity of 10 m/s. It gains speed and the velocity becomes 34 m/s after 10 minutes . Find its acceleration (2mks)

SECTION B:

Construct without using a protractor the triangle ABC so that BC=10cm, angle ABC = 60⁰ and

BCA = 45⁰

On the diagram , measure length of AC
Draw the circumference of triangle ABC
Construct the locus of a set of points which are equidistant from A and B.
Hence mark a point P such that APB = 45⁰and AP = PB
Mark a point Q such that angle AQB = 45⁰ and AB = AQ

(a) A quadrilateral ABCD has vertices A(0,2) , B(4,0) , C(6,4) and D(2,3). This is given a

transformation by the matrix -2 0 to obtain its image A^I B ^I C^I D^I. under a second transformation

0 – 2

which has a rotation centre (0,0) through –90⁰ , the image A^II B^II C^II D^II of A^I B^I C^I D^I is

obtained. Plot the three figures on a cartesian plane (6mks)

(b) Find the matrix of transformation that maps the triangle ABC where A (2,2) B (3,4) C (5,2)

onto A B C where A( 6,10) B (10,19 ) C ( 12, 13). ( 2mks)

19.

In the triangle OAB, OA = 3a , OB = 4b and OC = 5/3 OA. M divides OB in the ratio 5:3

Express AB and MC in terms of a and b
By writing MN in two ways, find the ratio in which N divides
AB
MC

In the figure below, SP = 13.2 cm, PQ = 12 cm, angle PSR = 80^O and angle PQR = 90⁰. S and Q are the centres (8mks)

Calculate:

The area of the intersection of the two circles

The area of the quadrilateral S P Q R

The area of the shaded region

In an experiment the two quantities x and y were observed and results tabled as below

X	0	4	8	12	16	20
Y	1.0	0.64	0.5	0.42	0.34	0.28

By plotting 1/y against x, confirm that y is related to x by an equation of the form

Y = q

P + x

where p and q are constants. (3mks)

(b) Use your graph to determine p and q (3mks)

(c ) Estimate the value of (i) y when x = 14

(ii) x when y = 0.46 (2mks)

A racing cyclist completes the uphill section of a mountain course of 75 km at an average speed of v km/hr. He then returns downhill along the same route at an average speed of (v + 20) km/hr. Given that the difference between the times is one hour, form and solve an equation in v.

Hence

Find the times taken to complete the uphill and downhill sections of the course.
Calculate the cyclists average speed over the 150km.

In the diagram below, X is the point of intersection of the chords AC and BD of a circle. AX = 8 cm, XC = 4cm and XD = 6 cm
Find the length of XB as a fraction
Show that XAD is similar to XBC
Given that the area of AXD = 6cm², find the area of BXC
Find the value of the ratio

Area of AXB

Area of DXC

A town B is 55 km on a bearing of 050⁰. A third town C lies 75km due south of B. Given that D lies on a bearing of 255⁰from C and 170⁰ from A, make an accurate scale drawing to show the positions of the four towns. (3mks)

(scale 1cm rep 10 km)

From this find,

(a) The distance of AD and DC in km (2mks)

(b) The distance and bearing of B from D (2mks)

MATHEMATICS I

PART 1

MARKING SCHEME (100MKS)

No. Log

= 3.6502

0.3681 2.5660

0.3682 1.6427 + -4 = 1.6502 = 2.8251

0.2087 Logs 2

361.8 2.5585 + – v ans (4) 6.6850 x 10 ^-2

3.6502 = 0.06685

2 x² – 5x + 1 = 0

x²– 5 x + ½ = 0

x² – 5 x = ^– ½

x – 5x + ^–5 ² = ^–½ + ^–5 ² (m)

2 4 4

= x – 5 = ^–½ + 25 = 17 (3)

4 16 16

= x – 5/4 = 17/16 = 1.0625

x – 5/4 ± 1.031

X₁= -1.031 = 1.25 = 0.2192

X₂= 1.031 + 1.25 = 1.281

A₁= P(1 + R/100)²= 6000 x 113/100 x 113/100 = Sh. 7661.40

A₂ = P + PRT/100 = 6000 + 15 X 2 = 6000 + 1800

100

= Shs. 7800

Amount by simple interest is more by Shs. (7800 – 7661. 40)

Shs. 138.60

Let a plate be p and a cup c.

3p + 4c = 380 x 5 15p + 20c = 1900

4p + 5c = 490 x 4 16p + 20c = 1960

-p -60 (m)

p = Shs 60

3(60) + 4 c = 380

4c = 380 –180 = 2000 (3)

c= Shs. 50

Plate = Shs. 60 , Cup = Shs. 50 (A both)

Ratio of juice to water = 1 : 7

In 1 glass = 1/8 x 200 = Sh 25

3 litres = 300 ml (undiluted concentrate) (3)

No. of glasses =v 3000 = 120 glasses

Cos (2 θ + 120) = 3/2 = 0.866

Cos 30 , 330, 390, 690, 750 ….

2 θ + 120 = 330

2 θ = 210 , = 105⁰(3)

2 θ = 390 – 120 = 270⁰ , θ₂ 135⁰

2 θ = 690 – 120 = 570⁰ , θ₃ 285⁰ (for 4 ans)

θ₄= 315^o ( for >2)

2 θ = 750 – 120 = 630⁰,

P = k 4 = K/4 (substitution)

Q² 9

K = 4 X 4 = 16

9 9

P = 16 v when Q = 4

9Q²

P = 16 = 1/9 (A) (3)

9x4x4

The perimeter = (3.6 + 2.8 ) x 2 = 12.8 cm

Max perimeter = (3.65 + 2.85) x 2 = 23 cm Expressions

% error = 13 –12.8 x 100 m = 0.2 x 100 (3)

12.8 12.8

= 1.5620% (A)

1 1/6 x ¾ – 11/12 = (7/6 x ¾ ) -11/12 = 7/8 – 11/12 = 21-22

½ of 5/6 ½ of 5/6 5/12 5/12

= -1/24 = -1 x 12 = -1

5/12 24 5 10 (3)

Volume of rod = П r²h = 22/7 x 4 x 14 = 704cm³ (m)

Volume of each cube = 2x2x2 = 8 cm³ A

No. of cubes = 704 /8 = 88 cm³ A

< AOB = 360 = 45⁰

Tan 67.5 = h

h = 4 x 2.414 A

= 9.650cm

Area of 1 triangle = ½ x 8 x 9.656 x 8 cm = 38.628 x 8 vm

Octagon area = 38.628 x 8 m

= 309.0 cm² (A)

3 2 -1 2

2 1 -1 y

3 – x = 2 (1) x = 1 (2)

2 – 1 = y y = 1 (A)

x ² + y² – 6x + 8y – 11 = 0

x² – 6x + (-3)² + y² + 8y + (4)² = 11 + (-3)² + (4)² (completing the square)

(x – 3)² + (y+4)² = 11 + 9 + 16 = 36

(x – 3)² + (y + 4)² = 6²

Centre is (3, -4)

Radius = 6 units As (3)

14.

Figs A C B and D C E are similar

AB = AC = and AB = BC

DE DC DE CE

10 = 6 + x

3 6

= 10 = 15 + y, m

3 y 60 = 18 + 3x

10y = 15 + 3y 3x = 42

7y = 15 x = 14

y = 15/7 (A) (3)

A (4 , 3) B(8,13)

(i) gdt = change in y = 13-3 = 10 = 5

change in x 8-4 4 2

(ii) Mag AB = 8 -4 4 =

13 -3 10

Length = Ö4²+ 10² = Ö116 = 10.77 units

(iii) Mid point = 4 +8 , 3 + 3

2 2

= (6, 8) (mid point) (5 mks)

gdt of perpendicular to AB = -ve rec. of 5/2

-2/5

Eqn is y = -2/5 x + c

8 = -2/5 x 6 + c = 40 = -12 + 5c

= c = 52/5

y = -2/5 x + 52/5 (A)

Acceleration = Change in velocity

Time

= (34 – 10) m/s = 24 m/s

60 x 10 600

= 0.04m/s_2-(2)

17.

Triangle (8)

AC = 9cm

Circumference Centre

Circle

Perpendicular bisector of AB

(b) a b 2 3 5 6 10 12

c d 2 4 2 10 19 13

2a +2b = 6 x 2 = 49 + 4b = 12

3a + 4b = 10 3a + 4b = 10

a = 2 4 + 2b = b

2c + 2d = 10×2 = 4c + 4d = 20 2 b = 2 b = 1

3c + 4d = 19 3c + 4d = 19

c = 1

2 (1) + 2d = 10

2d = 8 Matrix is 2 1 (A)

d = 4 1 4

OC = 5/3 (31) = 5A

19.

(a) = AO + OB MC = MO + OC

= -3a = 4b = -5/8 (4b) + 5

= 5A – 5/2 b

(b) MN = 5 Mc = 3(5a – 5/2 b)

= 5 s a – 5/2 s b

MN = BN + BN

= 3/8 (4 b) + (1 – t) (-BA)

= 3/8 (4 b) + (1 – t)(3a – 4 b)

= 3/2 b + 3 ta –4b + 4tb

= (3-3t) a (4t – 5/2)b

MN = MN

= 5 s a – 5/2 sb = (3-3t)a + (4t – 5/2 )b

= 5 a = 3 – 3t = 5s + 3t =3

= -5/2 s = 4t –5/2 v 5s + 8t = 5

-5t = -2 t = 2/5

5 s = 3 – 3(2/5)

= 3 – 6/5 = 9/5

s = 9/25

(i) AN : NB = 2 : 3

(ii) MN : 9 : 16

20.

θ x pr²

360

Area of sector SPR = 80/360 x 13.2 x 13.2 x 3.142

= 121.6

Area of triangle SPR ½ x 13.2 x 13.2 x sin 80

= 85.8 cm²

(m of area of ) A (at least one)

(m of area) A(at least one)

Area of segment = 121.6 – 85.8

= 35.8 cm²

Area of sector QPR = 90/360 x 3.142 x 12 x12

Area of PQR = ½ x 12 x 12 = 72²

Area of segment = 113.1 – 72

= 41.1cm²

Area of intersection = (35.8 + 41.1) = 76.9 cm²

b). Area of quadrilateral = Area of PQR + SPR

= 85.8 + 72 = 157.8cm²

Area of shaded region = Area of Quadrilateral – Area of sector SPR

= 157.8 – 121.6

= 36.2 cm²

y = q p + x = q 1 = x + p

p + x y y q q

Gradient = 1/q at (0, 0.95) (8,2.0) (8,2.0) gradient = 2.0 – 0.95 = 1.05

8 8

1 = 0.1312

q = 1 = 7.619

0.1312

q = 7.62.

y(1/y) Intercept p = 0.95 P = 0.95

q 7.62

p = 7.62 x 095 = 7.27

at x = 14, y = 2.7

at y = 0.46, 1/y = 2.174

x = 9.6.

a) Distance = 75km uphill speed = vkm/h

uphill Time = 75/v hrs

Downhill speed = ( + 20) km/h

Downhill Time = 75 hrs.

v + 20

Takes larger uphill

75 – 75 = 1

v v+20

75 (v+20) – 75v = 1

v(v + 20) 1

75v + 1500 – 75v = v(v + 20) = v² + 20v.

v² + 20v – 1500 = 0

v = – 20 + 20² – 4(1) (-1500)

2(1)

v = –20 + 400 + 6000 = –20 + v6400

2 2.

V₁ = –20 + 80 = 30km/hr

V₂ = – 20 – 80 X impossible

speed uphill = 30 km /hr, T = 75 time = 2 ½ hrs

speed downhill = 50 km /hr Time = 75 Time = 2 ½ hr

Average speed = Total distance = 150km = 37.5 km/ hr

Total time 4hrs

X	0	4	8	12	16	20
Y	1.0	0.64	0.5	0.42	0.34	0.28
1/y	1.0	1.56	2.0	2.38	2.94	3.57

D C

A x X x C = BX . XD

8 x 4 = 6BX

BX = 8 x 14² = 16

6 3

X AD = XBC

XA = 8 = 24 = 3

XB 16 16 2

XD = 6 = 3

XC 4 2

< AXD = BXC (vertically opposite <_s))

SAS holds : they are similar.

LSF = 3/2 ASF = (3/2)² = 9/4

Area A x A = 6cm₂ Area B x C = 6 x 9 = 27 = 13.5cm²

24.

a) AD = 50km

DC = 35km

BD = 90km

Bearing is 020⁰

Bearing is 134^o (8mks)

MATHEMATICS I

PART II

SECTION (52 MARKS)

Without using tables, simplify

1.43 x 0.091 x 5.04

2.86 x 2.8 x 11.7 (3mks)

Make x the subject of the formula if

y = a/x + bx (3mks)

Give the combined solution for the range of x values satisfying the inequality

2x + 1< 10 – x < 6x – 1 (3mks)

A man is employed at a KShs. 4000 salary and a 10% annual increment. Find the total amount of money received in the first five years (4mks)

A town A is 56 km from B on a bearing 062⁰. A third town C is 64 km from B on the bearing of 140^o. Find

(i) The distance of A to C (2mks)

(ii) The bearing of A from C (3mks)

Expand (x + y)⁶hence evaluate (1.02) to 3d.p. (3mks)

Rationalise the denominator in (2mks)

Ö 3

1 – v3

The table below shows daily sales of sodas in a canteen for 10 days.

Day	1	2	3	4	5	6	7	8	9	10
No. of	52	41	43	48	40	38	36	40	44	45

Calculate the 4 day moving averages for the data (3mks)

Find the image of the line y = 3x = 4 under the transformation whose matrix is.

3mks

2 1

-1 2

Three points are such that A (4 , 8), B(8,7), C (16, 5). Show that the three points are collinear (3mks)
Write down the inverse of the matrix 2 – 3 hence solve for x and y if

4 3

2x – 3y = 7

4x + 3y +5 (3mks)

Use the table reciprocals to evaluate to 3 s.f. 3mks

1/7 + 3/12 + 7/0.103

Given that O is the centre of the circle and OA is parallel to CB, and that angle

ABC = 107⁰, find

(i) Angles AOC, (ii) OCB (iii) OAB (3mks)

Two points A and B are 1000m apart on level ground, a fixed distance from the foot of a hill. If the angles of elevation of the hill top from A and B are 60^o and 30^o respectively, find the height of the hill (4 mks)
Two matatus on a dual carriageway are moving towards a bus stop and are on level 5 km from the stop. One is travelling 20 km/hr faster than the other, and arrives 30 seconds earlier. Calculate their speeds. (5mks)
If log x = a and log y = b, express in terms of a and b

Log x ³

VY (2mks)

SECTION B:

The table below gives the performance of students in a test in percentage score.

Marks	0-9	10-19	20-29	30-39	40-49	50-59	60-69	70-79
No. of Students	2	4	7	19	26	15	12	5

Using an assumed mean of 44.5, calculate

The mean
The standard deviation
Find the median mark

Draw the graph of y = 2x² – x – 4 for the range of x -3 = x = 3. From your graph

State the minimum co-ordinates

Solve the equations
2x² – x – 4 = 0
2x² – 3x – 4 = 0

Two concentric circles are such that the larger one has a radius of 6cm and the smaller one radius of 4 cm. Find the probability that an item dropped lands on the shaded region 4mks

Two unbiased dice are thrown. Find the probability of obtaining (4mks)
A product of 6
A sum of 8

iii. The same number showing (4mks)

Two pulley wheels centers A and B are joined by a rubber band C D E F G H C round them. Given that larger wheel has radius of 12 cm and AB = 20 cm, CD and GF are tangents common to both wheels and that CBA = 60^o), Find

BD (Length)
CD

iii. Arc length CHG and DEF, hence find the length of the rubber.

V A B C D is a right pyramid with a square base A B C D of side 5 cm. Each of its four triangular

faces is inclined at 75⁰ to the base. Calculate

The perpendicular height of the pyramid
The length of the slant edge VA
The angle between edge VA and base A B C D
The area of the face VAB

Plot the graphs of y = sin x^o and y = cos 2x^o on the same axes for –180 £ x £180^o.

Use your graphs to solve the equation 2 sin x = cos 2x

The depth of the water in a rectangular swimming pool increases uniformly from 1M at the shallow

end to 3.5m at the deep end. The pool is 25m long and 12m wide. Calculate the volume of the pool

in cubic meters.

The pool is emptied by a cylindrical pipe of internal radius 9cm. The water flows through the pipe at speed of 3 metres per second. Calculate the number of litres emptied from the pool in two minutes to the nearest 10 litres. (Take II = 3.142)

A rectangle A B C D is such that A and C lie on the line y = 3x. The images of B and D under a

reflection in the line y = x are B¹ (-1, -3) and D¹ (1,3) respectively.

Draw on a cartesian plane, the line y = x and mark points B¹ and D¹
Mark the points B and D before reflection
Draw the line y = 3x hence mark the points A and C to complete and draw the rectangle ABCD.

State its co-ordinates, and these of A¹ and C¹.

Find the image of D under a rotation, through – 90⁰, Center the origin.

MATHEMATICS I

PART II

MARKING SCHEME.

1.43 X 0.091 X 5.04 X 100000 91 X 504 = 7/10³

2.86 X 2.8 X 11.7 10⁵ 2 x 28 x 117 x 10³

(3)

= 0.007 (A)

y = a/x + bx yx = a + bx²

Either

bx²– yx + a = 0

x = y ± v y² – 4ab

2b (3)

2x + 1£ 10 – x £ bx -1

2x + 1 £ 10 – x 10 –x £ 6x –1

3x £ 9 11£ 7x

x £ 3 x £ 11/7 (3)

11/7 £ x £ 3

a = 4000 r = 110/100 = 1.1 ( 4000, 4000 + 4000, 4400 + 0/100 (4400——)

(a and r)

Sn = a(r ⁿ– 1)

R -1 1.1 Log = 0.04139

X 5

0.20695

0.1 (4)

= 4000 (1.1⁵–1) (any)

1.1 –1 4000 (1.6 – 1)

0.1

A = 4000 ( 0.6105)

0.1

= Sh. 2442 = Sh. 24,420 (A) (4)

0.1

(i) b²= a² + b² – 2ab Cos B

= 64² + 56²– 2(64) (56) cos 78

= 4096 + 3136 – 7168 (0.2079)

= 7232 – km 1490.3

b² = 5741.7 = 5.77 km (5)

(ii) b a

Sin B Sin A

75.77 = 64

Sin 78 sin A Sin A = 64 x 0.9781

75.77

Sin A = 0.08262

A = 55.7⁰ (or B = 46.3⁰)

Bearing = 90 – 28 – 55.7

= 0.06.3⁰ (A)

(x + y) ⁶ = 1 (x) ⁶ (y)⁰ + 6 (x)⁵(y)¹+15(x)⁴ (y)² + 20x³y³ + 15x²y⁴ + 6xy⁵ + y⁶

(1.02)⁶ = (1 +0.02)⁶ x = 1

y = 0.02

(1.02)⁶= 1+6 (0.02) + 15 (0.02)² + 15(0.02) + 20(0.02)³ + 15 (0.02)⁴

= 1 + 0.12 + 0.006 + 0.00016

= 1.12616

= 1.126 (to 3 d.p) (3)

3(1 + 3) = 3 + 3 3 + v3

(1- 3)(1+ 3) 1-3 2

Moving averages of order 4

M₁ = 52 + 41 + 43 + 48 184 = 146

4 4

M₂= 184 – 52 + 40 = 172 = 43 for 7

4 4 for > 4

M₃= 172– 40 + 38 = 170 = 42.5

4 4

M₄= 170 – 38+36 = 168 = 42

4 4

M₅ = 168 – 36 + 40 = 173 = 43 (3)

4 4

M₆= 172 – 40 + 44 = 176 = 44

4 4

M₇= 176 – 44 + 45 = 177 = 44.25

4 4

y = 3x + 4

A(0,4) B (1,7) Object points

A B A B

2 1 0 1 4 9

-1 2 4 7 8 13

Y = Mx + C

M = 13 – 8 = 5 = 1

9-4 5 1

y = x+c y = x + 4

8 = 4 + c c = 4

AB = 8 -4 4 BC = 16 – 8 -8 for either

7 -8 -1 5 – 7 -2

AB = ½ BC and AB and BC share point B.

A,B,C are collinear. (3)

2 -3

4 3 det. = 6 + 12 = 18

Inv.= 1 3 3

-4 2

1 3 3 2 -3 x 1 3 3 7

18 18

-4 2 4 2 y -4 2 5

x 36

y 18 -18 (3)

x = 2, y = -1 (A)

1/7 + 3/12.4 + 7/0.103

1/7 + 3/1.24 x 10^-1 + 7/1.03 x 10^-1

0.1429 + 3(0.8064) + 7 x 10 (0.9709)

= 0.1429 + 0.2419 + 67.96 (3)

=70.52 (A)

(i) ADC = 2×73

= 146⁰

(ii) OCB = x = 180 – 146 = 34

(iii) 360 – 107 – 146 – 34

= 73 ⁰

Tan 30⁰ = y/x y = x tan 30

Tan 60⁰ = 1000 + y ; y = x tan 60 – 1000

X tan 30⁰ = x tan 60 – 1000

0.5773 x = 1.732x – 1000

1.732x – 0.577 = 1000

1.155x = 1000

x = 1000

1.155 = 866.0 m (A) (4)

5 km Slower speed = x km/hr

Time = 5/x

Faster = (x+20) k/h

Time = 5/x=20 T₁ – T₂ = 5/x – 5/x+20 = 30/3600

5 (x+20) –5x 1

x(x+20) 120

120 (5/x + 100 – 5x) = x² + 20x (5)

x² + 20x – 12000

x = –20 400 + 48000

x = -20 ± 220

Spd = 100 km/h

And x = 120 km/h (A)

Log x = a log y = b

Log x³ = Log x³ – log y ^½

= 3 Log x – ½ Log y

= 8a – ½ ab

SECTION B

17.

Marks	Mid point (x)	d = x-44.5	F	E = d/10	Ft	T²	Ft²v
0-9	4.5	-40	2	-4	-8	16	32
10-19	14.5	-30	4	-3	-12	9	36
20-29	24.5	-20	7	-2	-14	4	28
30-39	34.5	-10	19	-1	-19	1	19
40-49	44.5	-0	26	0	0	0	0
50-59	54.5	-10	15	1	15	1	15
60-69	64.5	20	12	2	24	4	48
70-79	74.5	30	5	3	15	9	45

=90 =1 =223

(a) Mean = (1 / 90 x 10) + 44.5 = 44.5 + 0.111

= 44.610

(b) Standard deviation = 10 233/90 – (1/90)²

10 2.478 – 0.0001 (8)

10 2.478

10 x 1.574 = 15. 74 (A)

39.5 + 5.192 (A)

44.69

(a) The probability = Shaded area

Large circle area

Shaded area = ПR² – П r²

= 22/7 (4²– 3²) v = 22/7 x 7 = 22

Large area = 22/7 x4x4 = 352/7 (A)

Probability = 22 = 22 x 7 = 7

352/7 352 16

(b)

	1	2	3	4	5	6
1	1,1	1,2	1,3	1,4	1,5	1,6
2	2,1	2,2	2,3	2,4	2,5	2,6
3	3,1	3,2	3,3	3,4	3,5	3,6
4	4,1	4,2	4,3	4,4	4,5	4,6
5	5,1	5,2	5,3	5,4	5,5	5,6
6	6,1	6,2	6,3	6,4	6,5	6,6

(M)

(i) P(Product of 6) = P((1,6) or (2,3) or (3,2) or (6,1))

= 4/36 = 1/9

(4)

(ii) P (sum of 8) = P( (2,6) or (3,5) or (4,4) or (5,3) or (6,2) )

= 5/36 (A)

(iii) P (same number) = P (1,1) or (2,2) or (3,3) or (4,4) or (5,5) or (6,6)

6/36 = 1/6 (A)

(i) Cos 60 = x/20 x = 20 x 0.5 = 10 cm

BD = 12 – 10 = 2 cm

(ii) CD = y Sin 60 = y/20 y = 20x 0.8666

CD = 17.32 cm

(iii) CHG = 120 reflex = 240⁰

CHG = 240/360 x 2 x p x r

= 50.27

DBF = 120⁰/360 x 2 x П x r = 1/3 x 2 x 3.142 x 2

= 4.189 (A)

Length C D E f G H C = 50.27 + 2(17.32) + 4.189

= 89.189 (A)

(a) From the diagram, XO = 5/2 = 2.5

Tan 75⁰ = VO/2.5 v m

VO = 2.5 x 3.732

Perpendicular height = VO = 9.33 cm

2 (A)

Diagonal of base = 5² + 5² = 50

Length of diag. 50 = 7.071 = 5.536

VA²= AO² + VO² (m)

3.536² + 9.3²

12.50 + 87.05

= 99.55 = 9.98 cm² (A) (8)

(c ) = VAO Tan = 9.33 = 2.639

3.536

VAO = 69.24⁰ (A)

(d) Cos VBA = = 2.5 /9.98 = 0.2505

VBA = 75.49⁰

Area VBA = ½ x 5 x 4.99 x sin 75.45 m ( or other perimeter)

= 5 x 4.99 x 0.9681

= 24.15 cm² (A)

Volume = cross – section Area x L

X-sec Area = (1 x 25) + (½ x 25 x 2.5)

= 25 + 31.25 = 56. M

Volume = 56.25 x 12

= 675 m³

Volume passed / sec = cross section area x speed

= П r²x l = 3.14 x 9/100 x 9/100 x 3 (8)

= 0.07635 m³/sec v (M)

Volume emptied in 2 minutes

= 0.07635 x 60 x 2

= 9.162 m² (A)

1 m³ = 1000 l

= 9.162 litres

= 9160 litres (A)

24.

MATHEMATICS II

PART I

SECTION A (52 MARKS)

Use tables to evaluate

³Ö 0.0912² + Ö 3.152 (5mks)

0.1279 x 25.71

Simplify (a – b)²

a² – b² (2mks)

The gradient function of a curve that passes through point: (-1, -1) is 2x + 3.

Find the equation of the curve. (3mks)

Find the value of k for which the matrix k 3

has no inverse. (2mks) 3 k

Without using tables, evaluate log 128 – log 18

log 16 – log 6 (3mks)

Find the equation of the locus of points equidistant from point L(6,0) and N(-8,4). (3mks)
The value of a machine is shs. 415,000. The machine depreciates at a rate of 15% p.a. Find how many years it will take for the value of the machine to be half of the original value. (4mks)
Use reciprocal tables to evaluate to 3 d.p. 2 – 1

0.321 n2.2 (4mks)

Using the trapezium rule, estimate the area bounded by the curve y = x², the x – axis and the co-ordinates x = 2 and x = 5 using six strips. (4mks)
Solve the equation for 0⁰ £ q £ 360⁰and Cos²q + ½ Cosq = 0 (3mks)
Point P divides line MK in the ratio 4:5. Find the co-ordinates of point P if K is point (-6,10) and M is

point (3,-8) (3mks)

How many multiples of 3 are there between 28 and 300 inclusive. (3mks)
The line y = mx – 1, where m is a constant , passes through point (3,1). Find the angle the line makes with the x – axis. (3mks)
In the figure below, AF is a tangent to the circle at point A. Given that FK = 3cm, AX = 3cm, KX = 1.5cm and AF = 5cm, find CX and XN. (3mks)

Make X the subject of the formula (3mks)

V = ³Ö k + x

sk – x

Write down the inequalities that describe the unshaded region below. (4mks)

0.5 2 x

-1.5

-2

SECTION B (48 MARKS)

Draw the graph of y = -x² + 3x + 2 for –4 £ x £ 4. Use your graph to solve the equations

(i.) 3x + 2 – x² = 0 (ii) –x² – x = -2 (8mks)

The marks obtained by Form 4 students in Examination were as follows:

Marks	0-9		10-19		20-29		30-39		40-49	50-59
No. of students	2		8		6		7		8	10
Marks		60-69		70-79		80-89		90-99
No. of Students		9		6		3

Using 74.5 as the Assumed mean, calculate:

(i) The mean mark

(ii) The standard deviation (8mks)

In the figure below, a and b are the position vectors of points A and B respectively. K is a point on

AB such that the AK:KB = 1:1. The point R divides line OB in the ratio 3:2 and point S divides OK in

the ratio 3:1.

B K

0 a A

(a) Express in terms of a and b

(i) OK (iii) RS

(iii) OS (iv) RA

(b) Hence show that R,S and A are collinear. (8mks)

The figure below is the roof of a building. ABCD is a rectangle and the ridge XY is centrally placed.

Calculate:

(i) The angle between planes BXC and ABCD.

(ii) The angle between planes ABXY and ABCD. (8mks)

On the same axis, draw the graph of y = 2cosx and y = sin ½x for 0⁰ £ x £ 180⁰, taking intervals of 15⁰

(6mks)

From the graph, find:

(a) The value of x for which 2cosx = sin ½ x (1mk)

(b) The range of values of x for which –1.5 £ 2cos x £ 1.5 (1mk)

Two towns T and S are 300km apart. Two buses A and B started from T at the same time travelling towards S. Bus B travelled at an average speed of 10km/hr greater than that of A and reached S 1 ¼ hrs earlier.

(a) Find the average speed of A. (6mks)

(b) How far was A from T when B reached S. (2mks)

P and Q are two ports 200km apart. The bearing of Q from P is 040⁰. A ship leaves port Q on a bearing of 150⁰ at a speed of 40km/hr to arrive at port R 7 ½ hrs later. Calculate:

(a) The distance between ports Q and R. (2mks)

(b) The distance between ports P and R. (3mks)

A farmer has 15 hectares of land on which he can grow maize and beans only. In a year he grows maize on more land than beans. It costs him shs. 4400 to grow maize per hectare and shs 10,800 to grow beans per hectare. He is prepared to spend at most shs 90,000 per year to grow the crops. He makes a profit of shs 2400 from one hectare of maize and shs 3200 from one hectare of beans. If x hectares are planted with maize and y hectares are planted with beans.

(a) Write down all the inequalities describing this information. (13mks)

(b) Graph the inequalities and find the maximum profit he makes from the crops in a year. (5mks)

MATHEMATICS II

PART II

Use logarithm tables to Evaluate

³Ö 36.5 x 0.02573

1.938 (3mks)

The cost of 5 shirts and 3 blouses is sh 1750. Martha bought 3 shirts and one blouse for shillings 850. Find the cost of each shirt and each blouse. (3mks)
If K = ( y-c )^1/2

a) Make y the subject of the formula. (2mks)
b) Evaluate y, when K = 5, p = 2 and c = 2 (2mks)
Factorise the equation:

x + 1/x = 10/3 (3mks)

DA is the tangent to the circle centre O and Radius 10cm. If OD = 16cm, Calculate the area of the shaded Region. (3mks)

Construct the locus of points P such that the points X and Y are fixed points 6cm apart and

ÐXPY = 60⁰. (2mks)

In the figure below, ABCD is cyclic quadrilateral and BD is diagonal. EADF is a straight line,

CDF = 68⁰, BDC = 45⁰and BAE = 98^0.

Calculate the size of: (2mks)

a) ÐABD b) ÐCBD
Otieno bought a shirt and paid sh 320 after getting a discount of 10%. The shopkeeper made a profit of 20% on the sale. Find the percentage profit the shopkeeper would have made if no discount was allowed? (2mks)
Calculate the distance:
i) In nautical miles (nm)
ii) In kilometres (km)

Between the two places along the circle of Latitude:

a) A(30⁰N, 20⁰E) and B(30⁰N, 80⁰E) (Take Radius of Earth = 6371Km). (2mks)
b) X(50⁰S, 60⁰W) and Y(50⁰S, 20⁰E) (Take Radius of Earth = 6371Km). (2mks)
A rectangular tank of base 2.4m by 2.8m and height 3m contains 3,600 litres of water initially. Water flows into the tank at the rate of 0.5m/s. Calculate the time in hours and minutes required to fill the tank. (4mks)
Expand (1 + a)⁵ up to the term of a power 4. Use your expansion to Estimate (0.8)⁵ correct to 4 decimal places. (4mks)
A pipe is made of metal 2cm thick. The external Radius of the pipe is 21cm. What volume of metal is there in a 34m length of pipe (p = 3.14). (4mks)
If two dice are thrown, find the probability of getting: a sum of an odd number and a sum of scoring more than 7 but less than 10. (4mks)
Find the following indefinite integral ò 8x⁵ – 3x dx (4mks)

x³

The figure below represents a circle of radius 14cm with a sector subtending an angle of 60⁰ at the centre.

Find the area of the shaded segment. (3mks)

Use the data below to find the standard deviation of the marks.

Marks (x )

Frequency (f)

(4mks)

SECTION II (48MKS)

The figure below shows a cube of side 5cm.

Calculate:

a) Length FC (1mk)
b) Length HB (1mk)
c) Angle between GB and the plane ABCD. (1mk)
d) Angle between AG and the Base. (1mk)
e) Angle between planes AFC and ABCD. (2mks)
f) If X is mid-point of the face ABCD, Find angle AGX. (2mks)
Draw on the same axes the graphs of y = Sin x⁰ and y = 2Sin (x⁰ + 10⁰) in the domain 0⁰£ x⁰ £ 180⁰
i) Use the graph to find amplitudes of the functions.
ii) What transformation maps the graph of y = Sin x⁰ onto the graph of : y = 2Sin (x⁰ +10⁰).
The table below shows the masses to the nearest gram of 150 eggs produced at a farm in Busiro

country.

Mass(g)	44	45		46		47		48		49		50		51		52		53		54	55
Freq.	1	2		2		1		6		11		9		7		10		12		16	16
Mass(g)	56		57		58	59	60		61		62		63		64		65		70
Freq.	10		11		9	7	5		3		4		3		3		1		1

Make a frequency Table with class-interval of 5g. Using 52g as a working mean, calculate the mean mass. Also calculate the median mass using ogive curve.

A shopkeeper stores two brands of drinks called soft and bitter drinks, both produced in cans of same

size. He wishes to order from supplies and find that he has room for 1000 cans. He knows that bitter

drinks has higher demand and so proposes to order at least twice as many cans of bitter as soft. He

wishes however to have at least 90cans of soft and not more than 720 cans of bitter. Taking x to be

the number of cans of soft and y to be the number of cans of bitter which he orders. Write down the

four inequalities involving x and y which satisfy these conditions. Construct and indicate clearly by

shading the unwanted regions.

Two aeroplanes, A and B leave airport x at the same time. A flies on a bearing 060⁰ at 750km/h and B flies on bearing of 210⁰ at 900km/h:
a) Using a suitable scale draw a diagram to show the positions of Aeroplanes after 2hrs.
b) Use your graph to determine:
i) The actual distance between the two aeroplanes.
ii) The bearing of B from A.

iii) The bearing of A from B.

The Probabilities that it will either rain or not in 30days from now are 0.5 and 0.6 respectively. Find the probability that in 30 days time.
a) it will either rain and not.
b) Neither will not take place.
c) One Event will take place.
Calculate the Area of each of the two segments of y = x(x+1)(x-2) cut off by the x axis. (8mks)
Find the co-ordinates of the turning point on the curve of y = x³ – 3x² and distinguish between them.

MATHEMATICS II

PART I

MARKING SCHEME:

0.0912² = (9.12 x 10^-2)² = 0.008317

Ö 3.152 = 1.776

³Ö 1.776 + 0.008317

0.1279 x 25.91

= ³Ö 1.784317 No. log

0.1279 x 25.91 1.784 0.2514

0.1279 -1.1069

25.71 1.4101 +

0.5170

-1.7344

x 1/3

10^-1 x 8.155(6) 1^-1.9115

Or 0.8155(6)

(a – b)(a – b) = a – b

(a – b)(a + b) a + b

dy = 2x + 3

y = x² + 3x + c

-1 = 1 – 3 + c

c = 1 ; E.g y = x² + 3x + 1

K² – 9 = 0

K = ± 3

log 128 = log 64

18 9

log 16 log 8

6 3

= 2 log (8/3)

log (8/3)

= 2

Midpoint -8 + 6, 4 + 0 (-1, 2)

2 2

Gradient of LN = 4/-14 = -2/7

Gradient of ^ bisector = 7/2

y – 2 = 7/2

x + 1

y = 7/2X + 11/2

207,500 = 415,000(1 – 15 )ⁿ

100

0.5 = ( 85 )ⁿ

100

0.5 = 0.85ⁿ

log 0.5 = n log 0.85

log 0.5 = n

log 0.85

n = –1.6990 = -0.3010 = 4.264yrs

-1.9294 -0.0706

2 x 1 = 1 . x 20 = 0.3115 x 20 = 6.230

3.21 x 10^-1 3.21

1 = 1 = 0.5807 = 0.005807

172.2 1.722 x 10² 100

6.230 – 0.005807 = 6.224193

= 6. 224(3d.p)

X	2	2.5	3	3.5	4	4.5	5
y	4	6.25	9	12.25	16	20.25	25

h = ½

Area= ½ x ½[29+2(6.25+9+12.25+16+20.25+25)]

= ¼ [29 + 127.5]

= ¼ x 156.5 = 39.125 sq. units.

Cos q (cos q + ½ ) = 0

cos q = 0 cos q = -0.5

q = 90⁰, 270⁰ q = 120⁰, 240⁰

\ q = 90⁰, 120⁰, 240⁰, 270⁰

MP = 4 MK MK = -9

9 -18

MP = 4 ( -9 ) = ( -4 )

9 -18 8

\ P is ( -1,0 )

a = 30 d = 3 l = 300

300 = 30 + 3 (n – 1 )

300 = 30 + 3n – 3

300 – 27 = 3n

273 = 3n

91 = n

y = mx – 1

1 = 3m – 1

m = 2/3 = 0.6667

tan q = 0.6667 ; q = 33.69⁰

FK x FC = FA²

FC = 25/3 = 8 ^1/3 cm

CX = 8^1/3 – 9/2 = 23/6 = 3^5/6 cm

CX x XK = XA x XN

3^3/6 x 3/2 = 3 x XN

\ XN = 1^11/12 cm

V³ = k + x

k – x

V³k – V³x = k + x

V³k – k = x + V³x

V³k – k = x( 1 + v³)

V³k – k = x

1 + V³

(i.) x = 2 Þ x £ 2

(ii) y = -2 Þ y > -2

(iii)pts. (0.5,0)

(0,-1.5)

m = -1.5 – 0 = 3

0 – 0.5

Eq. Y = 3x – 1.5 y < 3x – 1.5

SECTION B

X	-4	-3	-2	-1	0	1	2	3	4
Y	-26	-16	-8	-2	2	4	4	2	-2

(i) Roots are x = -0.5 x = 3.6

(ii) y = -x²+ 3x + 2

0 = -x² – x + 2

y = 4x (-2, -8) (1, 4)

Roots are x = -2, x = 1

class x f d=x-74.5 fd d² fd²

0 – 9 4.5 2 – 70 – 140 4900 9800

10 – 19 14.5 8 – 60 – 480 3600 28,800

20 – 29 24.5 6 – 50 – 300 2500 15,000

30 – 39 34.5 7 – 40 – 280 1600 11,200

40 – 49 44.5 8 – 30 – 240 900 7,200

50 – 59 54.5 10 – 20 – 200 400 4,000

60 – 69 64.5 9 – 10 – 90 100 900

70 – 79 74.5 6 0 0 0 0

80 – 89 84.5 3 10 30 100 300

90 – 99 94.5 1 20 20 400 400

Sf = Sfd = Sfd² = 77,600

60 -1680

(i) Mean = 74.5 + -1680

= 74.5 – 28 = 46.5

(ii) Standard deviation = Ö 77600 – ( –1680 )²

60 60

= Ö 1283.3 – 784

= Ö 499.3 = 22.35

a (i.) OK = OA + AK = ½ a + ½ b

(ii) OS = ¾ OK = 3/8 a + 3/8 b

(iii)RS = RO + OS = 3/8 a – 9/40 b

(iv) RA = RO + OA = – 3/5 b + a

RA = a – 3/5 b RS = 3/8 a + 9/40 b

= 3/8( a – 3/5 b)

\ RS = 3/8 RA

The vectors are parallel and they have a common

point R \ point R, S and A are collinear

KB = 3m NK = 1.5m XB = 5m

(i) XK = Ö 5² – 3² = Ö 16 = 4m

let ÐXKN = q

cos q = 1.5 = 0.375

q = 67.97(8)⁰

(ii) In DXNK

XN = Ö 4² – 1.5² = Ö 13.75 = 3.708

In D SMR; MR = KB = 3m

SM = XN = 3.708m

Let ÐSRM = a

tan a = 3.708 =1.236

a = 51.02(3)⁰

21.

21.

	0	15⁰	30⁰	45⁰	60⁰	75⁰	90⁰	105⁰	120⁰	135⁰	150⁰	165⁰	180⁰
Y =2cosX	2.00	1.93	1.73	1.41	1.00	0.52	0.00	-0.52	-1	-1.41	-1.73	-1.93	-2.00
Y = sin ½ X	0.00	0.13	0.26	0.38	0.50	0.61	0.71	0.79	0.87	0.92	0.97	0.99	1.00

(a) X = 73⁰ ± 1⁰

(b) Between 40.5⁰ and 139.5⁰

300km

T S

Let the speed of A be X km/hr

Speed of B = (X + 10) km/hr

Time taken by A = 300 hrs

Time taken by B = 300 hrs

X + 10

300 – 300 = 5

x x + 10 4

300(x + 10) – 300x = 5

x(x + 10) 4

300x + 300 – 300x = 5

x² + 10x

x² + 10x – 2400 = 0.

x = 44.25

X = -54.25 N/A

(b) Distance covered by A in 1 ¼ hrs = 44.25 x 5/4 = 55.3 km

Distance of A from T is 300 – 55.3 = 244.7 km

(a) Distance = 15 x 40 = 300km

(b)

PR² = 200² + 300² –2x 200 x 300 cos70⁰

= 130,000 – 41040 = 88,960

PR = 298.3 km

sin 70⁰ sin a

sin a = 300 sin 70⁰

298.3

= 0.9344

a = 69.1⁰

Bearing of R from P is

40 + 69.1 = 109.1⁰

(i.) X > y

(ii) 4,400X + 10,800Y £ 90,000

Simplifies to 11X + 27y £ 225

(iii) X + y £ 15

X > 0; y > 0

Boundaries

x = y pts (6,6) (12,12)

11x + 27y = 225 pts (13,3) (1,8)

X + y = 15 pts (0,15) (8,7)

Objective function

2400 x 3200y

(pt (2,1)

2400X + 3200y = 8000

Search line ® 3X + 4y = 10

Point that give maximum profit is (12,3)

\ maximum profit

= 2400 x 12 + 3200 x 3 = 38,400 shs.

MATHEMATICS II

PART II

MARKING SCHEME

No log.

36.5 1.5623

0.02573 –2.4104 +

-1.9727

1.938 0.2874 –

-1.6853

-3 + 2.6853

3 3

-1 + 0.8951

1.273(4) ¬ 0.1049

= 1.273(4)

Let shirt be sh x,

let blouse be sh. y.

5x + 3y =1750 (i.)

3x + y = 850 (ii)

mult (ii) by 3

9x + 3y = 2550 (iii)

Subtract (iii) – (i.)

– 4x = -800

Subt for x

= 250

Shirt = sh 200 ; Blouse = sh 250

(a) K²= y – c

y – c = 4pK²

y = 4pK² + c

(b) y = 4 x 2 x 25 + 2 ; y = 202

x² + 1 – 10x = 0

3x²– 10x + 3 = 0

3x (x – 3) – 1(x – 3) = 0

(3x – 1) (x – 3 ) = 0

x = 1/3 or x = 3

Area D OAD pyth theorem AD =12.49cm

½ x 12.49 x 10 = 62.45cm²

Cos q = 10/16 = 0.625

q = 51.3⁰ 62.5

Sector 57.3⁰ x 3.14 x 100 40.2 –

360 = 22.3

ÐXPY = 60⁰

\ÐXC₁Y = 120⁰

B1\ÐC₁XY = ÐC₁YX

= 180⁰ – 120⁰= 30⁰

Construct 30⁰ angles

at XY to get centres

B1 C₁and C₂ mojar arcs drawn

2 on both sides with C₁X and C₂X

as centres.

DAB = 180⁰– 98⁰ = 82⁰

ADB = 180 – (68 + 45 ) = 67⁰

ABD = 180 – (67 + 82)

= 31⁰

(a) 180⁰– (67 + 82)⁰= 31⁰

ÐABD = 31⁰ Opp = 180⁰

(b) (180 – 82)⁰ = 98⁰82 + 98 = 180⁰

180⁰ – (98⁰– 45⁰) =

ÐCBD = 37⁰ 180 – (98 + 45)⁰

= 37⁰

10 x 320

100 Discount = sh 32

Sold at sh 288

If no Discount = ( 320 x 20 ) % = 22.7%

288

(a) Dist along circle of lat.

Long diff x 60 x cos q nm

100 x 60 x Cos 50⁰

100 x 60 x 0.866

5196nm = 100 x 2pR Cos 50⁰

360

100 x 2 x 3.14 x 6371

360 = 5780Km

(b) 80 x 60 Cos 50 = 3895 Km

Vol =2.8 x 2.4 x 3 = 20.16m³

1m³ = 1000 L

20.16m³ = 20160 L

20160

3600

16560 L to fill

0.5 L – 1 sec

16560 L – ?

165600

5 x 3600

33120 hr

3600 @ 9.41 hrs ; @ 564.6 min.

1⁵ + 5.1⁴a + 10.1³.a² + 10.1²a³ + 5.1.a⁴

a = -0.2

1 + 5(-0.2) + 10(-0.2)² + 10(-0.2)³+ 5 (-0.2)⁴

1 – 1.0 + 0.4 – 0.08 + 0.008 = 0.3277 (4d.p)

Area of metal : Material – Cross section.

p(R² – r²)

3.14 (21 –19)

Vol 6.28cm² x 3400cm

= 215.52m³

Possibility space:

. 1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

P(odd) = 3/6 = ½

P(Sum > 7 but < 10) = 9 /36

\ P(odd) and P(sum > 7 but < 10 )

= ½ x 9/36 = 9/72 = 1/8

ò( 8x⁵/x³– 3x/x³) d₄

ò( 8x² – 3x^-2) d₄

16x³/3 + 6x^-3/-3 + C

16x³/3 – 2/x³+ C

Area of DAOB

½ x 14 x 14 x 0.866 = 84.866cm²

Area of sector = 60 x3.14 x 14 x14 = 10.257

360

Shaded Area

84.666 – 10.257 = 74.409cm²

Marks	F	Fx	fx²
5	3	15	75
6	8	48	288
7	9	63	441
8	6	48	384
9	4	36	324

åx = åf=30 åfx=210 1512

S.d = Ö åfx² – ( åfx )²

åf åf

= Ö 1512 – (210)²

30 30

= Ö 50.4 – 49

= Ö 1.4 = 1,183

SECTION II .

(a) FC = Ö 5² + 7.07² = Ö 50 = 7.071

(b) HB = Ö 5² + 7.07²= Ö 75 = 8.660

(d) b = Tan^-1 5/7.071 = Tan^-10.7071 = 35.3⁰

(e) y = Tan^-1 5/3.535 = Tan^-1= 54.7⁰

(f) ÐAGX = 19.4⁰

y = Sin x

x⁰	0⁰	30⁰	60⁰	90⁰	120⁰	150⁰	180⁰
sin x⁰	0	0.50	0.66	1.00	0.866	0.500	0

y = 2 Sin (x⁰+ 10⁰)

X⁰	0⁰	30⁰	60⁰	90⁰	120⁰	150⁰	70⁰
2 Sin(x +10⁰)	0.3472	1.286	1.8794	1.286	0.3472	-0.3472	-1.8794

Amplitudes for y = Sin x⁰ is 1

For

y = Sin(x+10⁰) is 2.

c.f	X	F
61	53	12
16		54
93	55	16
103	56	10
11		57
123	58	9
130	59	7
135	60	5
138	61	3
142	62	4
145	63	3
148	64	3
149	65	1
150	70	1

Mean = x + 52 + -4

150

52 – 0.02

= 51.08

Median = 51.4g.

class interval 59

Class interval	mid point	Freg.	c.f
44-48	46	12	12
49-53	51	49	61
54-58	56	64	125
59-63	69	22	147
64-68	66	3	130
69-73	71	1	150

X + Y £ 1000

X £ 2Y

Y < 720

X > 90

21.(a) 1cm = 200Km/h

A = 200 x 7.5 = 1500 Km

B = 200 x 9 = 1800Km.

(b) (i.) 15.8cm x 200 (ii) Bearing 224⁰

= 3160 Km. (iii) Bearing 049⁰

(a) P(R) x P(R)¹ (b) P(R)¢ x P(R) (c) P(R) x P(R’)

= 0.5 x 0.6 0.5 x 0.4 P(R)’ x P(R)

= 0.3 = 0.2 0.5 x 0.6 = 0.3

0.5 x 0.4 = 0. 2= 0.5

y = x(x + 1)(x – 2)

= x³ – x² – 2x

A₁ = ò(x³ – x² –2x) d₄

^-1[¼ x⁴ – 1/3 x²]_-1

= 0 – ( ¼ + 1/3 – 1) = 5/12

A₂ = ²ò(x³ – x² –2x) d₄

= ⁰ò ¼ x⁴– 1/3 x³ – x²)^-2₀

= ( ¼ .16 – 1/3 .8 – 8 )

= 4^-0– 8/3 – 4 = – 8/3

A₁ = 5/12= A₂ = 2 ²/₃

y = x³ – 3x²

dy = 3x² – 6x

At stationary

Points dy = 0

i.e 3x²– 6x = 0

3x(x – 2) = 0

x = 0 or 2

Distinguish

dy = 3x² – 6x

d²y = 6x – 6

dx²

(i) x = 0 dy² = 6x – 6 = -6 (ii) x = 2

dx² d²y = 6

-6 < 0 – maximum. dx²

\ (0,0) Max Pt. 6 > 0 hence

Minimum Pt.

x = 2, y = 8 – 12 = -4

(2, -4) minimum point.

MATHEMATICS II

PART I

SECTION 1 (52 Marks)

Without using tables evaluate:

Ö7.5625 x ³Ö3.375

15 (5 mks)

Make k the subject of the formula.

y = 1 Ök + y

T² k (3 mks)

If A = (x, 2) and xB = x and if AB = (8), find the possible values of x.

-2 (3 mks)

Simplify completely. (3 mks)

rx⁴ – r

2xr – 2r

Solve the equation. (3 mks)

Log ₃ (8-x) – log ₃ (1+x) = 1

Under an enlargement scale factor -1, A(4,3) maps onto A¹ (4,-5). Find the co-ordinates of the centre of enlargement. (3 mks)

Find the equation of the line perpendicular to the line 4x-y = -5 and passing through the point (-3,-2). (2 mks)
Find the standard deviation of the data below:

3,5,2,1,2,4,6,5 (4 mks)

What is the sum of all multiples of 7 between 200 and 300? (4 mks)

Solve the equation.

½ tan x = sin x for -180⁰ £ x £ 360⁰. (3 mks).

Expand (1-2x)⁴. Hence evaluate (0.82)⁴ correct to 5d.p. (4 mks)

The line y = mx – 3 passes through point (5,2). Find the angle that the line makes with the x-axis. (2 mrks)
A two digit number is such that 3 times the units digit exceed the tens digit by 14. If the digits are reversed, the value of the number increases by 36. Find the number (4 mks)

In the figure below, O is the centre of the circle, OA = 7 cm and minor arc AB is 11 cm long. Taking P = ²²/₇, find the area shaded. (3 mks)

A box contains 36 balls, all identical except for colour. 15 of the balls are black, 15 are brown and the rest are white. Three balls are drawn from the box at random, one at a time, without replacement. Find the probability that the balls picked are white, black and brown in that order. (2 mks)

Find the inequalities that describe the unshaded region R below. (4 mks)

SECTION 2 (48 Marks)

Draw the graph of y = x² + x – 6 for -4 £ x £

Use your graph to solve the equations.

(i) x²+ x – 6 = 0 (ii) x² + 2x – 8 = 0 (8 mks)

The diagram below represents a bucket that has been placed upside down. The radius of the top surface is 15cm and that of the bottom is 40cm. The vertical height of the bucket is 50cm.

Determine:-

The volume of the bucket.
The curved surface area of the bucket. (leave your answers in terms of p)

Draw, on the same axes, the graphs of y = cos q and y = 5 sin q for – 180⁰£ q £ 180⁰

From your graph, determine the amplitude of each wave.
For what value(s) of q is cosq – 5 sin q = 0 (8 mks)

A point P lies on a coast which runs from West to East. A ship sails from P on a bearing of 032⁰. When it reaches Q, 7km from P, a distress signal is observed coming from another ship at R. Given that R is N.E of P and on a bearing of 066⁰ from Q, calculate:

Ð
The distance QR, between the two ships.
The shortest distance from R to the shore. (8 mks)

A bag contains x red balls and y yellow balls. Four times the number of red balls is equal to nine times the number of yellow balls and twice the total number of balls exceeds the number of yellow balls by 44.

How many balls of each colour are three in the bag?
If two balls are drawn out of the bag at random one at a time with replacement what is the probability that the two balls are red? (8 mks)

A Kenyan businessman goes on a trip to West Germany through Italy and back to Kenya. In Kenya he is allowed to take Ksh. 67,000 for sales promotion abroad. He converts the Kenya currency into US dollars. While in Italy, he converts ²/₅ of his dollars into Italian lire, which he spends in Italy. While in West Germany, he converts ⁵/₈ of the remaining dollars into Deutsche marks which he uses up before coming to Kenya. Using the conversion rates 1 US dollar = 1.8 Deutsche marks = 16.75

Ksh = 1340 Italian lire. Answer the following questions:

How many US dollars did he take out of Kenya?
How many Italian lire did he spend in Italy?
How much money, in Deutsche marks did he spend in West Germany?
How much money in Ksh. did he have on his return to Kenya? (8 mks)

PQRS is a parallelogram in which PQ = r and PS = h. Point A is the midpoint of QR and B is a point on PS such that PS : PB = 4:3. PA and QB intersect at M.

Given that PM = kPA and BM = tBQ where k and t are scalars, express PM in two different ways and hence find the values of k and t.

Express PM in terms of r and h only. (8 mks)

Two variables T and X are connected by the equation T = ab^xwhere a and b are constants. The values of T and X are given in the table below:

T	6.56	17.7	47.8	129	349	941	2540	6860
X	2	3	4	5	6	7	8	9

Draw a suitable straight line graph and use it to estimate the values of a and b. (8 mks)

MATHEMATICS III

PART II

Section I: (52 Marks)

Use mathematical tables to evaluate:

8.67 (3 mks)

Ö 0.786 x (21.72)³

Simplify completely. (3 mks)

4 – 1

x² – 4 x-2

An Indian on landing at Wilson Airport changes Re 6000 into Kenya shillings when the exchange rate is Re = Ksh. 1.25. He spent Ksh. 5000 when in Kenya and converted the remaining amount to Rupees at the same rate as before. Find out how much the Indian is left with in Rupees. (3mks)

The last of three consecutive odd numbers is (2x+3). If their sum is 105, find the value of x. (4 mks)

a S b is defined by: a S b = (a + b)

If B S (2 S 3) = 4 S 1, Find B. (3 mks)

Find the value of M. (3 mks)

85⁰

160⁰

(a) Expand (1+2x)⁶ upto the term containing x³ . (2 mks)

(b) By putting x = 0.01, find the approximate value of (1.02)⁶ correct to 4 S.F. (2 mks)

Show that x is the inverse of : Y = 3 -3 1 X = 2 1 (3 mks)

-5 2 5 3

The probabilities of three candidates K, M and N passing an examination is ²/₃, ¾ and ⁴/₅ Find the probability that :

(a) All pass: (1 mk)

(b) At least one fails: (2 mks)

In the figure, PR is tangent to the circle centre O. If ÐBQR=30⁰, ÐQBC=27⁰,and ÐOBA=37⁰, find ÐBAC and Ð

C A

B P R

A frustrum of height 10cm is cut off from a cone of height 30cm. If the volume of the cone before cutting is 270cm³ , find the volume of the frustrum. (3 mks)

Evaluate 0 (2 mks)

( 3x²– 1 ) dx

4 x ²

If one litre of water has a mass of 1000g, calculate the mass of water that can be held in a rectangular tank measuring 2m by 3m by 1.5m. (give your answer in tonnes). (2 mks)
Write down the three inequalities which define the shaded region. (3 mks)

(3,2)

(2,1) (4,1)

The depth of sea in metres was recorded on monthly basis as follows:

Month	March	April	May	June	July
Depth (m)	5.1	4.9	4.7	4.5	4.0

Calculate the three monthly moving averages. (3 mks)

A number of women decided to raise sh. 6300 towards a rural project for bee keeping. Each woman had to contribute the same amount. Before the contribution, seven of them withdrew from the project. This meant the remaining had to pay more. If n stands for original number of women, show that the increase in contribution per woman was: 44100 (3 mks)

n(n-7)

SECTION II: (48 Marks)

Find the distance between points A(50⁰ S, 25⁰ E) and B(50⁰ S, 140⁰ E) in:

(i) Km (ii) nm (8 mks)

(take radius of earth to be 6400km, P = 3.14)

The distance S in metres, covered by a moving particle after time t in seconds, is given by :

S = 2t³ + 4t³– 8t + 3.

Find:

(a) The velocity at : (i) t = 2 (ii) t = 3

The instant at which the particle is at rest. (8 mks)

A car starts from rest and its velocity is measured every second for six seconds. (see table below).

Time (t)	0	1	2	3	4	5	6
Velocity v(ms ^-1)	0	12	24	35	41	45	47

Use trapezium rule to calculate the distance travelled between t = 1 and t = 6. (8 mks)

Using a pair of compass and ruler only, construct triangle ABC such that AB=9cm, BC=14cm and ÐBAC = 120⁰ . Draw a circle such that AB, BC and AC are tangents. What is the radius of this circle? (8 mks)
The marks scored by 100 students in mathematics test is given in the table below:

Marks	10-19	20-29	30-39	40-49	50-59	60-69	70-79
No. of students	8	15	15	20	15	14	13

(a) Estimate the median mark. (2 mks)

(b) Using 44.5 as the assumed mean, calculate:-

(i) The mean mark: (2 mks)

(ii) The variance: (2 mks)

(iii) The standard deviation: (2 mks)

(a) On the same axes, draw the graphs of : y = sin x ; y = cos x

y = cos^x + sin X for 0⁰ Ð X Ð 360⁰.

(b) Use your graph to deduce

(i) The amplitude

(ii) The period of the wave y = cos x + sin x.

Cos x = – sin x for 0⁰ Ð X Ð 360⁰ .

Given a circle of radius 3 units as shown in the diagram below with its centre at O(-1, 6). If BE and DE are tangents to the circle where E (8,2). Given further that Ð DAB = 80⁰.

A E

(a) Write down the equation of the circle in the form ax² + bx + cy² + dy + e = 0 where a, b, c, d, e are constants. (2 mks)

(b) Calculate the length DE. (2 mks)

(d) Calculate the value of angle DCB. (2 mks)

A building contractor has to move 150 tonnes of cement to a site 30km away. He has at his disposal 5 lorries. Two of the lorries have a carrying capacity of 12 tonnes each while each of the remaining can carry 7 tonnes. The cost of operating a 7 tonne lorry is sh. 15 per km and that of operating a 12 tonne lorry is sh. 25 per km. The number of trips by the bigger lorries should be more than twice that made by smaller lorries. (8 mks)

(a) Represent all the information above as inequalities.

How should the contractor deploy his fleet in order to minimise the cost of moving the cement? (8 mks)

MATHEMATICS III

PART I

MARKING SCHEME

	SOLUTION	MRK	AWARDING
1.	Ö7.5625 = 2.75 ³Ö3.375 = ³Ö3375 X ³Ö10^-3 = ³Ö3³x 5³x 10^-1 = 3 x 5 x 10^-1 = 1.5 = 2.75 x 1.5 = 2.75 = 0.275 1.5 x 10 10	1 1 1 1 1	Method for Ö7.5625 Square root Method for ³Ö ³Ö Answer
		5
2.	T²y = Ö k+y K T⁴y²k = k+y T⁴y²k – k = y K(T⁴y²-1) = y K = y T⁴y² – 1	1 1 1	Removal of square root Rearrangement of terms Answer
		3
3.	(x 2) x = (8) -2 x² – 4 = 8 x = +Ö12 = + 2Ö3 = + 3.464	1 1 1	Matrix equation Quadratic equation Answers in any form
		3
4.	r(x² – 1) 2r(x – 1) r(x² – 1)(x² + 1) 2r (x – 1) r(x – 1)(x + 1)( x² + 1) 2r (x – 1) = (x + 1)( x² + 1) 2	1 1 1	Complete factorisation of numerator Factorisation of denominator Answer
		3
5.	1 = log₃3 8 – x = 3 1+x -4x = -5 x = 5 4	1 1 1	Logarithic expression. Equation Answer
		3
6.	Let the centre be (a,b) 4-9 = -1 4-a -5-b 3-b 4-a = -4+9 -5-b = -3+b a = 4 b = -1 centre is (4,-1)	1 1 1	Equation Linear equations Centre
		3
7.	Y = 4x + 5 Gradient = 4 Gradient of ^ line – ¼ y + 2 = – 1 x + 3 4 4y + x = -11	1 1	Gradient of ^ line. Equation.
		2
8.	X = 28 = 3.5 8 standard deviation = Ö 22 = Ö2.75 = 1.658 8	1 1 1 1	Mean d values d² values Answer
		4
9.	a = 203 d = 7 L = 294 294 = 203 + 7(n-1) n = 14 S 14 = 14 (203 + 294) 2 = 7 x 497 = 3479	1 1 1 1	For both a and b Equation For n Sum
		4
10.	Sin x = 2 sin x Cos x Sin x = 2 cosx Sin x 2 cos x = 1 cos x = 0.5 x = 60⁰, 300⁰, -60⁰	1 1 1	Simplification Equation All 3 values
		3
11.	(1 +-2x)⁴ = 1-8x + 24x² – 32x³ + 16x⁴ (0.82)⁴ = (1 + -2 x 0.09)⁴ x = 0.09 (0.82)⁴ = 1 – 0.72 + 0.1944 – 0.023328 + 0.00119376 = 0.35226576 @ 0.35227 (5 d..p)	1 1 1 1	Expansion Value of x All terms Rounded
		4
12.	2 = 5m – 3 m = 1 tan q = 1 q = 45⁰	1 1	Value of m. Angle
		2
13.	Let the number be xy 3y = x + 14 10y + x = 10x + y + 36 = 9y – 9x Þ 36 3y – x = 14 9y – 9x = 36 y = 5 x = 1 the number is 15.	1 1 1 1	1^st equation 2^nd equation method of solving Answer
	S	4
14.	Let ÐAOB = q q x 2 x 22 x 7 = 11 360 7 q = 90⁰ Area shaded = 90 x 22 x 7 x 7 – 1 x 7 x 7 360 7 2 = 77 – 49 2 2 = 28 = 14cm² 2	1 1 1	Value of q Substitution Answer
		3
15.	P(WBb) = 6 x 15 x 15 36 35 34 = 15 476	1 1	Method Answer
		2
16.	Equation inequality L₁ y = x y £ x L₂ y = -2 y ³ -2 L₃ 2y + 5x = 21 2y + 5x < 21	1 1 1 1	1 mark for each inequality. Method for obtaining L³
	(i) roots are x = -3 x = 2 (ii) y = x² + x-6 0 = x² + 2x-8 y = -x + 2 roots are x = -4 x = 2	4	2 1 1 1 1 1	For all correct points. 1 for atleast five correct points. Correct plotting. Scale Smoothness of curve Both roots Linear equation Both roots
			8

18.	h = 15 h+50 40 h = 30cm H = 80cm (a) Volume = ¹/₃ p x 40 x 40 x 80 – ¹/₃ p x 15 x 15 x 30 = 128000 p – 6750 p 3 3 = 121,250p cm³ 3 (b) L² = 80² + 40² L = 15² + 30² = 6400 + 1600 = 225 + 900 = 8000 = 1125 L = 89.44 cm L = 33.54 cm Curved surface area of bucket = p x 40 x 89.44 p x15x33.54 = 3577.6p – 503.1p = 3074.5cm²	1 1 1 1 1 1 1 1	Expression Value of H Substitution Volume L L Substitution Area
		8
19.

20.	(i) ÐRPQ = 13⁰ ÐPQR = 32⁰+90⁰+24⁰ = 146⁰ ÐPRQ = 180⁰– (146⁰+ 13⁰) = 21⁰ (ii) P = 7 sin13⁰ sin 21⁰ P = 7 sin 13⁰ Sin 21⁰ = 4.394km P T (iii) Let PR = q q = 7 sin 146⁰ sin 21⁰ q = 7 sin 146⁰ sin 21⁰ q = 10.92 km sin 45⁰ = RT 10.92 RT = 10.92 sin 45⁰ = 7.72 km (2 d..p)	1 1 1 1 1 1 1 1	Fair sketch ÐPRQ Equation Method Equation Distance PR Equation RT
		8
21.	(a) 4x = 9y 2(x+y) = y+44 Þ 2x + y = 44 4x – 9y = 0 4x + 2y = 88 11y = 88 y = 8 x = 18 (b) P(RR) = 18 x 18 = 81 26 26 169	1 1 2 1 1 1 1	Equation Equation Method of solving Value y Value x Method Answer
		8
22.	(a) 67,000 Ksh = 67,000 US dollars 16.75 = 4,000 dollars (b) 2 x 4,000 = 1600 US dollars 5 1600 US dollars = 1600 x 1340 = 2,144,000 Italian lire (c) Remainder = 2400 US dollars 5 x 2400 = 1500 US dollars 8 1500 US dollars = 1500 x 1.8 = 2700 Deutche marks (d) Remainder = 900 US Dollars 900 US Dollars = 900 x 16.75 Ksh. = 15,075 Ksh.	1 1 1 1 1 1 1 1	Method Answer Method Answer For 1500 Answer Method Ksh.
		8
23.	PM = kPA = k(r + 1h) 2 = kr + 1kh 2 PM = PB + BM 3h + t BQ 4 = 3h + t(-3h + r) 4 4 = 3h – 3t h + tr 4 4 = 3 – 3t h + tr 4 4 t = k 3 – 3t = 1k 4 4 2 3 – 3t = 1 t 4 4 2 5t = 3 4 4 t = 3 + 4 4 5 = 3 5 \ k = 3 5 \ PM = 3r + 3h 5 10	1 1 1 1 1 1 1 1	PM PM PM simplified Both equations method Value of k Value k PM
		8

24.	YLogT Log T = log a + x log b Log T Þ 0.82, 1.25, 1.68, 2.11, 2.54, 2.97, 3.40, 3.84 y – intercept = log a = 0 a = 1 gradient = 3.84 – 0.82 = 3.02 9 – 2 7 = 0.4315 log b = 0.4315 = 0.4315 b = antilog 0.4315 b = 2.7	1 1 1 2 1 1 1 8	Plotting Labeling of axis Linear All correct logs Value of a Method of gradient Value of b

MATHEMATICS III

PART II

MARKING SCHEME

SOLUTION MARKS AWARDING

1.	No log 8.69 0.9390 0.786 1.8954 21.72 1.3369 1.2323 1.7067 – 2 2 + 1.7067 2 2 – 1 + 0.8533 0.7134 x 10 ^-1 = 0.07134	M1 M1 A1	ü reading to 4 s.f Rearranging
		3
2.	4 – 1 (x-2)(x+2) (x-2) – x+2 (x-2(x+2) – (x-2) (x-2(x+2) -1 x+2	M1 M1 A1
		3
3.	Re6000 = Ksh. 75000 Spent 5000 Rem 2500 Rem 2500 1.25 Re 2000	M1 M1 A1
		3
4.	2x – 1 , 2z + 1 , 2x + 3 6x + 3 = 105 6x = 102 x = 17	M1 M1 A1 A1	Allow M1 for us of different variable.
		4
5.	4 * 1 = 5 4 2 * 3 = 5 6 A * 5 = 5 6 4 A + 5 = 5 x 5A 6 4 6 A + 5 = 25 A 6 24 A = 20	M1 M1 A1 3
6.	180 – M + 20 + 95 = 180 295 – M = 180 – M = – 115 M = 115	B1 B1 A1
		3
7.	1 + 2x + 60x² + 160x³ + 1 + 0.2 + 0.006 + 0.00016 = 1.20616 = 1.206	M1 M1 M1 A1 4	Only upto term in x³. Correct substitution Only 4 s.f.
8.	3 -1 2 1 = I -5 2 5 3 6 -5 3 -3 -10 +10 -5 + 6 1 0 0 1	M1 M1 A1	Matrix multiplication gives : I 1 0 0 1
		3
9.	(a) 2 x 3 x 4 = 23 4 5 5 (b) 2 x 3 x 1 + 2 x 1 x 4 + 1 x 3 x 4 3 4 5 3 2 5 3 4 5 1 + 4 + 1 10 15 5 = 17 10	M1 M1 A1
		3
10.	ÐQCB = 30⁰ 180 – (27 + 30) = 123⁰ \ BAC = 57⁰. OBA = 37⁰ OAB = 37⁰ AOB = 106⁰ \ ACB = 53⁰	M1 M1 A1	Isosceles triangle. Angle at centre is twice angle at circumference.
		3
11.	V = 1 x 3.14 x r ² x 10 = 270 L.S.F. 20 = 2 30 3 V.S.F = 2 ³ = 8 3 27 Vol. of cone = 8 x 270 27 = 80cm³ \ Vol. Of frusturm = (270 – 80) = 190cm³	M1 M1 A1
		2
12.	3x ³ – x ^{-1 2} 3 -1 ₁ x ³ + 1 ² x ₁ 8 + 1 – ( 1 – 1) 2 8 1 – 2 = 6 1 2 2	M1 A1 2
13.	(2 x 3 x 1.5) volume 9 m³ 1L º 1000 cm³ 1000 L = 1 m³ 9000 L = 9 m³ 1000 L = 1 tonne 9000 L = 9 tonnes.	M1 A1
		2
14.	y ³ 1 (i) y < x – 1 (ii) y < 5 – x (iii)	B1 B1
		3
15.	M₁ = 5.1 + 4.9 + 4.7 = 4.9 3 M₂ = 4.9 + 4.7 + 4.5 = 4.7 3 M₃ = 4.7 + 4.5 + 4.0 = 4.4 3	M1 M1 M1
		3
16.	Original contribution per woman = 6300 N Contribution when 7 withdraw = 6300 (n-7) Increase – Diff. 6300 – 6300 n-7 n 6300n – 6300(n-7) n(n-7) 6300n – 6300 + 44100 n(n-7) 44100 n(n-7)	M1 M1 1 3
SECTION II (48 Marks)
17.	(i) 115⁰ A B Centre of circles of latitude 50⁰ S. R Cos 50⁰ AB = 115 x 2p R Cos 50^o 115 x 40192 x 0.6428 360 = 8252.98 km (ii) Arc AB 60 x 115 Cos 50 nm 60 x 115 x 0.6428 nm 4435 nm	M1 M1 M1 A1 M1 M1 M1 A1	No. log 60 1.7782 1+5 2.0607 0.6428 1.8080 4435nm 3.6469
		8
18.	(a) V = ds = 6t²+ 8t – 8dt (i) t = 2 V = 6×4 + 8×2 – 8 = 32 ms^-1 (ii) t = 3 V = 6×9 + 8×3 – 8 = 70ms^-1 (b) Particle is at rest when V = 0 6t² + 8t – 8 = 0 2(3t – 2) (t+2) = 0 t = 2 t = -2 3 particle is at rest at t = 2 seconds 3		Do not accept t = -2. Must be stated.
		8
19.	Area under velocity – time. graph gives distance. A = { h ½ (y₁ + y₆ ) + y₂ + y₃ + y₄ + y₅ )} = 1 { ½ ( 12+47) + 24 + 35 + 41 + 45)} = 29.5 + 14.5 = 174.5m	B1 B1 M1 M1 B1 B1 A1	Trapezium rule only accepted. Formula. Substitution into formular.
		8
20.	Drawing actual Scale 1cm = 2cm Radius 1cm = 2cm	M1 M1 M1 M1 M1 M1 M1 M1	Bisect ÐA Bisect Ð B Intersection at centre of inscribed circle. Draw circle. Measure radius. Arcs must be clearly shown.
		8
21.	mean = 44.5 + 130 100 = 44.5 + 1.3 = 45.8 (b) Variance S (x – A) 2 = 2800 Sf 100 = 28 S.D. = Ö 28 = 5.292	M1 A1 M1 A1 M1 A1
		8
22.	y = sin x x 0 60 120 180 240 30 360 sin x 0 0.866 0.866 0 -0.866 -0.866 0 y = cos x x q 60 120 180 240 300 360 cos x 1 0.5 -0.5 -1.0 -0.5 0.5 1.0 y = cosx + sinx x q 60 120 180 240 30 360 cosx + sinx 1 1.366 0.366 -1 -1.366 -0.366 1.0 (c) Cos x = – sin x x = 45⁰ , 225⁰
23.	(i) amplitude = 1.366 (ii) Period = 300⁰ (a) (x+1) ² + (y-6)² = 3² x² + 2x + 1 + y² – 12y + 36 = 9 x² + 2x + y² – 12y + 28 = 0 (b) cos 10 = OD DE = 3 DE 0.9848 DE = 3.046 (c) Twice ÐOED 10⁰ x 2 = 20⁰ (d) DAB = 80⁰ \ DCB = 100⁰	M1 A1 M1 A1 M1 A1 M1 A1	Formular (x-a)² + (y-b)² = r² Cyclic quad.
		8
24.	Let number of trips by 12 tonne lorry be x. Let number of trips by 7 tonne lorry be y. (a) x > 0 ; y > 0 24x + 21y £ 150 12 x 25 x X + 15 x 7 x y £ 1200 300x + 105y £ 1200 x > 2y (b) Ref. Graph paper. Minimising: 3 – 12 tonne lorry and 2 – 7 tonne lorries should be deployed.	B1 B1 B1

MATHEMATICS IV

PART I

SECTION 1 (52MKS)

Evaluate using logarithms 3Ö7.673 – 15.612

12.3 (4mks)

Solve x – 3x – 7 = x – 2 (3mks)

3 5 5

In the given figure CD is parallel to BAC, calculate the values of x and y. (3mks)

C D

B A

The surface area and volume of a sphere are given by the formulars S = 4pr² and V= ⁴/₃ pr³.

Express V in terms of S only. (3mks)

A line perpendicular to y = 3-4x passes through (5,2) and intercepts y axis at (0,k)

Find the value of K. (3mks)

An alloy is made up of metals P,Q,R, mixed in the ratio 4:1: 5: A blacksmith wants to make 800g of the

alloy. He can only get metal P from a metallic ore which contains 20% of it. How many Kgs of the ore

does he need. (3mks)

The co-ordinate of point A is (2,8) vector AB = 5 and vector BC = 4 Find the

-2 3

co-ordinate of point C. 3mks)

Two buildings are on a flat horizontal ground. The angle of elevation from the top of the shorter building to the top of the taller is 20⁰ and the angle of depression from the top of the top of the shorter building to the bottom of the taller is 30⁰. If the taller building is 80m, how far apart are they

(4mks)

The given figure is a quadrant of a piece of paper from a circle of radius 50cm. It is folded along AB

and AC to form a cone . Calculate the height of the cone formed.

(4mks)

5Ocm

50cm

Express 3.023 as a fraction (2mks)
Point A (1,9), Point B(3,5) and C (7,-3). Prove vectorically that A,B and C are collinear. (4mks)
A salesman gets a commission of 4% on sales of upto shs 200,000 and an additional 2% on

sales above this. If in January he got shs 12,200 as commission, what were his total sales (4mks)

Water flows through a cylindrical pipe of diameter 3.5cm at the rate of 2m/s. How long to the nearest minute does it take to fill a spherical tank of radius 1.4m to the nearest minute? (4mks)
Rationalize the denominator in Ö3

Ö 7 – 2

Leaving your answer in the form Öa + Öb

Where a ,b, and c are integers (3mks)

For positive values of x, write the integral solutions of 3£ x² £ 35 (4mks)
8 girls working 5 hours a day take 12 days to drain a pool. How long will 6 girls working 8 hours a day take to drain the pool?( Rate of work is equal) (2mks)

SECTION II (48 mks)

In the given circle centre O , A,E,F, is target to the circle at E. Angle FED = 30⁰ <DEC = 20⁰ and <BC0 = 15⁰

A F

Calculate (i) <CBE (3mks)

(ii) <BEA (2mks)

(iii) <EAB (3mks)

The sum of the 2^nd and third terms of a G.P is ⁹/₄If the first term is 3,

(a) Write down the first 4 terms of the sequence . (5mks)

(b) Find the sum of the first 5 terms using positive values of the common ratio (r)

(3mks)

E and F are quantities related by a law of the form E = KFⁿWhere k and n are

constants. In an experiment , the following values of E and F were obtained .

E	2	4	6	8
F	16.1	127.8	431.9	1024

Use graphical method to determine the value of k and n (Graph paper provided) (8mks)

In the domain –2 £ x £ 4 draw the graph of y = 3x² + 1 –2x .Use your graph to solve the equation. 6x² 4x + 4 = 0 (graph paper provided) (8mks)
A solid sphere of radius 18cm is to be made from a melted copper wire of radius 0.4mm . Calculate the length of wire in metres required to make the sphere. (5mks)

(b) If the density of the wire is 5g/cm³. Calculate the mass of the sphere in kg. (3mks)

A right cone with slant height of 15cm and base radius 9cm has a smaller cone of height 6cm chopped off to form a frustum. Find the volume of the frustum formed (8mks)

9cm

PQRS are vertices of a rectangle centre. Given that P(5,0) and Q and R lie on the line x+5 = 2y, determine

(a) The co-ordinates of Q,R,S, (6mks)

(b) Find the equation of the diagonal SQ (2mks)

A tap A takes 3 hours to fill a tank. Tap B takes 5 hours to fill the same tank. A drain tap C takes 4 hours to drain the tank. The three taps were turned on when the tank was empty for 1½ hours. Tap A is then closed. Find how long it takes to drain the tank.

(8mks)

MATHEMATICS IV

PART II

SECTION I (52MKS)

Without using mathematical tables, evaluate (3mks)

Ö 0.0784 x 0.27 (leave your answer in standard form)

0.1875

A father is three times as old as his son. In ten years time , the son will be half as old as the father . How

old are they now? (3mks)

A,B,C,D, is a parallelogram diagram. ADE is an equilateral triangle. AB and CD are 3cm apart.

AB = 5cm. Calculate the perimeter of the trapezium ABCE (3mks)

E D C

A B

Given that a = -2, b = 3 and c = -1, Find the value of a³ – b – 2c² (2mks)

2b² – 3a²c

The exchange rate in January 2000 was US $ 1 = Ksh 75.60. and UK £1 = Ksh 115.80. A tourist came to Kenya with US $ 5000 and out of it spent ksh.189,000. He changed the balance in UK £ . How many pounds did he receive? (4mks)

ABC is a cross – section of a metal bar of uniform cross section 3m long. AB = 8cm and AC = 5cm.

Angle BAC = 60⁰ . Calculate the total surface area of the bar in M². (4mks)

The bearing of a school chapel C, from administration block A, is 250⁰ and 200m apart.

School flag F is 150m away from C and on a bearing of 020⁰. Calculate the distance and

bearing of A from F. (5mks)

A box has 9 black balls and some white balls identical except in colour. The probability of picking a white ball is ²/₃

(i) Find the number of red balls (2mks)

(ii) If 2 balls are chosen at random without replacement, find the probability that they are of different colour. (2mks)

Under an enlargement of linear scale factor 7, the area of a circle becomes 441.p

Determine the radius of the original circle. (3mks)

A circle has radius 14cm to the nearest cm . Determine the limits of its area. ( 3mks)
Expand (1 + 2x)⁵ up to the term with x³. Hence evaluate 2.04⁵ to the nearest 3 s.f. (4mks)
The n^th term of a G.P is given by 5 x 2 ^n-2

(i) Write down the first 3 terms of the G.P (1mk)

(ii) Calculate the sum of the first 5 terms (2mks)

3 bells ring at intervals of 12min, 18min and 30min respectively. If they rang together at 11.55am, when will they ring together again. (3mks)
On a map scale 1:20,000 a rectangular piece of land measures 5cm by 8cm. Calculate its actual area in hectares. (3mks)
It costs Maina shs. 13 to buy 3 pencils and 2 rubbers; while Mutiso spent shs.9 to buy one pencil and 2 rubbers. Calculate the cost of a pencil and one rubber (3mks)

Three angles of a pentagon are 110⁰, 100⁰ and 130⁰. The other two are 2x and 3x respectively. Find their values . (2mks)

SECTION II (48MKS)

Members of a youth club decided to contribute shs 180,000 to start a company. Two members withdrew their membership and each of the remaining member had to pay shs. 24,000 more to meet the same expense. How many members remained? (8mks)
A box contains 5 blue and 8 white balls all similar . 3 balls are picked at once. What is the probability that

(a) The three are white (2mks)

(b) At least two are blue (3mks)

A rectangular tennis court is 10.5m long and 6m wide. Square tiles of 30cm are fitted on the floor.

(a) Calculate the number of tiles needed. (2mks)

(b) Tiles needed for 15 such rooms are packed in cartons containing 20 tiles. How many cartons are

there in total? (2mks)

sell each carton to make 20% profit ? (4mks)

The following was Kenya`s income tax table in 1988.

Income in K£ P.a Rate (Ksh) £

1 – 2100 2

2101 – 4200 3

4201 – 6303 5

6301 – 8400 7

(a) Maina earns £ 1800 P.a. How much tax does he pay? (2mks)

(b) Okoth is housed by his employer and therefore 15% is added to salary to make taxable income. He

pays nominal rent of Sh.100 p.m His total tax relief is Shs.450. If he earns K£3600 P.a, how much

tax does he pay? (6mks)

In the given figure, OA = a , OB =b, OP: PA =3:2, OQ:QB = 3:2

B
R

O A

(a) Write in terms of a and b vector PQ (2mks)

(b) Given that AR = hAB where h is a scalar, write OR in terms h, a. and b (2mks)

(d) Calculate the value of k and h (3mks)

A transformation P = and maps A(1,3) B(4,1) and C(3,3) onto A¹B¹C¹. Find the

co-ordinates of A¹B¹C¹ and plot ABC and A¹B¹C¹on the given grid.

Transformation Q maps A¹B¹C¹onto A¹¹(-6,2) B¹¹(-2,3) and C¹¹(-6,6). Find the matrix Q and plot

A¹¹B¹¹C¹¹on the same grid. Describe Q fully. (8mks)

By use of a ruler and pair of compasses only, construct triangle ABC in which AB = 6cm,

BC = 3.5cm and AC = 4.5cm. Escribe circle centre 0 on BC to touch AB and

AC produced at P and Q respectively. Calculate the area of the circle. (8mks)

The following were marks scored by 40 students in an examination

330 334 354 348 337 349 343 335 344 355

392 341 358 375 353 369 353 355 352 362

340 384 316 386 361 323 362 350 390 334

338 355 326 379 349 328 347 321 354 367

(i) Make a frequency table with intervals of 10 with the lowest class starting at 31 (2mks)

(ii) State the modal and median class (2mks)

(iii) Calculate the mean mark using an assumed mean of 355.5 (4mks)

MATHEMATICS IV

PART 1

MARKING SCHEME

Ö – 7.939

12.3

= No log

7.939 0.8998

12.3 1.0899

T.8099 ¹/₃ = 3 + 2.8099 T.9363 3

= -0.8635

A1

Subtraction

Logs

Divide by 3

Ans

5x – 3 (3x –7 ) = 3(x – 2 )

5x – 9x + 21 = 3x – 6

-7x = -27

x = 3⁶/₇

A1

Multiplication

Removal ( )

Ans

3x +5y + x = 180

9x = 180

x = 20

y = 60

B1

Eqn

. r = 3v ¹/₃

. r = S ^½

\ 3V ¹/₃ = S ^½

4P 4P

3V = S ³/₂

4P 4P

V = 4P S ³/₂

3 4P

A1

Value r

Equation

Expression

Grad line = ¼y – 2 = ¼

x – 5

y = ¼ x + ¾

k = ¾

P in Alloy = ⁴/₁₀ x 800

= 320g

= 100 x 320

= 3.2 kg

M 1

A 1

Equation

P in alloy

Expression

Ans

B (a,b) , C (x ,y)

.a – 2 = 5

.b – 8 -2

.a = 8 b = 6 B(8, 6 )

x – 8 = 3

y – 6 4

x = 11, y = 10 c(11,10)

A1

B conduct

Formular

80 – x

.h = x tan 70

h = (80 – x ) tan 60

\ x tan 70 = 80 tan 60-x tan 60

2.7475x + 1.732x = 138.6

4.4796 x = 138.6

.h = 138.6 x tan 60

4.4796

= 53.59

A1

Expression for h both

Equation

Expression for h

Ans

2pr = 90 x 2p x 50

360

r = 12.5

h = Ö2500 – 156.25

= Ö2343.75

= 48.41 cm

A1

Equation

expression for h

ans

10.

100 n = 302.323

n = 3.023

99n = 299.3

n = 2993

990

= 3²³/₉₉₀

A1

Equation

Ans

11.

AB = 3-15-9

= 2

-4

BC = 4

-8

AB = ½ BC

\ AB // BC

But B is common

\ A,B,C are collinear.

B1

A B & BC

Both

12.

4% of 200,000 = 8000/=

balance = 4200/=

6% of x = 4200/=

x = 4200 x 100

= 70,000

sales = sh. 270,000

B1

Both

Expression

Extra sales

Ans

13 .

Time = ²²/₇ x ^3.5/₂x ^3.5/₂ x 200 hrs

²²/₇x 140x140x 140x 3600

= 8960

3600

= 2 hrs 29min

A1

Vol tank

Div x 3600

Tank

14.

Ö3 = Ö3 Ö7 + Ö2

Ö7Ö2 Ö2Ö2 Ö7+ Ö2

= Ö3 Ö7 + Ö2

= Ö21 + Ö6

A1

Multi

Expression

Ans

15.

3 £ x ² x²£ 35

±1.732 £x x £ ± 5.916

1.732 £ x £ 5.916

integral x : 2, 3, 4, 5

B1

Lower limit

Upper limit

Range

Integral values

16.

No of days = ⁸/₆ x ⁵/₈ x 12

= 10 days

A1

Expression

days

17.

(i) ÐCED = ÐECD = 30

Ð CDE = 180 – 60

= 120

Ð CBE = 180-120

=60

(ii) Ð AEC = 90+30

= 120

Ð EAB = 180-(120+45)

= 15⁰

(iii) ÐBEO = 90-45

= 45

B1

ÐA EB = 45⁰

ÐBEO

18.

.ar + ar² = ⁹/₄

3r + 3r² = ⁹/₄

12r² + 12r – 9 = 0

4r² + 3r – 3 = 0

4r² + 6r – 2r –3 = 0

(2r – 1) (2r + 3) = 0

r = ½ or r = -1¹/₂

Ss = 3(1- (1/2 )⁵)

1 – ½

= 3 (1-¹²/_{3 2})

= 6 ( ³¹/₃₂)

= 6 ³¹/₃₂

A1

19.

LOG E. 0.3010 0.6021 0.7782 0.9031

LOG F 1.2068 2.1065 2.6354 3.0103

Log E =n log F + Log K

.n = gradient = 2 2.4 – 1.4 = 12 = 3

Log k. = 0.3 0.7 – 0.3 4

.k = 1.995

¾ 2

E = 2F ³

B1

Log E

Log F

Scale

Plotting

Line

Gradient

.x -2 -1 0 1 2 3 4

.y 17 6 1 6 9 22 41

.y = 3x 2 – 2x + 1 –

0 = 3x 2 – 3x – 2

y = x + 3

B1

All values

At least 5

Line

Scale

Plotting

Smooth curve

Line drawn

Value of r

21.

.h = ¾ p x 18 x 18x 18

p x 0.04 x 0.04

= 24 x 18x 18x 18

0.04 x 0.04 x 100

= 48,600m

density = ⁴/₃ x ²²/₇ x 18 x 18x 18x 15 kg

1000

= 122.2kg

A1

N of wire

¸ to length in cm

¸ for length

conversing to metres

length

expression for density

conversion to kg

ans

22.

H = Ö15² – 9²

= Ö144

= 12

^X/₆ = ⁹/₁₂

X = 4.5

Volume = ¹/₃ x ²²/₇x (81 x 12 –20.25×6 )

= ²²/₂₁ (972 – 121 -5)

= 891 cm³

A1

Method

Radius

Small vd

Large vol

Subtraction of vol.

Ans

23.

R(-a , b) , Q (c,d), S(x , y) ,P (5,0)

PR is diagonal

(a) Mid point PR (0,0)

a + 5 = 0

.a = -5

b- 0 = 0

2
b = 0

R (-5,0)

Grad PQ = -2

Grad RS = -2

.d – 0 = -2

c –5

.d – 0 = ½

c+5

.d+ 2c = 10

2d – c = 5×2 –

4d – 2c = 10

5d = 20

d = 4

c = 3

Q (3, 4)

x + 3 , y+4 = (0,0)

2 2

x = -3 , y = -4 \ s(-3 -4)

(b) y – 4 = 8

x – 3 6

3y = 8x – 12

A1

Ans .

Expression both correct

Equation

Ans

Expression

Equation

MATHEMATICS IV

PART II

MARKING SCHEME

1.	784 X 27 = 187500 Ö 784 x 9 = 4 x 7x 3 62500 250 = 42 125 = 0.336	M1 M1 A1	Factors for Fraction or equivalent C.A.O
		3
2.	Father 3x , r son = x 2(x +10) = 3x + 10 2x +20 = 3x + 10 x = 10 father = 30	M1 A1 B1	Expression
		3
3.	3 = sin 60AE AE = 3 Sin 60 = 3.464 perimeter = 5×2 + 3.464 x 3 = 10+10.393 = 20.39	M1 A1 B1	Side of a triangle Perimeter
		3
4.	.a³ – b-2c² = (-2)³ – 3 –2(-1)² 2b² – 3a²c 2(3)²–3(-2)²(-1) = -8 –3-2 18 + 12 = -13 30	M1 M1 A1	Substitution Signs C.A.O
		3
5.	Ksh 189,000 = $ 189,000 75.6 = $ 2500 balance = $ 2500 = Kshs. 189,000 Kshs. 189,000 = 189,000 115.8 Uk ₤1632	M1 A1 M1 A1 A1 4	Conversion Conversion
6.	Area of 2 triangles = 2 (½ x 8x 5 sin 60) = 40 sin 60 = 40x 0.8660 = 34.64 cm² Area of rectangle = 300 x 8 + 300 x 5 +300 x BC BC = Ö64 +25 – 2 x 40cos 60 = Ö89 – 80 x 0.5 = Ö89 – 40 = Ö49 = 7 Total S.A. = 300 (8+5+7) + 34.64 cm² = 6000 + 34.64 = 6034.64 cm²	M1 M1 M1 A1	Areas of D B.C. expression Area
		4
7.	AF² = 3²+4²+-2+12x cos 50 = 25 – 24 x 0.6428 = 25-15.43 = 9.57 AF = 3.094 x 50 AF = 154.7m Sin Q = 200 sin 50^o 154.7 = 0.9904 Q = 82.04⁰ Bearing = 117.96⁰	M1 A1 M1 A1 B1	Bearing
		5
8.	(i) No. of white = w w = 2 w+9 3 3w = 2w + 18 w = 18 (ii) p(different colour ) = p(WB N BW) = 2 x 9 + 9 x 18 3 25 27 25 = ¹²/₂₅	M1 A1 M1 A1
		4
9.	A.sf = 1 49 smaller area = 1 x 441 p 49 = 9p pr² = 9p r² = 9 r = 3	M1 M1 A1
		3
10.	Largest area = 22 x (14.5)² 7 = 660.8 cm ² smallest area = ²²/₇ x (13.5)² = 572.8 572.8 £ A £ 660.81	M1 M1 A1
		3
11.	(1 +2 x)⁵ = 1 + 5 (2x) + 10 (2x)² + 10 (2x)³ = 1 + 10x + 40x² + 80x³ 2.045⁵ = 1+2 (0.52)⁵ = 1+10 (0.52)+ 40(0.52)²+80(0.52)³ = 1+5.2 + 10.82 + 11.25 = 28.27	M1 A1 M1 A1
		4
12.	Tn = 5x 2^{n –2} (i) T₁ , T₂, T₃ = 2.5, 5, 10 (ii) S₅ = 2.5(2⁵-1) 2-1 = 2.5 (31) = 77.5	B1 M1 A1	All terms
		3
13.	12 = 2² x 3 18 = 2 x 3² 30 = 2x3x5 Lcm = 22 x 32x 5 = 180 min = 3hrs time they ring together =11.55 +3 = 2.55 p.m	M1 A1 B1
		3
14.	Map area = 40cm ² Actual area = 200x200x40m² = 200x200x40ha 100×100 = 320ha	M1 M1 A1	Area in m² Area in ha CAO
		3
15.	3p + 2r = 13 p + 2r = 9 – 2p = 4 p = sh 2 r = 3.50	M1 A1 B1
		3
16.	110 + 100+130+2x +3x = 540 5x = 200 x = 40⁰ 2x , 3x = 80 and 120⁰ res	M1 A A1 2
17.	Contribution / person = 180,000 X New contribution = 180,000 x – 2 180,000 – 180,000 = 24,000 x –2 x 180,000x – 180,000x +360,000 = 24,000(x-2)x 24,000x² – 48,000x – 360,000 =0 x² – 2x – 15 = 0 x² – 5x + 3x – 15 = 0 x (x – 5)+ 3 (x – 5) = 0 (x + 3 )(x – 5) = 0 x = -3 or = 5 remaining members = 5-2 = 3	B1 B1 M1 M1 A1 M1 A1 B1	‘C’ eqn mult eqn factor both ans remaining members
		8
18.	(a) P (3 white) = 8 x 7 x 6 = 28 13 12 11 143 (b) P(at least 2 blue)=p(WBBorBBWorBWB)orBBB = 8 x 5 x 4 + 5 x 4 x 8 13 12 11 13 12 11 + 5 x 8 x 4 + 8 x 7 x 6 13 12 11 13 12 11 = 204 429 = 68 143 (c) p(2 white and one blue )= p(WWB or WBW or BWW) = 8 x 7 x 5 + 8 x 5 x 7 + 5 x 8 x 7 13 12 11 13 12 11 13 12 11 = 3 x 8 x 7 x 5 13 x 12 x 11 = 70 143	M1 A1 M1 M1 A1 M1 M1 A1
		8
19.	(a) recourt area = 10.5 x 6 m² title area = 0.3 x 0.3 m² No of tiles = 10.5 x 6 0.3 x 0.3 = 700 (b) No of cartons = 700 x 15 20 = 52.5 (c) Cost of 525 cartons = 525 x 100 + 800 x 525 + transport 5 = 10,500+420,000 = 430,500 sale price = 120 x 4.30,500 100 = sh 516,600 s.p of a carton = 516,600 525 = sh. 984	M1 A1 M1 A1 B1 M1 M1 A1
		8
20.	(a) Maina`s tax dues = 1800 x 10 100 = 180 (b) Taxable income = 3600 x 115 – n rent 100 = 36 x 115 – 100 x 12 20 = 4140 – 60 = 4080 Tax dues = 10 x 2100 + 15 x 1980 100 100 = 210 + 297 = 507 Tax relief = 270- Tax paid = 237	M1 A1 M1 A1 M1 M1 A1 B1	1^st slab 2^nd slab
		8
21.	(a) PQ = –³/₅ a + ³/₁b = ³¹/₂ – ³/₅ a (b) OR = h a + h b = a – ha + hb = (1-h) a + h b (c) OR = ³/₅ a + k (³¹/₂ b – ³/₅a) = (³/₅ – ³/₅k)a +3k b (d) 1 – h = ³/₅ – ³/₅k (i) 3k = h (ii) Sub (i) 1 – 3k = ³/₅ – ³/₅k 5- 15k = 3-3k 12k = 2 k = ¹/₆ h = ½	B1 M1 A1 M1 A1 M1 A1 B1
		8
22.	P(ABC) = 0 – 1 1 4 3 = -3 -1 -3 1 0 3 1 3 1 4 3 A¹ (-3,1)B¹ (-1,4)C¹(-3,3) Q(A¹B¹C¹) = a b -3 –1 -3 = -6 –2 –6 c d 1 4 3 2 8 6 => -3a + b = -6 -3c + d = 2 -a + 4b = -2 x 3 -c + 4d = 8 x 3 – 3a + 12b = -6 – 3c + 12d = 24 11b = 0 -11d = -22 b = 0 d = 2 a = 2 d = 2 c = 0 Q = 2 0 0 2	M1 A1 M1 M1 A1 B1 B1 B1	A¹ B¹ C¹ L Q A¹ B¹ C¹ drawn A^ll B^II C^II Ploted Destruction
		8
23. 24.	R = 2.2CM ± 0.1 Area = 22 x 2.2 x 2-2 7 = 15.21cm² Ef =40 efd = -80 (ii) model class = 351- 360 modern class = 341 – 350 (iii) mean = 355.5 – 80 40 = 355.5 – 2 = 353.5	B1 B1 B1 B1 B1 B1 M1 1 1 8 B1 B1 M1 A1 B1 B1 B1 B1
		A1
		8

MATHEMATICS V

PART I

SECTION 1 (52 MARKS)

Use logarithms to evaluate 6 Cos 40 ^0.25
63.4 (4mks)
Solve for x in the equation (x + 3) ²– 5 (x + 3) = 0 (2mks)
In the triangle ABC, AB = C cm. AC = bcm. ÐBAD = 30^o and ÐACD = 25^o. Express BC in terms of b and c. (3mks)
Find the equation of the normal to the curve y = 5 + 3x – x³ when x = 2 in the form
ay + bx = c (4mks)
Quantity P is partly constant and partly varies inversely as the square of q. q= 10 and p = 5 ½ when q =20. Write down the law relating p and q hence find p when qs is 5. (4mks)
Solve the simultaneous equation below in the domain 0 £ x £ 360 and O£ y £ 360
2 Sin x + Cos y = 3
3 Sin x – 2 Cos y = 1 (4mks)
Express as single factor 2 – x + 2 + 1
x + 2 x² + 3x + 2 x + 1 (3mks)
By use of binomial theorem, expand (2 – ½ x )⁵ up to the third term, hence evaluate (1.96)⁵
correct to 4 sf. (4mks)
Points A(1,4) and B (3,0) form the diameter of a circle. Determine the equation of the circle and write it in the form ay² + bx² + cy + dy = p where a, b, c, d and p are constants. (4mks)
The third term of a GP is 2 and the sixth term is 16. Find the sum of the first 5 terms of the GP. (4mks)
Make T the subject of the formulae 1 – 3m + 2
T² R N (3mks)
Vectors, a = 2 b = 2 and c – 6
2 0 4
By expressing a in terms of b and c show that the three vectors are linearly dependent. (3mks)
A cylindrical tank of base radius 2.1 m and height is a quarter full. Water starts flowing into this tank at 8.30 a.m at the rate of 0.5 litres per second. When will the tank fill up? (3mks)
A piece of wood of volume 90cm³ weighs 54g. Calculate the mass in kilograms of 1.2 m³ of the wood. (2mks)
The value of a plot is now Sh 200,000. It has been appreciating at 10% p.a. Find its value 4 years ago.
(3mks)
12 men working 8 hours a day take 10 days to pack 25 cartons. For how many hours should 8 men be

working in a day to pack 20 cartons in 18 days? (2mks)

SECTION II (48MARKS)

The tax slab given below was applicable in Kenya in 1990.
Income in p.a.                           rate in sh
1 – 1980                                  2
1981 – 3960                              3
3961 – 5940                              5
5941 – 7920                              7
Maina earns Sh. 8100 per month and a house allowance of Sh. 2400. He is entitled to a tax relief of Sh.

800 p.m. He pays service charge of Sh 150 and contributes Sh 730 to welfare. Calculate Mwangis net

salary per month. (8mks)

OAB is a triangle with OA = a , OB = b. R is a point of AB. 2AR = RB. P is on OB such that
3OP = 2PB. OR and AP intersect at Y, OY = m OR and AY = nAP. Where m and n are scalars.    Express in terms of a and b.
(i) OR                                                                                                                    (1mk)
(ii)AP                                                                                                                    (1mk)(b) Find the ratio in which Y divides AP                                                                (6mks)
The table below gives related values of x and y for the equation y = axⁿ where a and n are constants

X	0.5	1	2	3		10
Y	2	8	32		200	800

By plotting a suitable straight line graph on the graph provided, determine the values of a and n.

20. Chalk box x has 2 red and 3 blue chalk pieces. Box Y has same number of red and blue

pieces. A teacher picks 2 pieces from each box. What is the probability that
(a)        They are of the same colour.                                                                            (4mks)
(b)        At least one is blue                                                                                           (2mks)
(c)        At most 2 are red                                                                                              (2mks)

21. Point P(50^oN, 10^oW) are on the earth’s surface. A plane flies from P due east on a parallel of

latitude for 6 hours at 300 knots to port Q.
(a) Determine the position of Q to the nearest degree. (3mks)
(b) If the time at Q when the plane lands is 11.20am what time is it in P. (2mks)
(c) The plane leaves Q at the same speed and flies due north for 9 hours along a longitude to

airport R. Determine the position of R.                                                                       (3mks)
22.       Using a ruler a pair of compasses only, construct :
(a)        Triangle ABC in which AB = 6cm, AC = 4cm and Ð ABC = 37.5^o.                                (3mks)
(b)        Construct a circle which passes through C and has line AB as tangent to the circle at A.             (3mks)
(c)        One side of AB opposite to C, construct the locus of point P such that ÐAPB = 90^o.             (2mks)
23.       A particle moves in a straight line and its distance is given by S = 10t² – t³ + 8t where S is

distance in metres at time t in seconds.
Calculate:
(i) Maximum velocity of the motion.                                                                             (4mks)
(ii) The acceleration when t = 3 sec.                                                                              (2mks)
(iii) The time when acceleration is zero.                                                                                   (2mks)

A rectangle ABCD has vertices A(1,1) B(3,1), C(3,2) and D(1,2). Under transformation

matrix M = 2 2 ABCD is mapped onto A¹B¹C¹D¹

1 3
under transformation M = -1 0 A¹B¹C¹D¹ is mapped onto A¹¹B¹¹C¹¹D¹¹. Draw on the given grid
0 –2

(a)       ABCD, A¹B¹C¹D¹ and A¹¹B¹¹C¹¹D¹¹                                                                  (4mks)
(b)        If area of ABCD is 8 square units, find area of A¹¹B¹¹C¹¹D¹¹.                              (3mks)
(c)        What single transformation matrix maps A¹¹B¹¹C¹¹D¹¹ onto A¹B¹C¹D¹               (1mk)

MATHEMATICS V

PART II

SECTION 1 (52 Marks)

Evaluate without using mathematical tables (2.744 x 15 ⁵/₈)¹/₃ (3mks)
If 4 £ x £ 10 and 6 £ y £5, calculate the difference between highest and least
(i) xy (2mks)
(ii) ^y/_x (2mks)
A 0.21 m pendulum bob swings in such a way that it is 4cm higher at the top of the swing than at the bottom. Find the length of the arc it forms. (4mks)
Matrix 1 2x has on inverse, determine x (3mks)
x +3 x²
The school globe has radius of 28cm. An insect crawls along a latitude towards the east from A(50^o, 155^oE) to a point B 8cm away. Determine the position of B to the nearest degree. (4mks)
The diagonals of triangle ABCD intersect at M. AM = BM and CM = DM. Prove that triangles ABM and CDM are Similar. (3mks)
Given that tan x = ⁵/₁₂, find the value of 1 – sinx
Sin x + 2Cos x, for 0 £ x £ 90 (3mks)

Estimate by MID ORDINATE rule the area bounded by the curve y = x² + 2, the x axis and the lines x = O and x = 5 taking intervals of 1 unit in the x. (3mks)
MTX is tangent to the circle at T. AT is parallel to BC. Ð MTC = 55^o and Ð XTA = 62^o. Calculate Ð (3mks)
Clothing index for the years 1994 to 1998 is given below.

Year	1994	1995	1996	1997	1998
Index	125	150	175	185	200

Calculate clothing index using 1995 as base year. (4mks)

A²digit number is such that the tens digit exceeds the unit by two . If the digits are reversed, the number formed is smaller than the original by 18. Find the original number. (4mks)
Without using logarithm tables, evaluate log₅ (2x-1) –2 + log₅ 4 = log₅ 20 (3mks)
Mumia’s sugar costs Sh 52 per kg while imported sugar costs Sh. 40 per kg. In what ratio should I mix the sugar, so that a kilogram sold at Sh. 49.50 gives a profit of 10%. (4mks)
The interior angles of a regular polygon are each 172^o. Find the number of sides y lie polygon. (2mks)
Evaluate 2x = 2 + 3
341 9.222 (2mks)
A water current of 20 knots is flowing towards 060^o. A ship captain from port A intends to go to port

B at a final speed of 40 knots. If to achieve his own aim, he has to steer his ship at a course of 350^o.

Find the bearing of A from B. (3mks)

SECTION II (48 MARKS)

3 taps, A, B and C can each fill a tank in 50 hrs, 25 hours and 20 hours respectively. The three taps are turned on at 7.30 a.m when the tank is empty for 6 hrs then C is turned off. Tap A is turned off after four hours and 10 minutes, later. When will tap B fill the tank? (8mks)
In the domain –5 £ x £ 4, draw the graph of y = x² + x – 8. On the same axis, draw the graph of y + 2x = -2. Write down the values of x where the two graphs intersect. Write down an equation in x whose roots are the points of intersection of the above graphs. Use your graph to solve. 2x² + 3x – 6 = 0. (8mks)
The average weight of school girls was tabulated as below:

Weight in Kg	30 – 34	35 – 39	40 – 44	45 – 49	50 – 54	55-59	60-64
No. of Girls	4	10	8	11	8	6	3

(a) State the modal class.                                                                                           (1mk)
(b) Using an assumed mean of 47,
(i) Estimate the mean weight                                                                                (3mks)
(ii) Calculate the standard deviation.                                                                      (4mks)

The table below shows values of y = a Cos (x – 15) and y = b sin (x + 30)

X	0	15	30	45	60	75	90	105	120	135	150
a Cos(x-5)	0.97				0.71	0.5				-0.5	-0.71
b sin(x+3)	1.00				2.00				1.00		0.00

(a) Determine the values of a and b                                                                               (2mks)
(b) Complete the table                                                                                                  (2mks)
(c) On the same axes draw the graphs of y = across(x – 15) and y = b sin(x + 30)            (3mks)
(d) Use your graph to solve ½ cos (x – 15) = sin(x + 30)                                                 (1mk)

21. The diagram below is a clothing workshop. Ð ECJ = 30^o AD, BC, HE, GF are vertical

walls. ABHG is horizontal floor. AB = 50m, BH = 20m, AD=3m

(a) Calculate DE                                                                                                           (3mks)
(b) The angle line BF makes with plane ABHG                                                              (2mks)
(c) If one person requires minimum 6m³ of air, how many people can fit in the workshop         (3mks)

To transport 100 people and 3500 kg to a wedding a company has type A vehicles which take 10 people and 200kg each and type B which take 6 people and 300kg each. They must not use more

than 16 vehicles all together.
(a) Write down 3 inequalities in A and B which are the number of vehicles used and plot them

in a graph. (3mks)
(b) What is the smallest number of vehicles he could use. (2mks)

hire charge for the whole function (3mks)

A circle centre A has radius 8cm and circle centre B has radius 3cm. The two centres are

12cm apart. A thin tight string is tied all round the circles to form interior common tangent. The tangents CD and EF intersect at X.

(a) Calculate AX (2mks)
(b) Calculate the length of the string which goes all round the circles and forms the tangent.
(6mks)

Airport A is 600km away form airport B and on a bearing of 330^o. Wind is blowing at a speed of

40km/h from 200^o. A pilot navigates his plane at an air speed of 200km/h from B to A.
(a)     Calculate the actual speed of the plane.                                                                (3mks)
(b)     What course does the pilot take to reach B?                                                          (3mks)
(c)     How long does the whole journey take?                                                                (2mks)

MATHEMATICS V

PART I

MARKING SCHEME

1	SOLUTION	MKS	AWARDING
	No Log 13.6 1.1335 + Cos 40 1.8842 1.0177 – 63.4 1.8021 1. 2156 (4 + 3.2156) ¹/₄ 1.8039 Antilog 0.6366	B1 M1 M1 A1	Log + divide by 4 C.A.O
		4
2.	(x + 3) (x + 3 – 5) = 0 (x +3)b (x – 2) = 0 x = -3 or x = 2	M1 A1	Factors Both answers
3	BD = C Sin 30 = 0.05 CD = b Cos 25 = 0.9063b BC = 0.9063b + 0.5 C	B1 B1 B1	BD in ratio from CD in ratio form Addition
		3
4	Dy = 3 – 3x² dx x = 2, grad = 1 9 Point (2,3) y – 3 = 1 x – 2 99y – 27 = x – 2 9y – x = 25	B1 B1 M1 A1	Grad equ Grad of normal Eqn Eqn
		4
5	700 = 100 + n 2200 = 400 + n1500 = 300m m = 5 n = 200 P = 5 + 200 q2 When q = 5 P = 13	M1 A1 B1 B2	Equan Both ans Eqn (law) Ans (P)
		4
6	4 Sin x + 2 cos y = 6 3 Sin x – 2 Cos y = 1 7 sin x = 7 Sin x = 1 X = 90 Cos y = 1 Y = 0^o	M1 M1 A1 B1	Elim Sub
7	2(x +1) – 1(x + 2) + x + 2 (x+2) (x +1) = 2x +2 – x – 2 + x = 2 (x +2) (x + 1) = 2x + 2 (x + 2) (x + 1) = 2 x + 2	M1 M1 A1	Use of ccm Substitution Ans
8	(-2 – ½ x)5 = 2⁵ – 5 (2)⁴ ( ½ x) + 10(2)³( ½ x)² = 32 – 40x + 20x² = 32 – 4 (0.08) + 20 (0.08)² = 32 – 0.32 + 0.128 = 3	M1 A1 M1 A1
		4
9.	Circle centre C = (3 +1, 0 + 4) 2 2 C( 2, 2) R =Ö (2 – 0)² + (2 – 3)² =Ö 5 (y – 2)² + (x – 2)² = Ö5 y² + x²– 4y – 4x = 8 + Ö5	B1 B1 M1 A1	Centre Radius
		4
10	ar² =2, ar⁵ = 16 a = 2 \ 2 r⁵ = 16 r² r² 2r³ = 16 r³ = 8 r = 2, a = ½ S₅= ½ (1 – ( ½ )⁵) ½ = 1 – ¹/₃₂ = ³¹/₃₂	M1 A1 M1 A1	Both Sub CAO
		4
11	NR – 3MT² = 2RT² T²(2R + 3M) = NR T2 = NR 2R + 3m T = ! Ö NR 2R + 3m	M1 M1 A1	X mult 7² ans
		3
12	2 = m 2 + n 6 2 0 4 2 = 2m + 6n 2 = 0 + 4n n = ½ m = – ½ \a = – ½ b + ½ c \a b c are linearly dep	M1 A1 B1
		3
13	Volume = 22 x 2.1 x 2.1 x 2 x ¾ m³ 7 Time = 11 x 0.3 x 2.1 x 3 x 1,000,000 500 x 3600 = 11.55 = 11.33 hrs time to fill = 8.03 pm	M1 M1 A1
		3
14	Mass = 54 x 1.2 x 1,000,000 90 1000 = 720kg	M1 A1
		2
15	V₃ = P P(0.9)³ = 200,000 P = 200,000 0.9³ = 200,000 0.729 = Sh 274,348	M1 M1 A3
		3
16	No of hours = 8 x 12 x10 x 20 8 x 18 x 25 = 19200 3600 = 5hrs, 20 min	M1 A1
		2
17	Taxable income = 8100 + 2400 = sh. 10,500 = ₤6300 Tax dues = Sh 1980 x 2 + 1980 x 3 + 1980 x 5 + 3670 x 7 12 = 22320 12 = Sh 1860 net tax = 1860 – 800 p.m. = Sh 1060 Total deduction = 1060 + 150 + 730 = 1940 Net salary = 10,500 – 1940 = Sh 8560 p.m.	B1 M1 M1 A1 B1 B1 M1 A1	Tax inc 2 2 net tax total dedu.
		8
18	OR = ²/₃ a + ¹/₃b or (¹/₃ (2a + b) AP = ²/₅ b – a OY = m OR = A + n (²/₅b – a) ²/₅m b + ma = (1 – n)a + ²/₅ n b ²/₅m = ²/₅n m = n \m = 1 – m 2m = 1 m = ½ = n ½ AP = Ay AY:AP = 1:1	B1 B1 B1 M1 M1 A1 A1 B1	EXP, OY Eqn M = n Sub CAO Ratio
		8
19	Log y = n log x + log a Log a = 0.9031 A = 8 Grad = 1.75 – 0.5 0.4 + 0.2 = 1.25 0.6 = 2.08 n = 2 \y = 8x² x = 3 y = 8 x 3² = 72 y = 200 x = 5	B1 B1 B1 B1 B1 S1 P1 L1	Log x Log y A N Missing x and y Scale Points Line
		8
20	P (same colour) = P (XRRrr orXBB or YXX or YBB) = ½ (²/₅ x ¼ + ³/₅ x ²/₄) x 2 = 2 + 6 20 20 = 8 20 = ²/₅ (b) P(at least 1B) = 1 – P(non blue) = 1 – P (XRR or YRR) = 1 – ½ (²/₅ x ^¼) x 2 = 1 – ¹/₁₀ = ⁹/₁₀ (c) P(at most 2 Red) = 1 – P (BB) = 1 – ½ (³/₅ x ²/₄)² = 1 – ⁶/₂₀ = ¹⁴/₂₀ or ⁷/₁₀	M1 M1 M1 A1 M1 A1 M1 A1	Any 2 Any 2 Fraction
		8
21	(a) PQ = 1800nm q = 1800 60 x 0.6428 = 46.67 = 47^o Q (50^oN, 37^oE) (b) Time diff = 47 x 4 60 = 3.08 Time at P = 9.12am (c) QR = 2700 nm x ^o = 2700 60 = 45^o R (85^oN, 133^oW)	M1 A1 M1 A1 M1 A1 B1
		8
22		B1 B1 B1 B1 B1 B1 B1 B1	Bisector of 150 Bisector 75 AB AC ^ at A Bisector AC Circle Ð AB Locus P with A B excluded
		8
24	A¹B¹ C¹D¹ 2 2 1 3 3 1 = 4 8 10 6 1 3 1 1 2 2 4 6 9 7 A¹¹ B¹¹ C¹¹ D¹¹ -1 0 4 8 10 6 = -4 –8 -10 -6 0 –2 4 6 9 7 -8 -12 -18 -14 NM = -1 0 2 2 0 –2 1 3 = -2 -2 -2 -6 (b) det = Asf = 12 – 4 = 8 Area A¹¹ B¹¹ C¹¹ D¹¹ = 8 x 8 = 64 U² (c) Single matrix = Inv N = ½ -2 – 0 0 –1 = -1 0 0 – ½	B1 B1 B1 M1 A1 B1	Product Product Det Inverse
		6
23	Ds = 20t – 3t² + 8 =0 Dt 3t²– 20t – 8 = 0 T = 20 ! Ö400 + 4 x 3 x 8 6 t = 7.045 sec max vel = 148.9 – 140.9 – 8 = 0.9 m/s d² s = 6t – 20 dt² when t = 3 a = -2m/s² 6t – 20 = 0 6t = 20 t = 3 ²/₃ sec	M1 A1 M1 A1 M1 A1 M1 A1
		8

MATHEMATICS V

PART II

MARKING SCHEME

No	Solution	Mks	Awarding
1	2744 x 125 ¹/₃ 1000 8 2744 ¹/₃ x 5^{3 1}/₃ 1000 2³ 2³ x 7³ ¹/₃ x 5 103 2 2 x 7 x 5 = 3.5 10 2	M1 M1 A1	Factor Cube root
		3
2	(i) Highest – 10 x 7.5 = 75 Lowest – 6 x 4 = 24 – 51 (ii) Highest = 7.5 = 1.875 4 Lowest = 6 = 0.600 10 1.275	M1 A1 M1 A1	Highest Fraction
		4
3	Cos q = 17 = 0.8095 21 q = Cos 0.8095 = 36.03^o Arc length = 72. 06 x 2 x 22 x 21 360 7 = 26.422cm	M1 A1 M1 A1	q
		4
4	x² – 2x(x +3) = 0 x² – 2x² – 6x = 0 -x² – 6x = 0 either x = 0 or x = 6	M1 M1 A1	Equ Factor Both A
		3
5	8 = x x 2 x 22 x 28 Cos 60^o 360 7 8 = x x 44 x 28 x 0.5 360 7 x = 8 x 360 x 7 44 x 28 x 0.5 = 32.73^o = 33^o	M1 M1 A1 B1	x exp
6	ÐDMC = Ð AMB vert. Opp = q ÐMAB = Ð MDC = 180 – q BASE Ls of an isosc. < 2 Ð MBA = Ð MAC 180 – q base angles of isos < 2 <’s AMC and < CDM are equiangle \ Similar proved	B1 B1 B1
		3
7	Tan x = ⁵/₁₂ h = Ö b² + 12² = Ö25 + 144 = Ö169 = 13 1 – Sinx = 1 – 5 sin x + 2 Cos x ⁵/₁₃ + 2 x ¹²/₁₃ ¹²/₁₃ = 12 x 13 = 12 ²⁹/₁₃ 13 29 29	M1 M1 A1	Hypo Sub
		3
8	Y = x ² + 2 Area = h (y₁, = y₂ +……..y_n) = 1(2.225 + 4.25 + 8.25 +14.25 + 22.25) = 51.25 sq units	B1 M1 A1	Ordinals
		3
9	ÐCBA = 117^o Ð ACD = 55 Ð BAC = 180 – (117 + 55) = 8^o	B1 B1 B1 3
10		B1 B1 B1 B1	1994 1996 1997 1998
		4
11.	Xy = 35 y = ³⁵/_x 9x – 9y = -18 Sub x² + 2x – 35 = 0 x² + 7x – 5x – 35 = 0 x (x + 7) – 5(x + 7) = 0 (x – 5) (x + 7) = 0 x = -7 x = +5 y = 7 Smaller No. = 57 = 75	B1 M1 A1 B1
		3
12	Log₅ (2x – 1 )4 = log₅5² 20 4(2x – 1) = 5² 20 2x – 1 = 25 5 2x – 1 = 125 2x = 126 x = 63	M1 M1 A1
		3
13	C.P = 100 x 49.50110 = 45/- 52x + 40y = 45 x + y 45x + 45y = 52x + 40 -7x = -54 ^x/_y = ⁵/₇ x : y = 5 : 7	B1 M1 M1 A1
		4
14	2n – 4 it angle = 172 n (2n – 4) x 90 = 172n n 90 (2n – 4) x 90 = 172 n 180 n – 360 = 172n 180n – 172n = 360 8n = 360 n = 45	M1 A1 M1
		2
15	2 x = 2. 1 + 3. 1 6.341 9.22 2x = 2 x 0. 1578 + 3 x 0.1085 = 0.3154 + 0.3254 = 0.6408 x = 0.3204	B1 A1	Tables
		2
16	Bearing 140^o Sin q = 20 Sin 110 40 = 0.4698 = 228.02 Bearing of A from B = 198.42	M1 A1 B1
		3
17	Points that each tap fills in one hour A = 1 B = 1 C – 1 50 25 20 In one hour all taps can fill = 1 + 1 + 1 = 11 50 25 20 100 In 6hrs all can fill = 11 x 6 = 33 parts 100 50 taps A and B can fill = = 1 + 1 = 3 part in 1 hr 50 25 50 In 4 1 hrs, A and B = 25 x 3 + 1 6 6 50 4 Parts remaining for B to fill = 1 – 33 + 1 = 1 – 91 = 9 parts 50 4 100 100 Time taken = 9 x 25 hrs = 2 ¼ hrs 100 1 7.30 am 6. hrs 13.30 4.10 5.40pm 2.15 7.55 pm	M1 B1 B1 B1 M1 A1
18	x² + x – 8 = -2 – 2x y = x² + 3x – 6 Points of intersection (-4, 1.4) y = x² + x – 8 = 2x² + 3x – 6 x² + 2x + 2 y = x² + x – 8 x 2 2y = 2x² + 2x – 16 0 = 2x² + 3x – 6 2y = -x – 10 y = – 2.6 Ny = 1.2	8 B1 B1 B1 B1	Eqn Point of inter Line eqn Both
19	(a) Modal class = 45 – 49 (i) Mean = 47 + -55 50 = 47 – 1.1 = 45.9 (ii) Standard deviation = Ö 3575 – –55 ² 50 50 = Ö71.5 – 1.21 =Ö 70.29 = 8.3839	4 B1 B1 M1 A1 B1	fd fd²
		8
20	(a) a = 1 b = 2 ½ cos (x – 15) = Sin (x + 30) has no solution in the domain	B1 B1 B1 B1	All All A & b
		8
21	(a) O Cos 30 = 20 X X = 20 0.866 = 23.09 DE = Ö 50² + 23.09² = Ö 2500 + 533.36 = Ö 3033.36 = 55.076m (b) GB = Ö 20² + 50² = 53.85 Tan q = 14.55 53.85 = 0.27019 q = 15.12^o	B1 M1 A1 M1 A1
		8
	(c) Volume of air = 50 x 20 x 3 + ½ x 20 x 11.55 x 50 = 3000 + 5775 = 8775 No. of people = 8775 6 = 1462.5 j 1462	M1 M1 A1
		8
22	(a) A + B [ 16 5A + 3B ³ 50 2A + 3B [ 35 (b) 14 vehicles (c) A – 6 vehicles B – 8 Cost = 6 x 1000 + 8 x 1200 = 6000 + 9600 = 15,600/=	B1 B1 B1 M1 A1	In equation 3 Vehicles
		8
23	x = 8 12 – x 3 = 8.727 FBX = 3 = 0.9166 = 23.57 3.273 3FBX = 47.13 Reflex Ð FBD = 312.87 Reflex arc FD = 312.87 x 22 x 6 360 7 = 16.39cm Reflex Arc CE = 312.87 x 22 x 16 360 7 = 43.7cm FE (tangent) = Ö144 – 121 = Ö 23 = 4.796cm 2 FE = 9.592 Total length = 9.592 + 4.796 + 43.7 + 16.39 = 74.48 cm²	M1 A1 M1 A1 M1 A1 M1 A1
		8
24	(a) 200 = 40 Sin 50 Sin q Sin q = 40Sin 50 200 = 0.7660 5 =0.1532 q = 8.81^o Ð ACB = 180 – (50 + 8.81)^o = 121.19^o x = 200 Sin 121.19 Sin 50 x = 200 x Sin 121.19 Sin 50 = 200 x 0.855645 0.7660 = 223.36Km/h (b) Course = 330^o – 8.81^o = 321.19^o (c) Time = 600 321.19^o = 2.686 hrs	M1 A1 M1 M1 A1 B1 M1 A1 8

MATHEMATICS VI

PART I

SECTION I (52 MARKS)

Evaluate without mathematical tables leaving your answer in standard form

0.0171² X 3

855 X 0.531 (2 Mks)

Six men take 14 days working 8 hours a day to pack 2240 parcels. How many more men working

5 hours a day will be required to pack 2500 parcels in 2 days (3 Mks)

M In quadrilateral OABC, OA = 4i – 3j. OC = 2i + 7j

AB = 3OC. cm: mB = 2:3. Find in terms of i and j

C vector Om (3 Mks)

O A

By matrix method, solve the equations

5x + 5y = 1

4y + 3x = 5 (3 Mks)

In the given circle centre O, ÐABC = 126⁰.

Calculate ÐOAC (3 Mks)

A C

Solve the equation

2(3x – 1)²_– 9 (3x – 1) + 7 = 0 (4 Mks)

Maina, Kamau and Omondi share Shs.180 such that for every one shilling Maina gets, Kamau gets 50

Cts and for every two shillings Kamau gets, Omondi gets three shillings. By how much does Maina’s

share exceed Omondi’s (3 Mks)

Expand (2 + ¹/₂x)⁶ to the third term. Use your expression to evaluate 2.4⁶ correct to 3 s.f (3 Mks)
The probability of failing an examination is 0.35 at any attempt. Find the probability that

(i) You will fail in two attempts (1 Mk)

(ii) In three attempts, you will at least fail once (3 Mks)

Line y = mx + c makes an angle of 135⁰with the x axis and cuts the y axis at y = 5. Calculate the

equation of the line (2 Mks)

During a rainfall of 25mm, how many litres collect on 2 hectares? (3 Mks)
Solve the equation a – 3a – 7 = a – 2 (3 Mks)

3 5 6

The sum of the first 13 terms of an arithmetic progression is 13 and the sum of the first 5 terms is

–25. Find the sum of the first 21 terms (5 Mks)

The curved surface of a core is made from the shaded sector on the circle. Calculate the height of

the cone. (4 Mks)

20cm 125⁰ 20 cm

Simplify (wx – xy – wz + yz) (w + z) (3 Mks)

z² – w²

The bearing of Q from P is North and they are 4 km apart. R is on a bearing of 030 from P and on

a bearing of 055 from Q. Calculate the distance between P and R. (3 Mks)

SECTION II (48 MARKS)

In the given circle centre O, ÐQTP = 46⁰, ÐRQT = 74⁰ and ÐURT = 39⁰

U T P

S 39⁰

Calculate R

(a) ÐRST (1 Mk)

(b) ÐSUT (3 Mks)

(d) ÐPST (2 Mks)

The exchange rate on March 17th 2000, was as follows: –

1 US$ = Kshs.74.75

1 French Franc (Fr) = Kshs.11.04

A Kenyan tourist had Kshs.350,000 and decided to proceed to America

(a) How much in dollars did he receive from his Kshs.350,000 in 4 s.f? (2 Mks)

(b) The tourist spend ¼ of the amount in America and proceeded to France where he spend Fr

16,200. Calculate his balance in French Francs to 4 s.f (3 Mks)

Kshs. does he receive for his balance than he would have got the day he left? (3 Mks)

On the provided grid, draw the graph of y = 5 + 2x – 3x² in the domain -2 £ x £ 3 (4 Mks)

(a) Draw a line through points (0,2) and (1,0) and extend it to intersect with curve y = 5 + 2x – 3 x 2

read the values of x where the curve intersects with the line (2 Mks)

(b) Find the equation whose solution is the values of x in (a) above (2 Mks)

(a) Using a ruler and compass only, construct triangle PQR in which PQ = 3.5 cm, QR = 7 cm

and angle PQR = 30⁰ (2 Mks)

(b) Construct a circle passing through points P, Q and R (2 Mks)

The given Region below (unshaded R) is defined by a set of inequalities. Determine the inequalities (8 Mks)

2 R (3,3)

-3 5

The table below shows the mass of 60 women working in hotels

Mass (Kg)	60 – 64	65 – 69	70 – 74	75 – 79	80 – 84	85 – 89
No. of women	8	14	18	15	3	2

(a) State (i) The modal class (1 Mk)

(ii) The median class (1 Mk)

(b) Estimate the mean mark (4 Mks)

XY, YZ and XZ are tangents to the circle centre O

at points A, B, C respectively. XY = 10 cm,

YZ = 8 cm and XZ = 12 cm. (2 MKS)

.. B

A Y

(a) Calculate, length XA (2 Mks)

(b) The shaded area (6 Mks)

Maina bought a car at Kshs.650,000. The value depreciated annually at 15%

(a) After how long to the nearest 1 decimal place will the value of the car be Kshs.130,000 (4 Mks)

(b) Calculate the rate of depreciation to the nearest one decimal place which would make the value of

the car be half of its original value in 5 years (4 Mks)

MATHEMATICS VI

PART II

SECTION 1 (52 MARKS)

Simplify 32a¹⁰ ^-2/₅ ÷ 9b⁴ ^11/₂

b¹⁵ 4a⁶ (2 Mks)

Use logarithm tables to evaluate

Ö0.375 cos 75

tan 85.6 (4 Mks)

The marked price of a shirt is Shs.600. If the shopkeeper gives a discount of 20% off the marked price, he makes a loss of 4%. What was the cost of the shirt? (3 Mks)
The surface area (A) of a closed cylinder is given by A = 2pr² + 2prh where r is radius and h is height of the cylinder. Make r the subject. (4 mks)
In the circle centre O, chords AB and CD intersect at X. XD = 5 cm

XC = 1/4 r where r is radius. AX:XO = 1:2 Calculate radius of the circle. (3 mks)

A 5cm D

C O

Simplify 2 – 1 (3 mks)

5 – 2Ö3 5 + 2Ö3

P is partly constant and partly varies as q2. When q = 2, P = 6 and when q = 3, P = 16. Find q when P = 64 (4 mks)
The figure on the side is a tent of uniform cross-section A F

ABC. AC = 8m, BC = 8m, BD = 10m and (ACB = 120⁰. 8m

If a scout needs 2.5 m³ of air, how many scouts can fit 120^o C E

in the tent. 8m (4 mks)

B D

10m

The length of a rectangle is given as 8 cm and its width given as 5 cm. Calculate its maximum % error in its perimeter (3 mks)
ABCD is a rectangle with AB = 6 cm, BC = 4 cm AE = DH = 4 cm BF = CG = 12 cm. Draw a

labelled net of the figure and show the dimensions of the net

Expand (1 + 2x)⁶ to the 3rd term hence evaluate (1.04)⁶ (4 mks)
The eye of a scout is 1.5m above a horizontal ground. He observes the top of a flag post at an

angle of elevation of 20⁰. After walking 10m towards the bottom of the flag post, the top is observed at angle of elevation of 40⁰. Calculate the height of the flag post (4 mks)

A bottle of juice contains 405ml while a similar one contains 960ml. If the base area of the

larger Container is 120 cm². Calculate base area of the smaller container. (3 mks)

It takes a 900m long train 2 minutes to completely overtake an 1100m long train travelling at

30km per hour. Calculate the speed of the overtaking train (3 mks)

Okoth traveled 22 km in 2³/₄ hours. Part of the journey was at 16 km/h and the rest at 5 km/h.

Determine the distance at the faster speed (3 mks)

P and Q are points on AB such that AP:PB = 2:7 and AQ:QB = 5:4 If AB = 12 cm, find PQ

(2 Mks)

SECTION B (48 MARKS)

The income tax in 1995 was collected as follows:

Income in Kshs. p.a rate of tax %

1 – 39,600 10

39,601 – 79,200 15

79,201 – 118,800 25

118,801 – 158,400 35

158,401 – 198,000 45

Mutua earns a salary of Kshs.8,000. He is housed by the employer and therefore 15% is added to his salary to arrive at its taxable income. He gets a tax relief of Shs.400 and pay Shs.130 service charge. Calculate his net income (8 Mks)

The probability Kioko solves correctly the first sum in a quiz is ²/₅ Solving the second correct

is ³/₅ if the first is correct and it is ⁴/₅ if the first was wrong. The chance of the third correct is

²/₅ if the second was correct and it is ¹/₅ if the second was wrong. Find the probability that

(a) All the three are correct (2 Mks)

(b) Two out of three are correct (3 Mks)

A businessman bought pens at Shs.440. The following day he bought 3 pens at Shs.54. This

purchase reduced his average cost per pen by Sh.1.50. Calculate the number of pens bought earlier and the difference in cost of the total purchase at the two prices (8 mks)

In D OAB, OA = a, OB = b

OPAQ is a parallelogram.

ON:NB = 5:-2, AP:PB = 1:3

Determine in terms of a and b vectors

(a) OP (2 Mks)

(b) PQ (2 Mks)

(d) PN (2 mks)

A cylindrical tank connected to a cylindrical pipe of diameter 3.5cm has water flowing at 150

cm per second. If the water flows for 10 hours a day

(a) Calculate the volume in M³ added in 2 days (4 ms)

(b) If the tank has a height of 8 m and it takes 15 days to fill the tank, calculate the base radius

of the tank (4 mks)

A joint harambee was held for two schools that share a sponsor. School A needed Shs.15 million while

School B needed 24 million to complete their projects. The sponsor raised Shs.16.9 million while other

guest raised Shs.13.5 million.

(a) If it was decided that the sponsor’s money be shared according to the needs of the school

with the rest equally, how much does each school get (5 mks)

(b) If the sponsor’s money was shared according to the schools needs while the rest was in the ratio of

students, how much does each school get if school A has 780 students and school B 220

students (3 mks)

Voltage V and resistance E of an electric current are said to be related by a law of the form

V = KEⁿ where k and n are constants. The table below shows values of V and E

V	0.35	0.49	0.72	0.98	1.11
E	0.45	0.61	0.89	1.17	1.35

By drawing a suitable linear graph, determine values of k and n hence V when E = 0.75(8mks)

The vertices of triangle P,Q,R are P(-3,1), Q (-1,-2), R (-2,-4)

(a) Draw triangle PQR and its image P^IQ^IR^I of PQR under translation T = 3 on the provided grid 4 (2 Mks)

(b) Under transformation matrix m = 4 3 , P^IQ^IR^I is mapped on to P^IIQ^IIR^II. Find the

co-ordinates of P^IIQ^IIR^II and plot it 1 2 on the given grid (4 Mks)

MATHEMATICS VI

PART 1

MARKING SCHEME

171 X 171 X 3 X 10^-5 M1

855 X 531

= 2 X 10^-6 A1

No. of men = 6 X 14 X 8 X 2500 M1

2 X 5 X 2240

= 75 A1

Extra men = 75 – 6 = 69 B1

OM = 2i + 7j + 2/5 (4i – 3j + 6i + 21j – 2i – 7j) M1

= 2i + 7j + 2/5 (8i + 11j) M1

= 26 i + 57 j

5 5 A1

2 5 x = 1

3 4 y 5 M1

x ^-1/₇ ⁵/₇ 1

y = ³/₇ ^-2/₇ 5 M1

x = 3

y -1

x, 3, y = -1 A1

Reflex ÐAOC = 126 x 2 = 252⁰ B1

Obtuse ÐAOC = 360 – 252 = 108⁰ B1

= ¹/₂ (180 – 108)⁰

= 36⁰ B1

18x² – 39x + 18 = 0

6x² – 13x + 6 = 0 B1Ö equation

6x² – 9x – 4x + 6 = 0

3x(2x – 3) (3x – 2) = 0 M1

x = ²/₃ or A1

x =1 ½ B1

M : K : O = 4 : 2 : 3 B1Ö ratio

Maina’s = 4/9 X 180

= 80/- B1Ö Omondi’s

Omondi’s = 60/- and Maina’s

Difference = Shs.20/- B1 difference

(2 + ¹/₂x)⁶ = 2⁶ + 6(2⁵) (¹/₂x + 15 (2⁴) (¹/₂ x)² M1

= 64 + 96x + 60x² A1

2.4⁶ = (2 + ¹/₂ (0.8))⁶

= 64 + 96 (0.8) + 60 (0.64) M1

= 179.2

@179 to 3 s.f A1

P (FF) = ⁷/₂₀ X ⁷/₂₀

= ⁴⁹/₁₀₀ B1

P (at least one fail) = 1 – P (F^IF^I F^I)

= 1- ¹³/₂₀³ M1

= 1 – 2197 M1

8000

= 5803

8000 A1

grad = term 135

= -1 B1

y = mx + c

y = -x + 5 B1

Volume = 2 x 10,000 x 10,000 x 25 M1Ö x section area

1000 10 M1Ö conv. to litres

= 500,000 Lts A1

10a – 6(3a – 7) = 5(a -2) M1

10a – 18a + 42 = 5a – 10

– 13a = -52 M1

a = 4 A1

2a + 12d = 2

2a + 4d = -10 M1

8d = 12

d = 1¹/₂ A1

a = -8 B1

S₂₁ = ²¹/₂ (-16 + 20 X 3/2) M1

= 147 A1

2 p r = 120 x p x 40 M1

360

r = 6.667 cm A1

h = Ö 400 – 44.44 M1

= 18.86 cm A1

= (w (x – z) – y (x – z)) (w + z) M1Ö factor

(z – w) (z + w)

= (w – y) (x – z) (w + z) M1Ö grouping

(z – w) (z + w)

= (w – y) (x – z)

z – w A1

25⁰ B1Ö sketch

55⁰

Q 125 PR = 4 sin 125 M1

Sin 25

P 3

(a) <RST = 180⁰ – 74⁰ = 106⁰ B1

(b) < RTQ = 90⁰– 74⁰ = 16⁰ B1

< PTR = 46⁰ + 16⁰ = 62⁰ B1

< SUT = 62⁰ – 39⁰ = 23⁰ B1

= 180 – 32 = 148⁰ B1

Obtuse ROT = 360 – 148 = 212⁰ B1

(d) < PTS = 46 + 180 – 129 = 97⁰ B1

< PST = 180 – (97 + 39) = 44⁰ B1

(a) Kshs.350,000 = $ 350,000 M1

74.75

= $ 4682 A1

(b) Balance = ³/₄ x 4682

= $ 3511.5 B1

$3511.5 = Fr 3511.5 x 74.75 M1

11.04

= Fr 23780 A1

Expenditure = Fr 16 200

Balance = Fr 7580

= Kshs.97,024

Value on departure = Kshs.7580 X 11.04 B1 bothÖ

= Kshs.83 683.2

Difference = Kshs.97,024 – 83683.2 M1

= Kshs.13,340.80 A1

X	-2	-1	0	1	2	3
Y	-11	0	5	4	-3	-16

B1Övalues

S1Ö scale

8 — P1Ö plotting

6 — C1 Ö curve

4 —

-2 — 1 2 3 x

-4 —

-6 —

-8 — y=2x=2

-10 —

-12 —

-14 — x =-0.53 + 0.1 BI

-16 — Nx = 1.87+ 0.1

y = 5+2x-3x²=2-2x MI for equation

3x²-4x-4x-3=0 AI equation

x = -0.53 ± 0.1 B1

mx = 1.87 ± 0.1

y = 5 + 2x – 3x² = 2 – 2x M1 Ö for equation

\ 3×2 – 4x – 3 = 0 MA1 Ö equation

20.

B1 Ö 30⁰

R B1 Ö 2 ^ PQ, QR

B1 Ö 2 ^ bisectors

B1 Ö circle

9 Q

Radius = 4.2 ± 0.1 B1Ö radius

Area of circle = ²²/₇ x 4.22

= 55.44 ± 3 cm2

Area of D PQR = ¹/₂ x 3.5 x 7.5 sin 30 M1Ö D and circle

= 6.5625 cm²

Difference = 55.44 – 6.5625 M1Ö sub

= 48.88 cm² A1

Line (i) ^y/₂ + ^x/₅ = 1

5y + 2x = 10 B1Öequation

5y + 2x = 10 B1Ö inequality

Line (ii) ^y/₄ + ^x_/-3 = 1

3y = 4x + 12 B1Ö equation

3y < 4x + 12 or 3y – 4x < 12 B1Ö inequality

Line (iii) grad = ^-1/₃ y inter = 4

3y + x = 12 or 3y = -x + 12 B1Ö equation

3y + x < 12 B1Ö inequality

Line (iv) y – 3 = -3

x – 3 2

2y + 3x = 15 B1Ö equation

\ 2y + 3x £ 15 B1Ö equation

CLASS

60 – 64

65 – 69

70 – 79

75 – 79

80 – 84

85 – 89

496

938

1296

1155

246

174

Sf = 60

Sfx 3809

B1Ö x column

B1Ö f column

(a) (i) Modal class = 70 – 74 B1Ö model class

(ii) Median class = 70 – 74 B1Ö median

(b) Mean = 3809

60 M1

= 63.48 A1

S1Ö scale

B1 Ö blocks

59.5 – 64.5

64.5 – 69.5 e.t.c.

(c)

Histogram

20 —

15 —

10 – –

5 —

55 60 65 70 75 80 85 90

(a) XA = a, YA = 10 – a, YB = 10 – a, CZ = 10 – a = ZB

YZ = 10 – a + 12 – a = 8 M1

2a = 14

a = 7 cm A1

Cos X = 100 + 144 – 64

240 M1Ö any angle of the D

= 0.75

X = 41.41⁰

¹/₂ X = 20.70⁰ A1Ö ¹/₂ of the angle

r = OA = 7tan 20.7 B1 Ö radius

= 2.645 cm

Shaded area = ¹/₂X 10 X 12 sin 41.41 – ²²/₇ X 2.645² M1 Ö D & circle

= 39.69 – 21.99

= 17.7 cm² A1Ö

(a) 650,000 (0.85)ⁿ = 130,000 M1Ö formula

1.15n = 0.2

n = log 0.2 M1Ö

log 0.85

= 1.3010

1.9294

= – 0.6990 M1

– 0.0706

= 9.9 years A1

(b) 650,000 (1 – ^r/₁₀₀)⁵ = 325,000 M1

(1 – ^r/₁₀₀)⁵ = 0.5

1 – ^r/₁₀₀ = 0.5 ^1/5 M1

= 0.8706

^r/₁₀₀ = 0.1294 A1

r = 12.9 % B1

MATHEMATICS VI

PART II

MARKING SCHEME

SECTION I (52 MARKS)

= b¹⁵ ²_/5 X 4a⁶ ³_/2

32a¹⁰ 9b⁴ M1Ö reciprocal

= 2a⁵ A1

27 2

No. Log.

0.375 1.5740 +

cos 75 1.4130

2.9870 _

tan 85.6 1.1138

3.8732 = 4 + 1.8732

2 2

2.9366

0.0864

S. Price = 80 X 600

100

= Shs.480 B1

Cost Price = x

96x = 480 M1

100

x = Shs.500 A1

r² + hr = ^A/₂_p M1

r² + hr + (^h/₂)² = ^A/_2A + ^h/₄ M1

(r + ^h/₂)² = Ö 2A + h²

4p M1

r = ^-h/₂ ± Ö2A + h² A1

4p 4

(1²/₃r) (¹/₃ r) = (¹/₄ r) (5) M1

4r² – qr = 0

r(4r – q) = 0 M1

r = 0

or r = 2.25 A1

= 2 (5 + 2Ö3) – 1 (5 – 2Ö3) M1

(5 – 2Ö3) (5 + 2Ö3)

= 10 + 4Ö3 – 5 +2Ö3 M1

= 5 + 6Ö3 A1

13 3

P = Kq² + c

6 = 4k + c

16 = 9k + c M1 Ö subtraction

5k = 10

k = 2

c = -2 A1 Ö k and c

P = 64 2q² = 66

q = Ö33

= ± 5.745 A1

Volume = ¹/₂X 8 X 8 sin 120 X 10 M1 Ö area of x-section

No. of scouts = 32 sin 60 X 10 M1 Ö volume

2.5 M1

= 110.8

= 110 A1

Max. error = 2(8.5 + 5.5) – 2(7.5 + 4.5)

= 2 B1

% error = ²/₂₆X 100 M1

= 7.692% A1

G 3

B1 Ö net

H D G H B1 Ö dimen. FE must be 10cm

4cm 4cm

B1 Ö labelling

E 4cm A 12cm F 10cm E 3

4cm 12cm

(1 + 2x)⁶= 1 + 6(2x) + 15 (2x)² M1

= 1 + 12x + 60x² A1

(1.04)⁶= (1 + 2(0.02))⁶

= 1 + 12 (0.02) + 60(0.02)² M1

= 1.264 A1

BT = 10 cm B1

CT = 10 sin 40 M1

= 6.428 m A1

A₁ 10cm B C h = 6.428 + 1.5

1-5 = 7.928 B1

A.S.F = 405 ²_/3 = 27 ²_/3 = 9 B1

960 64 10

smaller area = 29 X 120 M1

164

= 67.5 cm² A1

Relative speed = (x – 30)km/h B1

2 km = 2 hrs

(x – 30)km/h 60 M1

2x – 60 = 120

x = 90 km/h A1

16 Km/h 5 Km/hr

x Km (22 – x) Km

x + 22 – x = 11

16 5 4 M1

5x + 352 – 16x = 220 M1Ö x-multiplication

11x = 132

x = 12 km A1

AP = ²/₉ x 12 = 2²/₃ cm B1 Ö both AP & AQ

AQ = ⁵/₉ x 12 = 6²/₃ cm

\ PQ = 6²/₃ – 2²/₃ = 4 cm B1 Ö C.A.O

Taxable income = ¹¹⁵/₁₀₀ x 8000 M1

= Shs.9200 p. m

= Shs.110,400 p.a A1

Tax dues = ¹⁰/₁₀₀ x 39600 + ¹⁵/₁₀₀ x 39600 + ²⁵/₁₀₀ x 31200 M1 Ö first 2 slabs

= 3960 + 5940 + 7800 M1 Ö last slab

= Shs.17,700 p.a

= 1475 p.m A1

net tax = 1475 – 400

= Shs.1075 B1 Ö net tax

Total deductions = 1075 + 130

= Shs.1205

net income = 8000 – 1205 M1

= Shs.6795 A1

(a) P (all correct) = ²/₃ x ³/₅ x ²/₅ M1

= ¹²/₁₂₅ A1

(b) P (2 correct) = ²/₅ x ³/₅ x ³/₅ + ²/₅ x ²/₅ X ¹/₅ + ³/₅ x ⁴/₅ x ²/₅

M1

= ¹⁸/₁₂₅ + ⁴/₁₂₅ + ²⁴/₁₂₅ M1

= ⁴⁶/₁₂₅ A1

= P(2 correct or 3 correct)

= ⁴⁶/₁₂₅ + ¹²/₁₂₅ M1

= 46 + 12 M1

125

= 58

125 A1

Old price/pen = 440

New price/pen = 494 B1Öboth expressions

x + 3

440 – 494 = 1.50

x x + 3 M1 Ö expression

440(x + 3) – 494x = 1.5x² + 4.5x M1Ö x-multiplication

x² + 39x – 880 = 0 A1 Ö solvable quad. Eqn

x² + 55x – 16x – 880 = 0 M1 Ö factors or equivalent

(x – 16) (x + 55) = 0

x = -55

or x = 16 A1 Ö both values

\ x = 16

difference in purchase = 19 X 1.50 M1

= Shs.28.50 A1

(a) OP = a + ¹/₄ (b – a) M1

= ³/₄ a + ¹/₄ b A1

(b) PQ = PO + OQ

= –³/₄ a – ¹/₄ b + ¹/₄ (a – b) M1

= –¹/₂ a – ¹/₂ b A1

= ¹/₄ (b – a) + ⁵/₃ b M1

= ²³/₁₂ b – ¹/₄ a A1

(d) PN = PB + BN

= ³/₄ (b – a) + ²/₃ b M1

= ¹⁷/₁₂ b – ³/₄ a A1

(a) Volume in 2 days = 22 x 3.5 x 3.5 x 150 x 20 x 3600 M1 Ö area of x-section

7 2 2 1,000,000 M1 Ö volume in cm³

= 103.95 m³ M1 Ö volume in m³

(b) 22 X r² x 8 = 103.95 x 15 x 7 M1

7 2

r² = 103.95 x 15 x 7 M1

2 x 2 2x 8

= 31.01 M1

r = 5.568 m A1

(a) Ration of needs for A:B = 5:8

A’s share = ⁵/₁₃ x 16.9 + ¹/₂ x 13.5 M1

= 13.25 Million A1

B’s share = (13.5 + 16.9) – 13.25 M1

= 13.25 M1

A’s share ⁵/₁₃ x 16.9 + ^39/₅₀ x 13.5

6.5 + 10.53

= 17.03 m A1

B’s share = 30.4 – 17.03 M1

= 13.37 Million A1

Log V = n Log E = log k

Log V	-0.46	-0.13	-0.14	-0.01	0.05
Log E	-0.35	-0.21	-0.05	0.07	0.13

B1Ö log V all points

B1Ö log E all points

S1 Ö scale

P1Ö plotting

Log V = n log E + log K L1 Ö line

Log K = 0.08

K = 1.2 ± 0.01 B1 Ö K

N = ^0.06/_0.06 B1 Ö n

= 1 ± 0.1

\ v = 1.2E B1Ö v

when E = 0.75, V = 0.9 ± 0.1 8

(a) T 3 PQR ® P^IQ^IR^I

4 P^I (0,5), Q^I (2,2) R^I (1,0)

P^I Q^I R^I P^II Q^II R^II

(b) 4 3 0 2 1 = 15 14 4 M1 Ö

1 2 5 2 0 10 6 1 A1 Ö

P^II (15,10), Q^II (14,6), R^II (4,1) B1Ö

= 5

area of P^II Q^II R^II = 5 (area P^IQ^IR^I)

= 5 X 3.5 M1Ö

= 16.5 cm² A1

Teachers' Resources

Lost Your KNEC Certificate? Here’s How to Quickly Obtain a Replacement

October 5, 2025 Maverick John Leave a comment

Lost Your KNEC Certificate? Here’s How to Quickly Obtain a Replacement

Misplacing a vital academic certificate can be quite distressing, particularly when it’s needed for job applications, further education, or professional licensing.

Fortunately, the Kenya National Examinations Council (KNEC) has simplified the process of obtaining a replacement for lost certificates through their online portal. This detailed guide will take you through each step, ensuring you receive your Certification Letter swiftly and without complications.

Overview of Steps to Follow

1. Create an Account
2. Log In
3. Choose “Lost Certificate”
4. Provide Exam Information
5. Upload Necessary Documents
6. Submit Your Application
7. Await Verification
8. Make Payment (Ksh 5,220)
9. Monitor Your Application
10. Collect Your Certification Letter

Reasons for Needing a Replacement Certificate

A KNEC certificate is an essential document that confirms your academic accomplishments. Whether it’s your KCSE, KCPE, or other national exam results, losing it can hinder your career or educational pursuits. Here’s why it’s important to act quickly:

– Job Applications: Employers require original certificates for verification.
– Higher Education: Colleges and universities need proof of qualifications.
– Professional Licensing: Organizations like ICPAK, EBK, or NCK require certified academic records.
– Immigration & Visa Processing: Some countries necessitate verified certificates.

Without a replacement, you risk missing significant opportunities.

Step-by-Step Instructions for Applying for a KNEC Certification Letter

KNEC has digitized the process, making it quicker and more efficient. Here’s how to apply:

1. Create an Account
Go to https://qmls.knee.ac.ke and set up an account using a valid email address and phone number.

2. Log In
Check your email for your login details. If you don’t receive it, look in your spam folder.

3. Choose “Lost Certificate”
After logging in, select the Lost Certificate option from the menu.

4. Provide Exam Information
Enter your Index Number, exam type (KCSE/KCPE), and the year of your exam. Click FIND to access your records.

5. Upload Necessary Documents

Attach the following:
– A copy of the Lost Certificate/Result Slip (if available, even a scanned photo is helpful).
– National ID (both sides) – a clear, high-resolution copy.
– Recent passport-size photo.
– Police Abstract – file a report at any police station to obtain this.
– Confirmation from the Registrar of Persons, which includes:
– A confirmation letter
– Fingerprint page
– Back page with personal details

Where to Obtain Registrar Confirmation:
– For Nairobi Applicants: Visit the NSsF Building, 7th Floor, Upperhill.
– For Other Regions: Go to your sub-county registrar of persons office.

6. Submit Your Application
Review all information carefully before submitting to prevent delays.

7. Await Verification
KNEC will assess your documents, which may take 3-7 working days.

8. Make Payment (Ksh 5,220)
Once verified, you’ll receive an email prompting you to log in and make the payment. You can use M-Pesa, debit/credit card, or bank transfer.

9. Monitor Your Application
Log in regularly to check the status. Processing typically takes 14-21 days.

10. Collect Your Certification Letter
Once it’s ready, KNEC will notify you via email. You can pick it up in person at their offices or request delivery (if available).

Common Challenges & Solutions

1. Missing Documents
Solution: Prepare all documents in advance. Use a scanner or the CamScanner app for clear copies.

2. Delayed Police Abstract
Solution: Report the loss promptly. Some stations can issue abstracts immediately.

3. Registrar of Persons Backlog
Solution: Arrive early to avoid long waits. Some offices allow online appointments.

4. Payment Issues
Solution: Ensure you have enough funds. If payment fails, reach out to KNEC’s helpline.

Benefits of KNEC’s Online System

– No Physical Queues: Apply from anywhere, at any time.
– Faster Processing: Digital submissions minimize manual errors.
– Transparent Tracking: Monitor your application in real-time.
– Secure Payments: Avoid risks associated with cash handling.

More related news:

FAQs About KNEC Certification Letters

– Q1: How long does the process take?
A: Usually 3-4 weeks, depending on verification speed.

– Q2: Can someone else collect the letter for me?
A: Yes, with an authorization letter and their ID.

– Q3: What if I don’t have a copy of my lost certificate?
A: A result slip or printout from the KNEC portal may be acceptable.

– Q4: Is the Certification Letter as valid as the original?
A: Yes, it is legally recognized for all official purposes.

– Q5: Can I apply if I lost my certificate years ago?
A: Yes, KNEC retains records indefinitely.

Final Tips for a Smooth Application

– Start Early: Don’t wait until the last minute.
– Keep Copies: Scan and save digital backups of all certificates.
– Follow Up: If there are delays, contact 0720741004 / 0732333566.

Conclusion

Losing your KNEC certificate can be stressful, but the replacement process is now easier than ever. By following these steps, you can obtain your Certification Letter without unnecessary delays. Act now to secure your academic future!

Need Assistance?
Contact KNEC at https://qmls.knee.ac.ke or call 0720741004.

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KNEC News

How to replace KNEC Lost Certificate: 10 Easy Steps

The Kenya National Examinations Council [KNEC] started implementing policy in replacement of certificates for the affected candidates who sat for…

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2 θ + 120 = 330

P = 16 = 1/9 (A) (3)

DE DC DE CE

v + 20

SECTION B:

MATHEMATICS I

MARKING SCHEME.

R -1 1.1 Log = 0.04139

Sin B Sin A

A B A B

2 1 0 1 4 9

Large circle area

Large area = 22/7 x4x4 = 352/7 (A)

Length of diag. 50 = 7.071 = 5.536

Volume passed / sec = cross section area x speed