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FORM 1 AGRICULTURE LESSON NOTES- EDITABLE

October 5, 2025 Maverick John Leave a comment

INTRODUCTION TO AGRICULTURE

The term agriculture comes from two Latin words:

Ager: meaning land or field

Cultura: meaning cultivation

Agriculture means field cultivation. But agriculture has continued to grow and expand that it can now be broadly be defined as:

The art and science of crop and animal production

Agriculture as an art

Agriculture is referred to as an art because it involves the following:

Tilling of land
Construction of farm structures
Measuring of distances
Machine operations
Harvesting of crops
Feeding and handling of livestock
Marketing of agricultural produce

Agriculture as a science

Agriculture is referred to as a science because it involves the following:

Crop pathology: study of crop diseases
Entomology: study of insects and their control
Soil science:
Genetics: plant and breeding
Agricultural engineering

Branches of agriculture

Crop production
Livestock production
Soil science
Agricultural economics
Agricultural engineering
Crop production

This is the production of crop on cultivated land.

Crop production is divided into:

a) Field crops

These are crops grown on fairly large area of land. May be annual or perennial crops.

b) Horticultural crops

The growing of perishable crops. It involves the following:

i) Floriculture: growing of flowers
ii) Olericulture: growing of vegetables
ii) Pomoculture: growing of fruits
Livestock production

This is the rearing of all types of animals. It involves:

a) Pastoralism (mammalian livestock farming)

Rearing of farm animals on pastures eg cattle, goats, sheep etc

b) Aquaculture

Rearing of aquatic animals eg fish farming (pisciculture)

c) Apiculture: keeping of bees
d) Aviculture: keeping of poultry
Soil science

This is a branch of agriculture that provides knowledge how soil is formed, how it works to sustain life and how it can be kept alive through many years

Agricultural economics

This branch deals with the utilization of scarce resources in the production of agricultural products.

Agricultural engineering

This branch deals with the use and maintenance of farm tools, machinery and structures.

Roles of agriculture in the economy

Provision of food
Source of employment
Provision of foreign exchange
Source of raw materials to the industries
Provision of market for industrial goods
Source of money or capital

FARMING SYSTEMS

This is how the farm and all the enterprises in it are organized. There are two main farming systems namely:

Extensive system
Intensive system
Extensive farming system

This is a farming system which involves the use of large tracts of land. Its characterized by:

Low capital investment
Low labour per unit area
Low yield per unit area

Intensive farming system

This is a system of farming which requires high capital and labour investment. Its characterized by:

High yield per unit area
Use of modern technology
High labour per unit area
High capital investment

N/B: Extensive and intensive farming systems can be practiced under:

a) Large scale farming
b) Small scale farming
a) Large scale farming

This involves the use of large tracts of land. Its features include:

Heavy capital investment
Use of skilled labour
High level of management
Products are for commercial purposes
Large tracts of land is used

Large scale farming can either be:

i) Plantation farming: growing of one type of crop (monoculture)
ii) Ranching: rearing of beef animals
b) Small scale farming

This is a type of farming which is practiced on small piece of land. The products are either for subsistence or commercial purposes.

Methods of farming

Pastoralism
Arable farming
Mixed farming
Shifting cultivation
Organic farming
Agro forestry
Pastoralism

This is the practice of rearing livestock on natural pasture can be:

Settled livestock farming
Nomadic Pastoralism
a) Nomadic Pastoralism

This is the practice of rearing livestock and moving with them from place to place in search of water and green pasture. Nomadic Pastoralism can only be practiced where:

Land is not a limiting factor
Land is community owned

Arable farming

This is the growing of crops on a cultivated land: can be,

Mono cropping
Mono culture
Mixed cropping
Inter cropping
a) Mono cropping

This is the growing of one type of crop per season. Its disadvantages include:

Cause soil erosion
Diseases spread easily
If the crop fails, the farmer suffers total loss
Leads to nutrient depletion in the soil

N/B: Mono cropping can be practiced under mono culture where only one crop is grown throughout as in plantation farming eg in Tea, Coffee plantations.

b) Mixed cropping

This is the practice of growing different crops on the same piece of land but on different plots or strips. Usually helps to control soil erosion.

c) Intercropping

This is the practice of growing different crops on the same piece of land per season.

Advantages of intercropping

If one crop fails, the farmer has the other crop to support him, ie does not suffer total loss
Helps to control soil erosion
If legumes are included, they will enrich the soil with nutrient
Also interrupts the spread of diseases
There is high yield per unit area of land
There is also proper utilization of land

Disadvantages of intercropping

Requires a lot of labour
Routine crop management practices difficult to carry out
Requires high capital investment

Mixed farming

This is the growing of crops and rearing of animals on the same piece of land.

Advantages of mixed farming

Animals benefit from crop residues /remains as food while crops benefit from animals wastes as manure
Gives farmers income throughout the year
Ensures proper utilization of labour and land throughout the year
In case one enterprise fails, the farmer will still depend on the enterprise

Disadvantages of mixed farming

Requires high initial capital investment
There is lack of specialization
Limited land area allowed for each enterprise
Requires a lot of labour

Shifting cultivation

This involves farming on a piece of land continually until its exhausted after which the farmer moves to a new fertile land. Shifting cultivation can be practiced where:

Land is abundant
Population is sparse
Land is communally owned
Low number of livestock units per area

Advantages of shifting cultivation

Low capital investment
No pests and diseases build up
Soil structure is regained
No land disputes as the land is owned communally

Disadvantages of shifting cultivation

Yield per unit area is low
A lot of time is wasted when the farmer shifts to new area and builds structure
Farmers have no incentive to develop and conserve water and soil
Cannot be practiced in areas where there is high population density

Organic farming

This is the growing of crops and rearing of animals without using agricultural chemicals. It can be practiced through:

Use of organic manures instead of artificial fertilizer
Use of medicinal plants instead of chemical
Mulching
Crop rotation, to control diseases

Importance of organic farming

Its environment friendly
Its cheap
Does not require special skills

Agro forestry

This is the growing of trees, crops and keeping of animals on the same piece of land.

Advantages of agro forestry

Trees help to conserve water and soil
High output per unit area
Helps to reduce soil erosion
Provides trees for building and fuel

FACTORS INFLUENCING AGRICULTURE

There are a number of factors which influence both crop and animal production, some of these factors include:

Human factors
Biotic factors
Climatic factors
Edaphic factors
HUMAN FACTORS

These are factors which are due to the behavior of human beings or how they do things and how they influence agriculture. These human factors are:

Level of education and technology
Health of the people
Economic conditions
Government policy
Transport and communications
Cultural beliefs and religion
Market forces
a) Level of education and technology

High level of education leads to:
Accuracy in applying inputs and assessing results
Helps in proper decision making and organization
Better problem solution
Better utilization of livestock feeds and fertilizers
Understanding of technical language used in agriculture
Development of skills for operating machines and their maintenance
Increase in efficiency and minimizes costs

b) Health of the people

Today the biggest threat to farming is the HIV/AIDS, ill health makes people do little or no work. The general effect of HIV/AIDS and ill health on agriculture includes:
Shortage of farm labour
Increase the cost of living through treatment, thereby lowering their purchasing power thus low demand for agricultural products
Low standards of living leads to lack of motivation to invest in agriculture, thus increasing poverty
Low food supply
A lot of funds used to control it, instead of being used to develop agriculture

c) State of the economy

Economic conditions which have affected agriculture include:

Collapse of cooperative societies which affected the sale of farm produce such as milk, sugar, cotton etc
Liberalization of the economy, which has led to dumping of cheap products from other countries, this has caused the drop in price of agricultural products leading to low income to farmers

N/B: Kenya can benefit from liberalization by:

Producing goods of high quality and selling them competitively
Diversification

d) Government policy

These are the laws which are put in place by the government that govern the production, marketing and distribution of agricultural products. The policies that the government can put in place which can encourage the agricultural production include:

Heavy taxation of imports to prevent dumping of cheap goods into the local market
Subsidizing the growing of local crops thus making them affordable to farmers
Enact policies to enforce the production of high quality products
Put in place, policies aimed at conservation of natural resources in order to sustain agriculture
Stepping up disease and pest control eg through quarantine, vaccination etc

e) Transport and communication

Transport and communication plays an important role in conveying agricultural products
Railway lines are goods for transporting bulky goods to long distances
Airways are also efficient for air lifting horticultural products
Weather roads are necessary to transport farm produce to factories

N/B: proper transport and communication therefore will promote the development of agriculture, the electronic media eg radio, TV, internet, all need to be cheap and affordable to all farming areas.

f) Cultural practices and religious beliefs

The society’s beliefs and culture may also effect agriculture eg Muslims do not eat pork and therefore may not see the need for rearing pigs even if pigs are very productive.
Pastoral communities also only keep animals and may find it difficult to diversify to livestock farming even if its profitable.

N/B: A combination of the above factors may retard agricultural development

g) Market forces:

The local demand and supply of agricultural produce will also affect the level at which farmers produce, also the international demand eg of Kenyan coffee, Tea will affect how much the farmers produce.

BIOTIC FACTORS

These are influences (factors) caused by living organisms, living both in and on the soil surface. These organisms include:

Pests
Parasites
Predators
Decomposers
Pathogens
Pollinators
Nitrogen fixing bacteria

Effects of pests

They feed on plants lowering both the quality and quantity of produce
They transmit diseases
Injure the plants, thus exposing them to secondary infection
Increases the cost of production eg through buying chemicals to control them

N/B: other effects of living organisms on agriculture include:

They decompose the organic matter in the soil eg the decomposers
Encourage aeration through burrowing into the soil
Cause nitrogen fixation and denitrification
Cause soil borne diseases
Acts as soil borne pests to growing crops
Mans activities eg cutting trees, earth moving etc affect soil formation
Some living organisms eg ticks also acts as parasites to animals thereby transmitting diseases
Some insects and birds also act as pollinators to flowering plants thus enabling cross pollination

CLIMATIC FACTORS

These are factors due to the changes in the climate. Climate is the weather condition of a place taken over a long period of time. These climatic factors include:

Rainfall
Temperature
Wind
Humidity
Light

a) Rainfall

Rainfall is very important in agriculture production as it ensures supply of water required by all life processes. Aspects of rainfall important in agriculture include:

i) Rainfall reliability
ii) Rainfall amount

iii) Rainfall distribution

iv) Rainfall intensity
v) Form of rainfall
i) Rainfall reliability

This is the assurance that rain will fall come the expected time eg there are two rainy seasons in Kenya. Long rains begin around march 15 – 20 of every year and short rains occur in October – November.

Reliability of rainfall determines:

Time of land preparation
Time of planting

N/B: when rainfall fails to follow the expected patterns, there is usually heavy crop failure and loss of livestock.

ii) Rainfall amount

Rainfall amount is the quantity of rainfall that falls in a given area within a year. Its measured in mm/year. Rainfall amount determines:

Type of crop to be grown
Type of animals reared

iii) Rainfall distribution

Rainfall distribution refers to how the rainfall was spread throughout the year. It determines the crop variety grown in an area

iv) Rainfall intensity

Rainfall intensity is the amount of rain that falls in an area within a period of 1 hour. Its measured in mm/hr.

High rainfall intensity causes: damage to crops, and also soil erosion

v) Form of rainfall

This is the form in which rainfall falls ie may be form of hailstones etc

b) Temperature

This is the hotness or coldness of a place measured in degrees Celsius or centigrade

N/B: All crops thrive well under certain range of temperature known as cardinal range. These crops require narrower ranges of temperature within the cardinal range this is called optimum range.

Effects of low temperature on crop production

Slow growth rate of crops as process like photosynthesis etc will be slow
High incidences of diseases infection to crops eg Elgon die back, CBD, hot and cold diseases in coffee
Quality of crops eg tea, pyrethrum improves with the lowering of temperature

Effects of high temperature on crop production

Increase evaporation leading to wilting in crops
Increase rate of growth or hasten the maturity of crops
Improve the quality of crops such as pineapples
Causes incidences of diseases infection eg leaf rust in coffee and pest infestation eg aphids in vegetables

c) Wind

Wind is air in motion. Wind influences agricultural production by:

Causing lodging in cereals and damage to crops
Blowing away and bringing in rain bearing clouds
Acting as an agent of seed dispersal
Acting as agent of pollination
Increasing the spraed of pest and diseases
Destroying farm structures by carrying away roof tops
Also causes a cooling effect

d) Humidity

Humidity is the amount of water vapour in the air at a given temperature.
Relative humidity is the amount of water vapour held in the air at a given temperature compared to what it would hold when saturated
Evaporation is the loss of water from the soil surface in form of water vapour
Transpiration is the loss of water vapour through the leaf pores
Evapotranspiration is the loss of water vapour both from the soil and leaf pores

N/B: humidity influences:

Rate of evapotranspiration
Temperature of a given area

e) Light

Light provides energy required for photosynthesis

Photosynthesis is the process by which carbon dioxide in the air and water in the soil are synthesized in the presence of light to form carbohydrates. The light is absorbed by green pigments called chlorophyll.

Carbon dioxide + water = glucose

3CO₂ + 6H₂O = C₆H₁₂O₆

Aspects of light important in crop growth are:

i) Light intensity
ii) Light duration

iii) Light wavelength

i) Light intensity

This is the strength in which light is harnessed by chlorophyll for the purposes of photosynthesis.

N/B: The rate of photosynthesis increases with increase in light intensity up to where other factors become limiting eg water.

ii) Light duration

This refers to the period during which light is available to plants per day. The duration is usually 12 hours in a 24 hour day. Plant varieties are classified into:

Short day plants: requires less than 12hrs eg soya beans, rice, tobacco

Long day plants: requires more than 12hrs of day light eg some wheat varieties

Day neutral plants: requires 12hrs of light eg coffee, maize, beans etc

iii) Light wavelengths

Chlorophyll only absorb certain wavelengths of light which are not present in artificial light a part from ultra violet or infra red light

N/B: light influences:

Rate of photosynthesis in green plants
Flowering of plants
Performance of livestock eg growth rate and laying % in poultry

EDAPHIC (SOIL) FACTORS

Soil is derived from latin word solum

Solum means floor

Soil is the natural, consolidated material that originates from weathered mineral rock and decomposing organic matter.

Importance of soil

It’s a natural medium on which seeds germinate and roots grow.
It supplies plants with the mineral nutrients necessary for crop growth
It provides water, air, and warmth for small animals, micro organisms and plant roots to sustain life
It provides anchorage to plants
It also shelters many micro organisms

SOIL FORMATION

Soil is formed through the process of weathering and decomposition of organic matter
Weathering is both chemical and physical transformation that take place in the rocks, converting the components minerals into soils
Decomposition is the decaying/rotting of organic matter.( remains of dead plants and animals) that break down to form soil

Types of weathering

Physical weathering
Biological weathering
Chemical weathering
a) Physical weathering

Agents of physical weathering include:

i) Water
ii) Moving ice

iii) wind

iv) temperature
i) Water

Running water wears away the rocks over which it flows by rolling stones and hand particles on them.
Rain water dissolves carbon dioxide and forms weak carbonic acid which falls into rocks and dissolve them
Moving ice also has a grinding effect
When it rains, the rain drops hit the ground with force
Rainfall erodes soil surfaces

ii) Wind

Strong winds carry rock dust which hit hard on the surface of rocks which then break down to form soil.

iii) Temperature change

Due to temperature changes taking place within the rocks, they crack and crumble to form soil.
Also in cold places, the water in rocks freezes and expands which then produces pressure on rocks then they break to small particles

b) Biological weathering

This is carried through plants, animals and mans activities
Large animals eg elephants, buffalloos, cattle etc when they move, cause pressure on the rocks causing them to break down
Mans activities like mining cultivation and construction of buildings, roads, reduce the size of rocks into smaller particles

c) Chemical weathering

This is weathering which takes place due to chemical decomposition or change in the chemical structure of the rocks

Types of chemical weathering

i) Carbonation
ii) oxidation

iii) Hydration

iv) Hydrolysis
v) Dissolution
i) Carbonation

When it rains, rain water combines with free carbon dioxide in the air to form a weak carbonic acid eg

Rainwater + carbon dioxide = carbonic acid

H₂O + CO₂ = H₂CO₃

The weak carbonic acid reacts with limestone found in the rocks to form calcium bicarbonate eg

Weak carbonic acid + Limestone = calcium bicarbonate

H₂CO₃+ CaCO₃ = Ca(HCO₃)₂

Calcium bicarbonate formed from this reaction is soluble in water and the process effectively dissolves the rock minerals

ii) Oxidation

This is common in rocks having iron. Oxygen reacts with iron which is in ferrous state. This process forms unstable crystal which is easily decomposed and disintegrated

iii) Hydration

Minerals in rock combine with water to form hydrated compounds. Hydrated compounds so formed are weaker than the original form and these are then acted upon by physical or mechanical agents of weathering

iv) Hydrolysis

this is the reaction of minerals with water which then undergoes weathering process through other agents.

v) Dissolution

The minerals in the rock dissolve in water leaving behind unstable rock, which can break easily.

FACTORS INFLUENCING SOIL FORMATION

Parent material
Climate
Topography
Time
Living organisms
Parent material

The texture of the parent material affects the rate of soil formation. Freely drained parent materials can develop soils faster than dense impermeable parent materials. Also minerals composition of the soil depends on the nature of the materials eg coarse grained soils are from granite which when fully disintegrated will separate into constituent minerals like feldspar, quartz and mica

Climate

Climate factors like rainfall, temperature, light and relative humidity and wind are all important in soil formation. Due to continuous weathering, rainfall for example provides water which is an important reactant in all forms of weathering high temperature also spend up most chemical reactions

Topography

Topography may either increase or delay the effects of climate on soil reaction eg factors like slope, degree of exposure or shelter may influences the degree of sol erosion which leads to shallow or deep soils.

Topography also affects the movement of products of weathering which consist of soluble and solid particles. It therefore affects the soil depth and type of vegetation

Time

The length of time over which the soil forming processes have been in action affects the age of the soil. Where the soil forming processes have been taking place for a long time, deep mature soils can be found. This is possible if other factors such as topography, parent materials climate etc, favour the development of deep soils. Where soils erosion has been severe because of topography there is a tendency for the soils to remain shallow and youthful with poorly differentiated profile.

Living organisms

Living organisms affect accumulation of organic matter and also profile mixing. The micro organisms eg rhizobium add nitrogen to the soil

Vegetation cover also reduces surface erosion and this in turn mineral removal is reduced. Therefore the nature and number of organisms growing on and in the soil play a big role in the kind of soil that develops

SOIL PROFILE

Soil profile is the vertical arrangement of soil layers. The layers are called horizons

There are four broad groups of horizons, namely: A,B, C, and D

Top soil ———– horizon A

Sub soil ———– horizon B

Substratum —— horizon C

Parent rock —— horizon D

Cross – section of soil profile

Superficial layer

This is a layer consisting of dry and decayed organic matter covering the soil surface

Top soil (horizon A)

This is the top layer of the soil. Its dark in colour because it contains humus in it. It has many living organisns and plant nutrients, this layer of the soil has goodcrumb structure and is quite permeable to air and water.

Sub soil (horizon B)

Its below the top soil
Has no humus and usually orange brown in colour
It has few living organisms and deeper growing roots of plants
It may have an impermeable layer called the hardpan

Causes of hardpan

Working the soil when wet with heavy machinery
Cultivation at the same depth throughout

Disadvantages of hardpan

Hinders air circulation in the soil
Prevent crop root penetration

Weathered rock (substratum)

This layer is found beneath the sub soil. Its made of partly weathered rolck with no humus. Its hard and therefore impermeable to water.

Parent rock

This is the bedrock. The soil formed from this rock. Ponds of water are often formed on this rock. Roots of some plants in very dry areas reach these ponds to absorb water

Transitional zone

This is a zone between any two bordering soil layers, whereby one layer gradually merges into the next one in the series

Influence of profile on crop production

Most plant nutrients are found in the top soil
The deeper or thicker the profile, the better its for crop production
Loosely packed soil allows for easy root penetration
The nature of the bed rock also determines the nutrients availability in the soil.

SOIL CONSTITUENTS

Soil is made up of the following:

Mineral matter
Soil water
Soil air
Organic matter
Living organisms
Mineral matter

These are inorganic compounds formed from the weathering of rocks. They differ in size ranging from an clay to gravel. They include:

Clay
Silt
Sand
Gravel

Influence of mineral particles on crop production

They make the main frame work of the soil

They hold plant roots firmly together

How to determine the mechanical composition of the soil

Using various sieves of different diameter

Soil water

Soil has water which comes from rainfall and also from irrigation in dry lands

Forms of soil water

Superfluous water
Capillary water
Hygroscopic water

Superfluous water

This is water which is held by gravity. Its also called gravity water.
Its easily lost because its loosely held by soil particles
Its readily available to plants but not useful because too much of it limits aeration

Capillary water

This is water occupying the micro pores. Its held by soil particles
It’s the water available to plants. Its also reffered to as available water

Hygroscopic water

This is water which forms a thin film around the particles. Its not available to plants

Functions of water to plants

Soil water maintains the life of plants
Its used as a raw material for protein for diffusion of mineral salts and oxygen into the root hairs and the mineral salts dissolved in water are conducted upwards to the leaves.
Its also acts as a solvent for the diffusion of other substances from one part ofplant to another
It makes protoplasm and cell sap of the growing plants
It keeps the cell turgid and thus supports plant
Also cools the leaves of the plant during transpiration

Experiment 1 to find the percentage of soil water content

Apparatus: – dish, stirring, weighing balance, soil sample and heater or oven

Procedure: –

Measure the mass of the dish
Pour soil in the dish and weigh
Half fill the dish with water
Heat upto a bout 105^oc
Cool the sol with a dessicater then reweigh – repeat the process until you get a constant mass

Soil air

The spaces between the soil particles are filled with air. These include

Oxygen —————– 20.6

Carbon dioxide ——- 0.6 – 0.6

Nitrogen ————– 78.6

Other rare gases.

The amount of air available in the soil is inversely proportional to the amount of water in rhe soil pore spaces.

Oxygen present in the air is essential for the respiration of roots and other living organisms in the soil

Nitrogen in the soil is converted into nitrates by the nitrogen fixing bacteria

Air is also needed by the micro organisms living in the soil

Excess carbon dioxide in the soil is poisonous to plants

Experiment 2: To find the percentage of air by volume in a soil

Apparatus

Small tin
Graduated cylinder
Knife and stirring rod

Procedure

Turn the empty tin upside down and press firmly into the ground until the tin is completely filled with soil
Turn the tin upright and level the soil to the brim of the tin with a ruler
Pour 250cm³ of water into a cylinder and scrap off soil into the water until no bubbles comes out
Record the final volume of soil and cylinder

Soil organic matter

Organic matter in the soil is the remains of the dead plants and animals plus their waste products
Humus is the decayed organic matter

Importance of organic matter

Decomposes to release nutrients to plants
Makes the soil lighter to cultivate
Also improves the soil structure

Experiment 3 To find the % of humus content in the soil

Apparatus
Dish
Garden soil
Tripod stand
Wire gauze
Bunsen burner

Procedure

Weigh the empty dish
Put the garden in the dish and reweigh
Place in an oven at about 105^oc
Cool in a dessicater and reweigh
Repeat the process several times until a constant weight is obtained
Note the difference weight

Soil living organisms

There are two types of living organisms in the soil namely:

Macro organisms

Micro organisms

Macro organisms are large organisms found in the soil eg rodents, earthworms, ants, termites, plant roots etc

Micro organisms are tiny organisms which can only be seen with the help of a microscope they include bacteria, fungi, protozoa etc.

Importance of soil living organisms

They barrow in the soil and aerate the soil and improve drainage
They help in the decomposition of organic matter
Some also fix nitrogen in the soil eg the nitrogen fixing bacteria

Experiment 4: To show the presence of living organisms in a soil sample

Apparatus

2 flasks
Rubber cork
Muslin bag
Heater
Lime water
Garden soil

Procedure

Put a handful of garden soil in two muslin bags labeled A and B
Heat the soil in muslin bag B strongly to kill the micro organisms
Suspend the two bags in the flasks also labeled A and B, the flasks should contain lime water
Leave the apparatus for 4hrs

Observation

Lime water in flask A turns milky
Lime water in flask B remains clear

Conclusion

Lime water in flask A turns milky because of the presence of carbon dioxide produced during respiration. Carbon dioxide turns lime water milky
Lime water in flask B remained clear since the living organisms were killed during heating so no respiration took place

Physical properties of soil

These include:

Soil structure
Soil texture
Soil colour

Soil structure

This is the way in which the individual soil particles are arranged

Types of soil structure

Single – grained structure
Crumby structure
Granular structure
Platy structure
Blocky structure

(a) Single – grained structure

In this structure, the particles are not cemented together. They exist as individual grain. They form no aggregates and are non porous.

They are mostly found in top soils of sandy soils and in arid climate and in alkaline soils

(b) Crumby structure

This type consists of small, soft porous aggregates of irregular shapes. They are not closely fitted together

(c) Granular structure

This is made of friable rounded aggregates of irregular shapes called granules. Its formed when particles co agulate and are cemented together to form rounded aggregates whose diameter is not more than 15cm

When wet it becomes porous since the spaces are not readily closed by swelling. The structure is found in top horizon in cultivated soils and in the sub- soil under grass. The structure is not porous and is usually affected by tillage.

(d) Prismatic structure

This is where the structure aggregates are arranged vertically. The primary particles are vertically oriented forming distinct columns which vary in length depending on the type of soil.

The structure is found in sub soil of arid and semi arid soils

N/B: If the tops are rounded, they are called columnar. But if the tops have clear cut edges, the its called Prismatic

Platy soil structure

In this structure, the aggregates are arranged on top of one another on thin horizontal plates. The plates overlaps and impair permeability and hence drainage and root penetration. The structure is found in top soils of clay soil and forested area.

(f) Blocky structure

Here the aggregates are in form of rectangular blocks. The aggregates easily fit together a long vertical edges

Influence of soil structure on crop production

A loosely packed structure ensures good air circulation in the soil
Good structure also ensures proper water holding capacity
Good structure also gives proper root anchorage
Good structure also reduces then soils liability to erosion

Factors that influence the soil structure

Parent material

The physical and chemical properties of the parent rock will determine the type of structure being formed

Soil forming processes

Processes which lead to soil formation will determine the type of structure being formed

Climate

In areas where a lot of rainfall is followed by dry periods cracks tend to form giving rise to good structure which is well aerated

Organic matter

Presence of organic will stabilize the soil structure

Living organisms

Living organisms also help to decompose organic matter which turn improve structure

Cultivation

The nature of cultivation eg digging channels results in a better structure

Inorganic compounds

Presence of compounds like iron oxide have binding properties and help in the formation of granules

Soil texture

This refers to the various mineral particles present in a soil sample.

Particles Diameter

Clay 002mm and below
Silt 002 —— 0.02
Fine sand 02 ——- 0.2
Coarse sand 2 ——– 2mm
Gravel 2 ———- 20mm
Stone 20mm and above

Determination of soil texture

Can be determined by:

Mechanical analysis
Chemical analysis

Mechanical determination of soil texture

Apparatus

Sieves of different diameter
Containers
Weighing balance

Procedure

Put a known amount of soil sample in a container
Pass the soil through a sieve of the smallest diameter and shake
Weigh the soil that remains in the sieve
Repeat the process using sieves of different diameter until all the soil I passed through

Observation

After every sieving it will be observed that a certain amount of soil remains in the sieve

Conclusion

Soil is made up of different sized particles of different diameter

Experiment 6: to show that soil is made up of different sized particles

Apparatus

Measuring cylinder
Sodium carbonate
Garden soil

Procedure

Put some soil sample in a measuring cylinder
Add about 4 times its volume of water with sodium carbonate to aid in dispersion of particles
Cover the mouth of the cylinder with the hand and shake vigorously for about 2min.
Place cylinder on the bench for about 1hr or more to allow the contents to settle down

Observation

At the end of the period, it will be seen that fractions have settled in layers
The heavy, coarse gravels settle first, then followed in succession by sand, silt and clay
The humus and organic matter remain floating in the water or on top of the clay

Conclusion

From the above observations, it can then be concluded that soil is a mixture of particles of different sizes.

Influence of soil texture on crop production

Coarse soils have poor water holding capacity
Very fine textured soils also have poor aeration

Soil colour

Soil colour depends mainly on the mineral composition of the soil
If the soil was made from a rock containing a lot of iron compounds, it tends to be brownish yellow, reddish or orange in colour
Humus content also gives dark brown colour
Soil colour influences temperature of the soil

Soil classification

Soil can be classified based on the following

Soil structure
Soil texture
Soil colour
Soil ph

According to structure, soils could be classified as granular, crumby, blocky, or platy soil structures

According to texture, a soil containing high proportion of sand particles is called sandy soils, if it contains high amount of clay then its called clay soils

In terms of colour, soils could be either dark coloured soils or light coloured soils

Types of soils

Sandy soils
Silty soils
Clay soils
Clay loams
Loamy soils

Sandy soils

They have bigger particles
Contains 50 – 80% sand, and 20 – 50% silt and clay
Organic matter content is 0.1 – 3%
Are well drained
Are more prone to soil erosion have low water holding capacity
They are slightly acidic
Easy to cultivate but less fertile

how to improve sandy sols

Add organic matter
Addition of fertilers

Silty loams

They contain 20 – 30% sand
Also contains 70 – 30% clay
Has 0.1 – 4% organic matter
They are fine textured, well drained and have a good water holding capacity
They have moderately acidic ph
Moderately fertile and aerated

Clay loams

They contain 20 – 50% sand
Clay and silt is 20 – 60%
Has organic matter content of 0.1 – 6%
They are fine textured
Poorly drained and aerated
Has capillarity and water retention
They are rich in plant nutrients
Are suitable for flood irrigation for rice growing
This soil can be improved through drainage

Clayey soils

Have clay content of more than 40%
Have high water holding capacity
Have crystalline and platy structure
Expand when wet
Crack when dry
Get water logged easily
Also suitable for flood irrigation
Have high capillarity

Loamy soils

They contain 30 -50% sand, 50 -70% silt and clay and 0.4% organic matter
Are moderately textured and drained
Are slightly acidic
Have good water holding capacity
Can be improved by planting cover crops and adding organic manures

Experiment 7: To compare the porosity and water holding capacity of sand, loam and clay

Apparatus

Measuring cylinder
Funnels
Cotton wool
Dry sand, loam and clay

Procedure

Place equal volumes of each soil in each funnel plugged with cotton wool
Tap all the funnels persistently until all visible air spaces are filled up
Stand each funnel in the open end of measuring cylinder and add 50cm³ of water into each funnel
Note the time taken for the first drop of water through into the cylinder

Observation

After some time, it will be seen that water level is high in sand than the rest

Conclusion

Sandy soil is more porous than the other 2

Clay soil has the highest water holding than the other 2

Experiment 8: To compare the capillarity of sand, loam and clay

Apparatus

3 long cylinders
Dry sand, clay and loam
Water trough
Clock
Ruler

Procedure

Close the lower end of each tube with a plug of cotton
Fill each tube with different soils
Tap the end of each tube gently in the bench to tightly pack the soils
Stand and clamp each tube with a clamp and put in an empty water trough
Poor water into the trough to a depth of 5cm
Measure the height of water in each tube after 3 – 5min
Take as many readings as much as possible
Record the readings

Observations

Water will be seen to be rising up the tubes
It rises very fast in sand and loam in the first 3 – 5min. but very slow in clay
After 2hrs water level will be higher in loam than in clay soil and least in sand
Water rise continues in clay soil but stops after some time in loam

Conclusions

Clay and loam have higher capillary action due to their fine pore spaces
Sand has poor capillary action due to their large pore spaces
Clay soil has the highest capillarity

Chemical properties of soil

Soil ph
Soil mineral content
Soil pH

This is the acidity or alkalinity of soil solution
Acidity is determined by hydrogen ion concentration while alkalinity is determined by hydroxyl ion concentration

Influence of soil ph on crop production

Soil ph affects the availability of various nutrients eg low ph makes P, and molybdenum less available and high ph makes Mn, K, Fe and zinc less available
Very low ph affects the activities of micro organisms eg nitrogen fixing bacteria
Different crop species require different ph ranges

Ways of modifying pH

Apply lime to raise the pH
Apply basic fertilizers
Apply sulphur to raise the pH
Apply acidic fertilizers to lower the Ph

FARM TOOLS AND EQUIPMENT

TOOL

A tool is any instrument held in the hand and used to do work

EQUIPMENT

This is something used for specific purpose

Why farmers use tools and equipment

To increase efficiency
To make farm operations easier
To minimize injuries
To enhance production

Precautions in handling tools and equipment

Proper maintenance
Proper use of tools
Proper storage
Use safety devices and clothing
Proper dressing
Skilful handling of tools

Categories of farm tools and equipment

Garden tools and equipment
Livestock production tools and equipment
Workshop tools and equipment
Plumbing tools and equipment
Masonry tools and equipment

Factors determining the choice of tools to use

The task to be performed
The tools efficiency
The level of knowledge and skill of user
Availability of the tools

General Maintance practices of farm tools

Sharpen the cutting edge
Grease the moving parts
Repair or replace worn out parts
Proper and safe storage
Clean after use
Tighten loose nut and bolts
Oil and paint before long storage

Reasons for maintaining farm tools and equipment

To durability
To improve efficiency
To avoid injury
Reduce production cost

CROP PRODUCTION I

LAND PREPARATION

Land preparation involves all those activities that make land suitable for planting eg ploughing, harrowing, ridging and rolling

Seed bed: this is apiece of land prepared ready for planting. To achieve good germination of seeds the following must be achieved:

Suitable size of clods
Good depth
Looseness of soil
Absence of weeds

Reasons for land preparation

To kill weeds
To incorporate manure and other organic matter in the soil
To destroy different stages of crop pest such as eggs, larva or adult stages by burying them and exposing them to the heat
To encourage the penetration of roots in the soil
To make subsequent operation easy
To encourage water penetration in the soil

Operations in land preparation

Land clearing
Primary cultivation
Secondary cultivation
Tertiary operations
Land clearing

This is the removal of vegetation cover from the surface before land is cultivated. Its done to prepare land for cultivation as well as a method of land reclaimation

Conditions that necessitate land clearing

When opening up virgin land
Where a stalk growing crop was previously planted
Where the interval between primary and secondary cultivation is long such that land is reverted back to its original virgin state
Where land was left fallow for a long time

Methods of land clearing

Tree felling
Burning
Slashing
Use of chemicals
a) Tree felling

This involves cutting down trees. Axes, pangas, are used and small power saws where the trees are few. Bulldozers and root rakers are used where trees are on large scale. After cutting down the trees, destumping or removal of stumps and disposal of trash is done.

b) Burning

here fire is set on the vegetation cover. Should be done when the speed of wind is low to avoid spread of fire to other fields. Burning should be discouraged because:

it destroys organic matter
kills soil micro organisms
also destroys plants nutrients

c) Slashing

Small bushes or grasses can be cleared by slashing. Slashers or pangas are used in a small area, while a tractor drawn mower can be used in large areas

d) Use of chemicals

Chemicals used to kill weeds are called herbicides. They kill weeds faster and more easily.

Primary cultivation

This is the initial opening of land either after land clearing or following a previous crop. Primary cultivation should be done well before the onset of rains to give time for all operations to be done in good time.

Importance of primary cultivation

To remove weeds
To burry organic matter for easy decomposition
To facilitate water infiltration and aeration
To destroy soil borne pests by exposing them to predators and sun
To make planting easy

Ways of carrying out primary cultivation

Hand digging
Mechanical cultivation
Use of ox plough
a) Hand digging

This is mainly the use of simple hand tools such as jembes, mattocks and fork jembes to cut and turn the soil slices.

b) Mechanical cultivation

Where large pieces of land is involved, farmers use tractor mounted implements which include mould board, disc ploughs. Also there is use of sub soilers to break the hard pan.

c) Use of an ox plough

This is use of ploughs drawn (pulled) by animals such as donkeys, camels, oxen etc. common in areas where such animals are available and the terrain is flat.

Aspects to be considered when carrying out primary cultivation

i) Time of cultivation
ii) Depth of cultivation

iii) Choice of implements

i) Time of cultivation

land preparation should be done early enough before the onset of rains.

Reasons for early cultivation

To give weeds and other vegetation enough time to dry up and decompose into organic matter
To allow carbon dioxide and other gases to diffuse out of the soil while being replaced by oxygen required in seed germination and growth of soil organisms
Also gives time for subsequent operations to be done giving way for early planting

ii) Depth of cultivation

factors that determine the depth of ploughing are:

The type of crop to be planted: Deep rooted crops require a soil which has been cultivated deeply, because it will facilitate easy root penetration. Shallow rooted crops may not need deep cultivation
The implements available: There are some implements which canot cut the soil beyond a certain depth. Such implements can be sharpened or weight be added
Type of soil: heavy soils are hard particularly when they are dry. Simple implements such as jembes tend to dig shallowly on such hard soils

iii) Choice of implements

Choice of implements used in primary cultivation is determined by:

The condition of the land: If the land has a lot of stones and stumps, it would be advisable for one to choose a disc plough which would not break easily when working on such land. A jembe cannot be used efficiently on land which has a lot of couch grass because it cannot pull all the rhizomes.
The type of tilth required: very fine tilth requires the use of different types of implements
The depth of cultivation needed: heavy implements are necessary when deep cultivation is needed and light implements are required when shallow cultivation is necessary

Secondary cultivation

These are operations which follow the primary cultivation and means seedbed refinement practices before planting, also called harrowing

Importance of secondary cultivation

To remove any weeds that might have germinated after primary cultivation
To break the soil clods into small pieces for easy planting
To level the field on order to achieve a uniform depth of planting
To incorporate organic matter into the soil in order to encourage decomposition before planting

Factors that determine the number of times of secondary cultivation

Size of planting materials: Big seeds such as those of groundnuts, maize etc require a fairly rough seedbed, and small seeds such as those of finger millets require fine seedbed
Slope of the land: When the land is very steep, less cultivation should be done to discourage soil erosion
The moisture content of the soil: In dry soils less cultivation are preferred so as to conserve the available moisture
Condition of the soil after primary tillage: where there is plenty of trash, more harrowing operations should be carried out to incorporate most of the trash into the soil

N/B: Implements used for secondary cultivation are: pangas, jembes, fork jembes, and garden rakes. Tractor drawn harrows eg discs, spike toothed and spring tine harrows

Tertiary operations

These are operations carried out to suit production of certain crops. They are carried out after land clearing primary cultivation and secondary tillage. They include:

Leveling
Rolling
Ridging
a) Leveling

This is the practice of making the soil surface flat and uniform so as to promote easy germination of small seeded crops such as wheat, grasses, and barley. It facilitates uniform germination of seeds.

b) Rolling

This is done to compact soil which is loose or fine tilth. Its done to prevent small seeds from being carried away by wind and to prevent soil erosion. Also increases seed soil contact. Implements used are: simple hand tools and heavy rollers

c) Ridging

This is the process of digging soil in a continuous line and heaping it on one side to form a bund ( ridge) and a furrow. The ridges are important for planting root crops like Irish potatoes, cassava etc. ridging helps in: tuber expansion and easy harvesting of root crops.

N/B: Other tillage operations include:

Sub soiling
Minimum tillage
Sub soiling

This is the process of cultivating the soil for the purpose of breaking up the hard pans which might have formed as a result of continuous use of heavy machinery in land preparation. Implements used in sub soiling are:

Sub soiler
Chisel ploughs
Cultivators

Importance of sub soiling

Helps to break up hard pans
Helps to facilitate gaseous exchange in the soil
Also brings to the surface, minerals which might have leached to the deeper layers

N/B: hard pan is an impervious layer of soil found within the sub soil.

Minimum tillage

This is the application of a combination of farming practices aimed at least disturbance to the soil.

Reasons for carrying out minimum tillage

To reduce the cost of cultivation or ploughing by reducing the number of operations
To control soil erosion, mulching and cover cropping greatly reduce chances of soil erosion
To maintain soil structure, continuous cultivation destroys soil structure hence its avoided
To conserve moisture, continuous cultivation exposes the soil to the heat of the sun thus enhance evaporation of available moisture
To prevent disturbance of roots and underground structures for example tubers and bulbs
To prevent exposure of humus to adverse conditions such as the suns heat that cause volatilization of nitrogen

Ways of carrying out minimum tillage

Application of herbicides in controlling weeds
Use of mulch on the soil surface. Mulch prevents weeds from growing by smothering them
Timing cultivation, late weeding of cotton crop, for example often produces a clean seedbed for finger millet to be sown without further cultivation
Restricting cultivation to the area where seeds are to be planted. Weeds in the rest of the field are controlled by slashing
Establishment of cover crop on the field
Uprooting or slashing weeds on perennial crops

WATER SUPPLY, IRRIGATION AND DRAINAGE

Sources of water

Surface water
Ground water
Rain water
Surface water

Sources of surface water are:

Rivers
Streams
Lakes

Ground water

Sources of ground water are:

Springs
Wells
Boreholes
a) Springs

Here water comes out of the ground as a result of an impervous layer meeting the ground surface.
Low wall can be constructed around the spring to increase the water volume for easier pumping
Also on higher ground, water can be conveyed to lower grounds by gravitational flaw

Diagram of a spring

b) Wells

Wells are holes dug in the ground until water table is reached. Can go up to 15m deep.
It’s advisable to dig the well during dry season to ensure that even during dry season water will be available
Fence around the well to avoid contamination
Construct a reinforced slab with a lockable lid to prevent contaminations and wearing of the top sides of the well. Water is lifted using buckets

Diagram of a well

c) Boreholes

These are deep holes drilled or sunk into the ground by use of drilling machines. The holes are usually sunk into the Parent rock to ensure continuous supply of water. The hole is of small diameter and usually lined with metal casing perforated at the bottom end to allow the water to rise up. Special pumps operated by either electricity or engines are used to lift water out of the hole.

Diagram of borehole

Rain water

Collected from roofs then stored in tanks. Ponds also constructed to store the run off. This is done during the rainy season.

Water collection and storage

Methods of water collection and storage include:

Dams
Weirs
Water tanks
Dams

This is a barrier constructed across a river or dry valley to hold water and raise its level to form a reservoir or lake. It has a spillway to allow excess water flow away. The accumulated water is then pumped to farms.

Weirs

A weir is a barrier constructed across the river to raise the water level, but still allow water to flow over it

Water tanks

Rain water, ground water and run off can be stored in tanks. The water storage structures (tanks) include:

Concrete tanks (overhead or underground)
Corrugated iron sheets
Steel tanks
Plastic tanks

Parts of a water tank

Funnel lid
Overflow pipe
Drainage pipe
Roof
Gutter
Outlet
Base

Diagram of water tank

Pumps and pumping of water

Water pumps

Types of water pumps include:

Centrifugal/Rotar dynamic pumps
Piston/Reciprocating pumps
Semi rotary pumps
Hydram pumps
a) Centrifugal pumps: These are made of metal discs with blades that rotate at high speed. They are powerful and can pump water for irrigation. Electric motors, diseal or petrol engines are used to operate them.
b) Piston pumps: Consist of pistons that move back and forth thereby pushing water through the pipes. Do not pump a lot of water thus suitable only for domestic and livestock use.

Diagram of a piston pump

c) Semi rotary pumps

These are operated by hand, and mostly used to pump water from wells for domestic and livestock use

d) Hydram pumps

these are operated by the force of flowing water. The higher the speed of water, the greater the pressure created in the pump. Cannot pump stationary water and only suitable for slopy areas, where water flows at high speed.

N/B: Pumping of water is the lifting of water from one point to another by use of mechanical force.

Conveyance of water

This is the process of moving water from one point mostly from storage to where its used or stored

Ways of conveying water

Piping
Use of containers
Use of canals
a) Piping

This is where water is moved through pipes

Types of water pipes

i) Metal pipes
ii) Plastic pipes

iii) Hose pipes

i) Metal pipes

These are two types: Galvanized iron and Aluminum pipes

Galvanized iron pipes are heavy and suitable for permanent installation of water system. Alumimium pipes are light and used for irrigation systems,

N/B: metal pipes are expensive but durable

ii) Plastic pipes

These are made of synthetic materials. There advantages include:

They are cheap
Easy to install
Durable when installed properly

Disadvantages include:

Become brittle when exposed to sun
Can burst under high pressure
Can be eaten by rodents

iii) Hose pipes

There are two types: rubber hose pipes and plastic hose pipes

Rubber hose pipes are more expensive but durable, hose pipes are used to convey water from taps to various areas eg irrigation areas or washing places

b) Use of containers

Water is drawn and put in containers such as drums, jerry cans, pots, tanks and buckets which are carried by animals, bicycles, human beings and vehicles

c) Use of canals

Water is conveyed from a high point to a lower appoint along a gradual slope to avoid soil erosion. Water conveyed in canals is mostly used for irrigation and livestock drinks

WATER TREATMENT

Water treatment is the process of making raw water from source safe for use in the farm.

Importance of treating water

To kill disease causing micro organisms such as cholera and typhoid bacteria which thrive in dirty water
To remove chemical impurities such as excess fluoride this may be harmful to humans
To remove smell and bad taste
To remove sediments of solid particles

The process of water treatment

Filtration at water intake
Softening of water
Coagulation and sedimentation
Filtration
Chlorination
Storage

Stage 1: Filtration at water intake

At the pint of water intake, water is made to pass through sieves before entering the intake pipe. This is to trap large impurities. Several sieves of different sizes are made.

Stage 2: Softening of water

The water in the pipe flows into the mixing chamber. This is a small tank where water circulates and is mixed with soda ash ( sodium bicarbonate) and alum ( aluminium sulphate) these chemicals are added into water in equal proportions. Soda ash softens the water, while alum helps to coagulate solid particles which finally settle down to the bottom

Stage 3: Coagulation and sedimentation

The softened water moves to the coagulation tank which is a circular and large solid particles such as silt and sand coagulate and settle down. The tanks is also open to allow in fresh air into the water. Water should stay in this tank for at least 30 hrs to kill bilharzias which cannot survive in water stored that long

Stage 4: Filtration

Water with very few impurities passes into a filtration tank where all the remaining solid particles such as silt are removed. The filtration tank has layers of different sizes of gravel and a top layer of sand. At its bottom is a layer of large pieces of gravel, this is followed by another layer of gravel but of fine texture. A layer of fine sand is placed on top of this fine gravel. These layers allow water to seep through very slowly leaving all the solid particles behind. When water leaves this tank, its clean.

Stage 5: Chlorination

The filtered water enters the chlorination tank. In this tank, small amount of chlorine solution is controlled by a doser and the amount added will depend on the volume of water to be treated and the outbreak of water borne diseases. Chlorine kills pathogens

Stage 6: storage

Water is then stored in large tanks, before distribution to consumers.

General uses of water in the farm

For domestic purposes eg washing, cooking etc
For watering livestock eg washing pigs
For diluting chemicals
For processing farm produce eg coffee etc
For construction of buildings
For irrigation

IRRIGATION

Irrigation is the artificial application of water to soil for the purpose of supplying sufficient moisture to crops.

Conditions that make it necessary for irrigation

In dry areas
During dry periods
In the growing of paddy rice
Soften the soil during transplanting
To effect the application of fertilizers and other chemicals

Types of irrigation

Surface irrigation
Sub surface irrigation
Overhead irrigation
Drip/Trickle irrigation

Factors that determine the type of irrigation to use

Capital availability
Topography of the land
Water availability
Type of soil
Type of crop to be irrigated

Surface irrigation

Here water is applied to the field by allowing it to flow on top of the ground surface.

Methods of surface irrigation

Flood irrigation
Furrow irrigation
Basin irrigation
Boarder irrigation
a) Flood irrigation

In flood irrigation, water is allowed to cover the whole field a few centimeters in depth. Its suitable for growing paddy rice fields.

Advantages of flood irrigation

Its cheap to establish and maintain
Does not require skills

Disadvantages of flood irrigation

There is uneven distribution of water in the field
A lot of water is wasted

b) Furrow irrigation

Here water is supplied by use of open ditches or furrows. Its suitable for all crops and application to most soils

Maintenance of furrows

Repair furrows when worn out or eroded
Remove weeds and silts

Advantages of furrow irrigation

Reduces chances of fungal diseases
Cheap to establish
Require little skills

Disadvantages of furrow irrigation

A lot of water is lost through evaporation and seepage
Erosion can occur if the furrows are not maintained
If water has high content of salt, it may have damaging effect on the plant roots

c) Basin irrigation

Basin irrigation involves the application od water into basins that have been checked by construction of banks or ridges. The basins may be rectangular ring shaped or have contour checks

This system is suitable in:

Relatively flat areas
Soils of low infiltration
For crops requiring large quantities of water
Soils that require leaching

Advantages of basin irrigation

Helps to control soil erosion
Retains rain water in the basins

Disadvantages of basin irrigation

Much land is occupied by water covering channels and ridges
There is no surface drainage
Requires precise land grading
Requires a lot of labour
Cannot be used in crops that require free draining soils
May result in accumulation of salts

Areas where basin irrigation is being practiced in kenya: mwea tebere, ahero, bunyala, west kano etc

d) Boarder irrigation

This is where parallel ridges guide a sheet of water that spread cover a relatively flat, but slanting piece of land. The ridges form long boarders. This method is applied where:

Soils have low to relatively high infiltration capacity
Crops are closely spaced, such as wheat, barley fodder crops as well as legumes

Advantages of boarder irrigation

Its easy and simple to operate
Requires less labour as compared to basin irrigation
Boarder ridges can be constructed economically with simple farm implements eg ox drawn ridgers
Large irrigation streams can be efficiently used

Sub surface irrigation

This is a system of irrigation where water is supplied to crops using underground perforated pipelines or any other porous medium that make water available from below the soil surface. Pipes sometimes referred to as conduits

The system is suitable in soils of high capillarity and water holding capacity

Advantages of sub surface irrigation

Little labour requirements
No need to construct dykes or soil grading
Can be practiced on both sloppy and flat land
Water does not cause soil erosion
Does not encourage fungal diseases
Economizes use of water
Minimizes theft of pipes

Disadvantages of sub surface irrigation

Its expensive method ie to buy pipes and to lay them
Pipes can be broken during weeding
Nozzles can get blocked

Overhead irrigation

This is the application of water above the crops by means of sprinklers or watering cans. Wind breaks should be constructed to avoid misdirecting the water.

Advantages of overhead irrigation

Water is evenly distributed over the required area
There is less water wastage than in furrow irrigation
It can be practiced on slopy grounds
Foliar fertilizers can be applied together with irrigation water thus reducing labour costs
Sprinkler systems can be easily be moved from one place to another

Disadvantages of overhead irrigation

Its expensive to install
Encourages fungal diseases eg blight, CBD
Causes soil erosion
Requires establishment of wind breaks

Sprinklers used are: oscillatory sprinklers, spring loaded sprinklers

Sprinklers can also be classified into: rotating head, perforated pipe system

Maintenance of sprinklers and pipes

Lubricate the rotating parts
Repair broken parts
Cleaning and unblock the nozzles

Drip/Trickle irrigation

Here pipes with tiny perforations are used. As water passes through the plastic pipes, water comes out through the holes in small amounts and drips to the ground.

Advantages of drip irrigation

Requires little amount of water
Can also use water of low pressure
Discourages fungal diseases eg blight, CBD
Does not encourage the growth of weeds
Can be used in sloppy topography

Disadvantages of drip irrigation

Pipes are expensive to buy and install
Require clean water, since dirty water will block the perforations

Factors to consider when choosing irrigation equipment

Capital availability
Topography
Availability of repair and maintenance
Type and source of power
Source

DRAINAGE

This is the method of removing excess water from water logged land. It’s a method of land reclaimation.

Land reclaimation is the process of bringing back waste land to agricultural production

Importance of drainage

Improves soil aeration: removal of excess water around the root zone allows for enough air for proper growth
Increases soil volume: increases the amount of soil around the roots
To raise the soil temperature: improves the rate at which soil worms up for better plant growth
Increases microbial activities: micro organisms in the soil increase in number due to proper aeration, they help to improve soil structure and make plant food more readily available
Reduce soil erosion: well drained soils have higher water holding capacity which helps to reduce water run off and increase infiltration
Remove toxic substances: due to water logging, soluble salts such as those of sodium increases in concentration to levels that are toxic to plants or may retard growth

Methods of drainage

Use of open ditches:

ditches are dug for the water to flow in by gravity to a water way thereby lowering the water table. May be U shaped or V shaped or trapezoidal

Underground drain pipes:

perforated pipes are laid underground. Water then seeps from the surrounding area into the pipes and flows to a water away. Such drains do not interfere with field operations. The pipes may be made of steel, clay or plastic materials

French drains:

ditches are dug, filled with stones and gravel, then covered with soil. Water from the surrounding area seeps into these drains and is carried into a water way

Cambered beds:

raised beds are constructed on the poorly drained soils

Pumping: where other methods of drainage are not possible, water is pumped out.

Areas where drainage has been carried out in kenya are: yala and bunyala to control flooding, ahero to control flooding of river nyando, loriaan region

WATER POLLUTION

This is the contamination of water by either chemical, industrial wastes, farm residues etc, making it unsafe for human beings and animals.

Agricultural practices that pollute water

Fertilizer and pesticides: chemicals compound found in the fertilizers and other pesticides do not decompose easily, hence they find their way into water sources through drainage, irrigation channels, erosion, seepage and leaching
Improper disposal of used farm chemicals: when containers contaminated with chemicals are disposed of into water sources, the result is water pollution
Damping of farm wastes: farm wastes such as slurry, manure used polythene, dead animals etc when improperly disposed of cause water pollution.
When land is cultivated or the soil is left bare erosion will easily occur leading to contamination through unwanted soil
Blockage of irrigation channels and water ways prevents free flow of water leading to stagnation of contaminated water
When pit latrines and sewage sites are located near water sources, they cause pollution
Other sources of pollution include industrial wastes and generalized contamination in the atmosphere and the environment

Methods of preventing water pollution

Practice organic farming
Safe disposal of used farm chemicals and industrial wastes
Proper location of pit latrines, sewage sites and waste dumps
Control of irrigation and establishment of grassed water ways to purify the water
Controlled use of fertilizers, manures and farm chemicals
Ensuring that the water source is free from contamination from the farm
Treating and piping water for farm use

SOIL FERTILITY I

This is the ability of the soil to produce and maintain high yields of crops for an indefinite period.

Characteristics of fertile soil

Should have good depth
Be well drained not water logged
Well aerated
Good water holding capacity
Supply nutrients needed by plants in correct amount and form available to plants
Correct soil pH for different crops
Free from crop pests and diseases

How soil loses fertility

Continuous growing of arable crops: continuous cultivation makes the soil loose and liable to erosion, this leads to lose of fertility.
Mono cropping: growing of crops every season leads to depletion of soil nutrients
Soil erosion: This leads to lose of top fertile soil
Leaching: leads to lose of soil nutrients into the lower horizons of the profile
Poor soil aeration: if soil is poorly aerated, the denitrifying bacteria increase in number and they make the infertile by converting nitrates into free nitrogen.
Poor drainage of the soil: If the soil poorly drained, the soil becomes flooded, forms acid soils which are useless for cultivation
Dry soils: If the soils are dry, the nutrients cannot be dissolved to be used by crops
Change of pH: soil pH influences the availability of certain nutrients eg low pH decreases solubility of phosphorus and high pH also decreases the availability of K, Mn etc
Accumulation of salts: certain salts usually become toxi if present in excess eg Mn, boron, fluorine etc
Burning of land: burning of land kills certain micro organisms and destroys certain nutrients

Ways of maintaining soil fertility

Control of soil erosion: control of erosion prevents loss of top fertile soil
Crop rotation : this ensures maximum utilization of crop nutrients. Also helps to control pest and diseases, will also add nutrients if legumes are included in the rotation
Maintaining soil pH: when soil pH is maintained at given ranges, particular nutrients will be available in the soils
Proper drainage: soil should be well drained to eliminate flooding
Weed control: control of weeds ensures no competition for nutrients, adequate space for crops and destroys alternate hosts for crop pests and diseases.
Minimum tillage: this helps to maintain soil structure and prevent erosion
Use of manures: manures supply a wide range of plant nutrients to the soil
Use of inorganic fertilizers: inorganic fertilizers supply specific plant nutrients

Organic matter, humus, and manures

Organic matter: this is the remains of dead plants and animals and their waste products

Humus: humus is the decayed organic matter ie the remains of plants and animals which have decomposed

Manure: manures are organic substances that are added to the soil to provide one or more plants nutrients. They have high matter content

Roles of organic matter in the soil

Increases water holding capacity and also water infiltration due to its colloidal nature
Releases a wide range of nutrients into the soil thus improves fertility
Provides food and shelter to micro organisms such as ants and rodents
Improves soil structure by binding soil particles
Buffers soil pH by avoiding rapid chemical changes due to the addition of lime and fertilizers
Reduces toxicity of plants poisons that have built up on the soil as a result of continous use of pesticides and fungicides

Classification of manures

Manures are classified according to: method of preparation and nutrients from which they are prepared.

There are 3 types of manures:

Farmyard manure
Compost manure
Green manure

Farmyard manure

Farmyard manure is a mixture of animal waste (urine and dung) and crop residues used as animal beddings.

Importance of farmyard manure

Increases yield of the crop
Adds organic matter into the soil and improves the texture and water holding capacity of thee soil
Adds useful bacteria to the soil

Factors influencing the quality of FYM

Type of animals: dung from fattening animals is richer in nutrients than farm growing animals which extract a lot of phosphorus from food eaten
Type of food eaten: the richer the food in terms of minerals the richer will be the manure
Type of litter used: wood shavings and saw dust are slow to decompose and contain no nutrients and absorb 1.5 times as much urine as their weight, while nappier grass provide both N and P, but has low absorptive capacity.
Method of storage: farmyard manure must be stored well in a place with a cemented floor and covered roof. The N and P are soluble and therefore can get leached by heavy rains
Age of FYM: well rotten manure is richer in nutrients and easier to handle and mix with the soil

Preparation of farmyard manure

A bedding of grass, wood shavings or saw dust is provided in the house of farm animals eg cattle, sheep
The animals deposit their droppings and urine on the bedding materials
After some time, ie daily, months or more as in poultry, the beddings are replaced with new ones
The discarded beddings are deposited in a specially prepared shaded place
New layers of used beddings are continuous added until a heap is formed
N/B: decomposition and mineralization of the materials take place through activities of certain bacteria resulting in a rich manure

Green manure

This is a type of manure prepared from green plants. The plants are grown for the purpose of incorporating into the soil when its green at the flowering stage for the purpose of improving soil fertility.

Characteristics of plants used for green manure

They should be highly vegetative or leafy
They should have faster growth rate
They should have high nitrogen content, thus preferably legumes
The plants must be capable of rotting quickly
The plants should be hardy ie can establish in poor conditions

Reasons why green manure is not commonly used

Most of the crops grown are food crops and its hard for people to use them as green manure
Green manure crops might use most of the soil moisture and leave very little for next main crop
Most of the nutrients are used up by micro organisms in the process of decomposing the green manure plant. These will only be released by micro organisms when they die.
It takes time for green manure crop to decompose and therefore planting is delayed

Preparation of green manure

The plant to be used is planted in the field
The plant is allowed to grow up to flowering stage
Its then incorporated into the soil by ploughing
Left for 2hrs to decompose after which the field is prepared for planting the main crop

Compost manure

Compost manure is the accumulation of plants residue, mixed with animal waste, piled together in a heap where conditions are conducive for decomposition, sometimes contain refuse and kitchen left over foods

Preparation of compost manure

There are two methods namely:

Indore method
Four heap system (stalk method)
Indore method

This was devised in a place called Indore in India

Procedure

A pit is prepared which is 1 – 2 m deep. The material to be made into compost is placed is placed.
The first layer 0.5m deep consist of fresh material to be made into compost eg grass, maize stalk refuse etc
The layer is followed by dung, old compost to provide micro organisms to decompose the fresh material
Artificial fertilizers eg SSP and Muriate of potash are added to increase the nutrient level of the compost

N/B: Nitrogenous fertilizers are not added because they are easily leached

The 2^nd layer is followed by a layer of top soil with micro organisms to decompose plant and animal remains
The same is repeated until the pit is full. Its also kept moist by applying water during dry season

Pit	Pit	Pit	Pit	Pit
1	2	3	4	5

Pit 1, 2, 3, and 4 are simultaneously filled and after 3 – 4 wks, the materials in pit 4 is taken to pit 5

This process is repeated until the material that was prepared 1^st is well rotten and taken to the field as composed.

The Four heap system

In this method, 4 – 7 heaps are used.
Materials used are crop residue, animal waste old manure FYM or onorganic fertilizers and top soil.
The materials is placed in heap X, then transferred to heap Y after 3 – 4 wks. After another 3 – 4 wks, the compost is taken to heap Z where it stays for another 3 – 4 wks then taken to the field

N/B: The manure heaps must be turned occasionally at least every 3 months to facilitate circulation within the heap, manure should be ready after 6 months.

Cross section through a compost heap

N/B: too much water runs the compost
Too little water stop the bacterial action
Always keep the compost under cover of grass and soil
Posts are fixed at a distance of 1.2 m a part to form the 4 corners of the heap, the post should be 2m high

Factors to consider when selecting a site for compost manure

Drainage of the site
Direction of prevailing wind
Size of the farm ie centrally placed
Accessibility

Problems associated with organic manures

Bulkiness
Laborious in application and transport
They spread diseases, pests and weeds
Loose nutrients when poorly stored eg through leaching
If not fully decomposed, crops will not benefit since it releases nutrients which can scotch the crops

AGRICULTURAL ECONOMICS I

Economics: this is the study of how man and society choose with or without money to employ scarce resources to produce goods and services over a period of time and eventually distribute them for consumption now and in the future.

Agricultural economics: this is an applied science that aims at maximizing out put while minimizing costs by combining the limited factors of production to produce goods and services for use by the society over a period of time.

Factors of production

Land
Capital
Labour
Management/entrepreneurship

Basic concepts of economics

These concepts include:

Scarcity
Preference and choice
Opportunity cost
Scarcity

Scarcity means limited in supply

The factors of production named above are scarce and the production needs are many therefore the need for choice

Preference and choice

Since the factors of production are limited, the farmer needs to make a choice on what to produce. This choice has to be guided by the needs of the society and the preference of the farmer on what he needs to produce.

Opportunity cost

When the farmer makes a choice on what to produce, he is forced to leave others due to scarcity of resources eg a piece of land is suitable for growing both rice and maize and he choose to grow maize, the value that could have been derived from rice becomes the opportunity cost.

Opportunity cost is the value of the best foregone alternative

FARM RECORDS

Farm records can be defined as the systemic entries and storage of information of various farm business activities and transactions in appropriate books and sheets.

Uses of farm records

Helps to compare the performance of different enterprises within a farm
Show the history of the farm
Guides a farmer in planning and budgeting of farm operations
Help to detect loses or theft on the farm
Help in the assessment of income tax to avoid over or under taxation
Helps to determine the value of the farm or to determine the assets and liabilities of the farm
Make it easy to share the profits and loses in partnership
Helps in settling dispute among heirs to the estate when a farmer dies without leaving a will
Help to show whether a farm business is making profits or loses
Helps in supporting insurance claims on death, theft etc
Provide labour information like terminal benefits eg NSSF

Types of farm records

Production records
Inventory records
Field operation records
Breeding records
Feeding records
Marketing records
Labour records
Inventory records

This is the physical count of everything that the farm owns and all that it owes others. There are two types of inventory records namely:

Consumable goods inventory
Permanent goods inventory
a) Consumable goods inventory

This is inventory showing a list of goods which normally are used up during a production process, therefore needs constant replacement. Such goods include:

Fertilizers
Livestock feeds
Planting materials eg seeds
Chemicals eg insecticides, herbicides
Sisal ropes and strings etc

Example of consumable goods inventory

RECIEPTS			ISSUES
DATE	ITEM	QUANTITY	DATE	ISSUED TO	QUANTITY	BALANCE IN STORE

b) Permanent goods inventory

This is inventory showing a list of goods which are permanent in nature ie the type of goods which will not get used up in the production process such goods include:

Farm machinery and implements
Farm equipment and buildings
Livestock such as breeding stock
Annual crops
Hand tools
Land (arable)

Example of permanent goods inventory

DATE	ITEM	QUANTITY	WRITTEN OFF	BALANCE IN STOCK	REMARKS

Production records

This is a record which show the total yield from each enterprise and also the yield per unit of the enterprise.

Example of production records (dairy milk production record)

Month…………………………………………………………year……………………………………………………

Name or no. of cow

Days of the month

31^st day

TOTAL

5AM

5PM

5AM

5PM

5AM

5PM

5AM

5PM

BETA

ZABAH

MOON

TOTAL

Example of production record ( crop)

Plot/field No. 15

CROP	SIZE OF FARM	SEED RATE	DATE OF PLANTING	DATE OF HARVESTING	YIELD IN BAGS

Field operation records

This record contains all the activities carried out in the production from land preparation, planting to harvesting. It contains the following information:

Date of land preparation
The size of field
Crop variety planted
Type and amount of fertilizer applied
Seed rate

Example of field operation record

SEASON……………………………………………………………….FIELD NO………………………………………………………

Crop grown………………………………………………………….Variety……………………………………………………………

Ploughing date……………………………………………………..Planting time…………………………………………………

INPUTS

Seed rate kg/ha…………………………………………………………………………………………………………………………….

Fertilizer at planting………………………………………………Amount………………………………………………………..

Top dressing………………………………………………………….Amount……………………………………………………….

Other treatment………………………………………………………………………………………………………………………….

Pests ……………………………………………………………………..Control………………………………………………………..

Diseases ………………………………………………………………..Control……………………………………………………….

Weeds ……………………………………………………………………Control………………………………………………………

Other treatment…………………………………………………………………………………………………………………………..

OUTPUT

Harvesting date…………………………………………………………..Method used…………………………………………

Yield /hac…………………………………………………………………….

Remarks…………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………….

Breeding records

These are records showing the breeding activities and programmes for different animals in the farm.

Importance of breeding records

Help the farmer to plan his breeding programmes
Help in selection of animals within a herd

Example of cattle breeding record

Name/No. of cow	Name of bull/sire	Date of service	Date of pregnancy diognosis	Expected date of calving	Actual date of calving	Sex of calf	Wt. of calf	rmks

N/B: students to draw sheep, pig, and sow breeding records

Feeding records

This is a record showing the type and amount of feeds used in the farm.

Example of feeding records

Month:……………………………………….

Enterprise…………………………………..

Type of feed……………………………….

Date

No. of animals

Amount received (kg)

Amount used (kg)

Balance in stock (kg)

remarks

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List of all Private Colleges in Kenya, Courses List and Fees

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List of all Private Colleges in Kenya, Courses List and Fees

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African Institute of Research and Development Studies
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Cascade Institute of Hospitality
Clastars College
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Equip Africa College of Medical & Health Sciences
Equip Africa Institute
Embu College
Gretsa Institute of Technical & Professional Studies
Hemland College of Proffessional & Technical Studies
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ICDL Africa Ltd
JFC Munene College
Joan School Of Nursing
Jodan College of Technology
KAN College Of Professional Studies
KCA Technical College
Kenya Aeronautical College
Kenya Christian Industrial Training Institute
Kenya Institute Of Management
Kenya Institute of Social Work and Community Development
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Kenya YMCA College of Agriculture & Technology
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Rongo University Course List, Fees, Requirements, How to Apply

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Rongo University Course List, Fees, Requirements, How to Apply

RONGO UNIVERSITY

PROGRAMMES ON OFFER FOR 2023/2024 CYCLE

#	PROGRAMME CODE	PROGRAMME NAME	INSTITUTION TYPE	YEAR 1 – PROGRAMME COST	2022 CUTOFF	2021 CUTOFF	2020 CUTOFF
1	1073101	BACHELOR OF ARTS	PUBLIC	KSH 122,400	27.808	–	–
2	1073109	BACHELOR OF SCIENCE (MATHEMATICS)	PUBLIC	KSH 224,400	19.914	20.100	20.048
3	1073112	BACHELOR OF SCIENCE (BIOCHEMISTRY)	PUBLIC	KSH 244,800	16.974	17.459	16.791
4	1073115	BACHELOR OF SCIENCE (COMPUTER SCIENCE)	PUBLIC	KSH 306,000	18.638	19.223	19.481
5	1073122	BACHELOR OF SCIENCE (AGRICULTURE)	PUBLIC	KSH 275,400	17.043	17.459	16.791
6	1073123	BACHELOR OF SCIENCE (MICROBIOLOGY)	PUBLIC	KSH 244,800	16.974	17.459	16.791
7	1073133	BACHELOR OF COMMERCE (B.COM)	PUBLIC	KSH 183,600	21.444	–	–
8	1073135	BACHELOR OF EDUCATION (ARTS)	PUBLIC	KSH 183,600	27.670	23.905	28.652
9	1073137	BACHELOR OF EDUCATION (SCIENCE)	PUBLIC	KSH 244,800	27.259	22.636	31.536
10	1073150	BACHELOR OF SCIENCE (INFORMATION SCIENCE)	PUBLIC	KSH 204,000	21.444	22.544	22.361
11	1073151	BACHELOR OF BUSINESS MANAGEMENT	PUBLIC	KSH 183,600	21.444	22.544	22.361
12	1073155	BACHELOR OF EDUCATION IN SPECIAL EDUCATION	PUBLIC	KSH 244,800	25.498	22.636	27.545
13	1073157	BACHELOR OF TOURISM MANAGEMENT	PUBLIC	KSH 204,000	24.248	22.544	22.361
14	1073182	BACHELOR OF SCIENCE (COMMUNICATION & JOURNALISM)	PUBLIC	KSH 204,000	27.901	25.295	27.735
15	1073185	BACHELOR OF SCIENCE (HORTICULTURE)	PUBLIC	KSH 275,400	17.043	17.459	16.791
16	1073189	BACHELOR OF SCIENCE IN HUMAN REOURCE MANAGEMENT	PUBLIC	KSH 183,600	25.574	23.052	22.361
17	1073213	BACHELOR OF SCIENCE (ENVIRONMENTAL SCIENCE)	PUBLIC	KSH 244,800	17.043	17.459	16.791
18	1073215	BACHELOR OF SCIENCE (AGRICULTURAL ECONOMICS & RESOURCE MANAGEMENT)	PUBLIC	KSH 275,400	16.974	17.459	16.791
19	1073217	BACHELOR OF SCIENCE (FASHION DESIGN AND TEXTILE TECHNOLOGY)	PUBLIC	KSH 204,000	24.279	17.459	16.791
20	1073222	BACHELOR OF HOTEL AND HOSPITALITY MANAGEMENT	PUBLIC	KSH 204,000	24.915	22.544	22.361
21	1073223	BACHELOR OF TOURISM AND HOSPITALITY MANAGEMENT	PUBLIC	KSH 204,000	22.353	–	–
22	1073247	BACHELOR OF SCIENCE IN SUPPLY CHAIN MANAGEMENT	PUBLIC	KSH 183,600	21.444	–	–
23	1073292	BACHELOR OF ARTS (GEOGRAPHY)	PUBLIC	KSH 153,000	24.656	19.921	24.111
24	1073295	BACHELOR OF ARTS IN INTERNATIONAL RELATIONS	PUBLIC	KSH 153,000	27.008	22.926	26.703
25	1073298	BACHELOR OF SCIENCE (ENTREPRENEURSHIP)	PUBLIC	KSH 183,600	21.444	22.544	22.361
26	1073324	BACHELOR OF ARTS (KISWAHILI)	PUBLIC	KSH 153,000	25.083	22.926	28.638
27	1073327	BACHELOR OF PROJECT PLANNING AND MANAGEMENT	PUBLIC	KSH 183,600	22.895	22.544	22.361
28	1073366	BACHELOR OF SCIENCE (PHYSICS)	PUBLIC	KSH 224,400	16.974	17.459	16.791
29	1073417	BACHELOR OF SCIENCE (AGRICULTURAL EXTENSION AND EDUCATION)	PUBLIC	KSH 275,400	22.531	17.459	16.791
30	1073428	BACHELOR OF SCIENCE (INFORMATICS)	PUBLIC	KSH 204,000	18.638	19.223	19.481
31	1073450	BACHELOR OF SCIENCE (COMMUNICATION AND PUBLIC RELATIONS)	PUBLIC	KSH 204,000	22.916	22.926	25.232
32	1073462	BACHELOR OF SCIENCE (BOTANY)	PUBLIC	KSH 244,800	16.974	17.459	16.791
33	1073464	BACHELOR OF SCIENCE (ZOOLOGY)	PUBLIC	KSH 244,800	16.974	17.459	16.791
34	1073500	BACHELOR OF SCIENCE (GRAPHIC, COMMUNICATION AND ADVERTISING)	PUBLIC	KSH 204,000	26.681	22.926	22.210
35	1073513	BACHELOR OF SCIENCE (CHEMISTRY)	PUBLIC	KSH 224,400	16.974	17.459	16.791
36	1073520	BACHELOR OF ARTS (LINGUISTIC)	PUBLIC	KSH 153,000	22.916	22.926	25.330
37	1073543	BACHELOR OF SCIENCE (HEALTH RECORDS AND INFORMATICS)	PUBLIC	KSH 204,000	23.913	17.459	22.361
38	1073550	BACHELOR OF ARTS (POLITICAL SCIENCE AND PUBLIC ADMINISTRATION)	PUBLIC	KSH 153,000	25.712	22.926	26.095
39	1073555	DIPLOMA IN SALES & MARKETING	UNIVERSITY TVET	KSH 59,420	–	–	–
40	1073557	DIPLOMA IN FASHION DESIGN & CLOTHING	UNIVERSITY TVET	KSH 59,420	–	–	–
41	1073587	BACHELOR OF SCIENCE (AGRIBUSINESS)	PUBLIC	KSH 204,000	16.974	17.459	16.791
42	1073601	DIPLOMA IN ACCOUNTING TECHNICIANS	UNIVERSITY TVET	KSH 59,420	–	–	–
43	1073621	BACHELOR OF ARTS (LINGUISTICS, MEDIA AND COMMUNICATION)	PUBLIC	KSH 204,000	25.134	22.926	25.195
44	1073646	BACHELOR OF ARTS (ECONOMICS)	PUBLIC	KSH 183,600	19.914	20.100	20.048
45	1073657	BACHELOR OF SCIENCE IN HEALTH SYSTEMS MANAGEMENT	PUBLIC	KSH 204,000	17.043	17.459	22.361
46	1073722	DIPLOMA IN GENERAL AGRICULTURE	UNIVERSITY TVET	KSH 59,420	–	–	–
47	1073729	DIPLOMA IN ELECTRICAL & ELECTRONICS	UNIVERSITY TVET	KSH 59,420	–	–	–
48	1073755	DIPLOMA IN HUMAN RESOURCE MANAGEMENT	UNIVERSITY TVET	KSH 59,420	–	–	–
49	1073758	DIPLOMA IN INFORMATION COMMUNICATION TECHNOLOGY (ICT)	UNIVERSITY TVET	KSH 59,420	–	–	–
50	1073771	DIPLOMA IN SOCIAL WORK AND COMMUNITY DEVELOPMENT	UNIVERSITY TVET	KSH 59,420	–	–	–
51	1073806	BACHELOR OF ARTS (SOCIOLOGY)	PUBLIC	KSH 153,000	22.916	22.926	24.492
52	1073876	DIPLOMA IN BUILDING TECHNOLOGY	UNIVERSITY TVET	KSH 59,420	–	–	–
53	1073901	DIPLOMA IN SECRETARIAL STUDIES	UNIVERSITY TVET	KSH 59,420	–	–	–
54	1073B61	BACHELOR OF SCIENCE (APPLIED STATISTICS)	PUBLIC	KSH 244,800	19.914	20.100	20.048
55	1073E69	DIPLOMA IN FOOD & BEVERAGE, PRODUCTION, SALES AND SERVICE MANAGEMENT	UNIVERSITY TVET	KSH 59,420	–	–	–

Academic Programmes

SCHOOL OF AGRICULTURE, NATURAL RESOURCES AND ENVIRONMENTAL STUDIES(SANRES)

PROGRAMME NAME

MINIMUM REQUIREMENTS

MODE OF STUDY

DOCTOR OF PHILOSOPHY IN:

– PLANT BREEDING

– ENVIRONMENTAL BIOLOGY

– ENVIRONMENTAL

PLANNING AND MANAGEMENT

Masters degree in related Fields

Full Time

MASTER OF SCIENCE IN:

– PLANT BREEDING

– HORTICULTURE

– ENVIRONMENTAL SCIENCE

– ENVIRONMENTAL BIOLOGY

– ENVIRONMENTAL PLANNING AND MANAGEMENT

A lower Second Class honors degree in a relevant field and two (2) years relevant work experience Research Publications

Full Time

BACHELOR OF SCIENCE IN HORTICULTURE	KCSE Mean grade of C+ with at least a C(plain) in Mathematics / Physics, C+ in Biology and C+ in Chemistry or KNEC Diploma in Horticulture or its equivalent.	Full Time
BACHELOR OF SCIENCE IN AGRICULTURAL EXTENSION AND EDUCATION	KCSE Mean grade of C+ with at least a C(plain) in Mathematics / Physics, C+ in Biology and C+ in Chemistry or KNEC Diploma in AGED or its equivalent.	Full Time
BACHELOR OF ENVIRONMENTAL SCIENCE	KCSE Mean grade of C+ with at least a C(plain) in Mathematics / Physics, C+ in Biology and C+ in Chemistry or KNEC Diploma in Environmental Science or its equivalent.	Full Time
BACHELOR OF SCIENCE IN AGRICULTURAL ECONOMICS & RESOURCE MANAGEMENT	KCSE Mean Grade of C+, C + in Mathematics and Business / Commerce / Accounting / Economics / C+ in Agriculture, KCSE mean grade C (plain) or a relevant diploma	Full time
BACHELOR OF SCIENCE IN AGRIBUSINESS MANAGEMENT	KCSE Mean Grade of C+ , C + in Mathematics and Business / Commerce / Accounting / Economics / C+ in Agriculture, KCSE mean grade C (plain) or a relevant diploma	Full Time

DIPLOMA IN AGRICULTURE

KCSE mean Grade of C- (minus) with at least

C-(Minus) in either mathematics/physics, C (Plain) in any two of the following subjects (Agriculture, Biology, Chemistry, Biological Science, Physical Science, Home Science & Geography)

A relevant certificate

Full Time

SCHOOL OF ARTS AND SOCIAL SCIENCES (SASS)

PROGRAMME NAME

MINIMUM REQUIREMENTS

MODE OF STUDY

DOCTOR OF PHILOSOPHY IN:

• KISWAHILI

• SOCIOLOGY

• LINGUISTICS

Relevant Masters

Weekend

MASTER OF ARTS IN:

• HISTORY

• KISWAHILI

• RELIGION

• LINGUSTICS

• SOCIOLOGY

• LITRATURE

• GEOGRAPHY

• ECONOMICS

• PUBLIC ADMINISTRATION AND PUBLIC POLICY

Bachelor degree with at least a Lower Second honours with two years relevant work experience

Research Publications

Weekend

BACHELOR OF ARTS IN ECONOMICS	K.C.S.E Mean grade C+ and C+ in Mathematics or a mean grade of C plain and a relevant diploma.	Full time& weekend
BACHELOR OF ARTS	K.C.S.E Mean Grade of C+ or C plain and a Relevant Diploma	Full time& weekend
BACHELOR OF ARTS IN INTERNTIONAL RELATIONS AND DIPLOMACY	K.C.S.E Mean Grade of C+ or a Relevant Diploma	Full time& weekend
DIPLOMA IN LEADERSHIP AND GOVERNANCE	K.C.S.E C-(Minus) or relevant certificate	Full time& weekend
DIPLOMA IN CRIMINOLOGY AND SECURITY STUDIES	K.C.S.E C- (Minus) or relevant certificate	ODEL
DIPLOMA IN SOCIAL WORK	K.C.S.E Mean Grade C- (Minus) or relevant certificate.	Full time& weekend

SCHOOL OF BUSINESS & HUMAN RESOURCE DEVELOPMENT (SBHRD)

PROGRAMME NAME	REQUIREMENTS	MODE OF STUDY
DOCTOR OF PHILOSOPHY IN BUSINESS ADMINISTRATION SPECIALIZATIONS: ACCOUNTING FINANCE PROCUREMENT AND LOGISTICS MANAGEMENT STRATEGIC MANAGEMENT HUMAN RESOURCE MANAGEMENT PROJECT PLANNING & MANAGEMENT	Master’s degree in related field.	Weekend
MASTER OF TOURISM AND HOSPITALITY MANAGE- MENT	A lower Second Class Honours Bachelors Degree or a Cumulative Grade Point Average of (GPA) 2.50 on a scale of 4.00 in a Business related Discipline from	Weekend

	an accredited institution with additional relevant training, evidence of research and relevant working experience OR Postgraduate diploma in a Business Related Discipline	Fulltime/Evening/ Weekend
BACHELOR OF SCIENCE IN HUMAN RESOURCE MANAGEMENT	KCSE Mean Grade of C+ (Plus), C (Plain) in English, C- (Minus) in Mathematics / Business Studies or its equivalent. Diploma in a related field or CPAK / CPS,	Fulltime/ Evening/ Weekend
BACHELOR OF SCIENCE IN PROJECT PLANNING & MANAGEMENT	KCSE Mean Grade of C+ (Plus) with at least a C+ (Plus) in English and C (plain) in Mathematics / Accounting / Commerce or Diploma in a related business course or Diploma in any other discipline from a recognized Institution	Fulltime/ Evening/ Weekend
BACHELOR OF HOTEL AND HOSPITALITY MANAGEMENT	KCSE Mean Grade of C+ (Plus) or A Diploma in a related discipline from a recognized institution or Higher National Diploma in a related discipline from a recognized Institution.	Fulltime/Evening/ Weekend
BACHELOR OF TOURISM MANAGEMENT	KCSE Mean Grade of C+ (Plus) or its equivalent or a Diploma in a related discipline from a recognized / accredited Institutions or Higher National Diploma in a related discipline from a recognized /institution	Fulltime/Evening/ Weekend

BACHELOR OF COMMERCE:

• ACCOUNTING

• FINANCE

• PROCUREMENT & SUPPLY CHAIN MANAGEMENT

• LOGISTICS AND TRANSPORT MANAGEMENT

• STRATEGIC MANAGEMENT

• HUMAN RESOURCE MANAGEMENT

• PROJECT PLANNING AND MANAGEMENT

KCSE Mean grade of C+ (plus) and a minimum of C (plain) in both Mathematics, business studies and English or a Diploma in a related discipline from a

recognized Institutions or Higher National Diploma in a related discipline from a recognized institution

Fulltime/Evening/ Weekend

BACHELOR OF BUSINESS MANAGEMENT:

• ACCOUNTING

• FINANCE AND BANKING

• PURCHASING AND SUPPLIES MANAGEMENT

• HUMAN RESOURCES MANAGEMENT

• BUSINESS ADMINISTRATION

KCSE Mean Grade of C+ (plus) and a minimum of C in both Mathematics or business studies and English, or a Diploma in a related discipline from a recognized Institutions or Higher National Diploma in a related discipline from a recognized institution.

Fulltime/Evening/ Weekend

BACHELOR OF SCIENCE IN SUPPLY CHAIN MANAGEMENT: • PROCUREMENT AND SUPPLIES MANAGEMENT • LOGISTICS AND TRANSPORT MANAGEMENT • PROJECT PLANNING AND MANAGEMENT	KCSE Mean Grade of C+ (plus) and a minimum of C (plain) in both Mathematics, business studies and English or business studies and English, or a Diploma in a related discipline from a recognized Institutions or Higher National Diploma in a related discipline from a recognized institution.	Fulltime/Evening/ Weekend
DIPLOMA IN HOTEL AND HOSPITALITY MANAGEMENT	KCSE Mean Grade of C- (Minus) with at least C- in English or Kiswahili and D+ in Mathematics or its equivalent	Full Time
DIPLOMA IN PROJECT PLANNING MANAGEMENT	KCSE Mean Grade of C- (Minus) with at least C- in English or Kiswahili and D+ in Mathematics or its equivalent	Full Time
DIPLOMA IN HUMAN RESOURCE MANAGEMENT	KCSE Mean Grade of C- (Minus) with at least C- in English or Kiswahili and D+ in Mathematics or its equivalent	Full Time
DIPLOMA IN TOURISM MANAGEMENT	KCSE Mean Grade of C- (Minus) with at least C- in English or Kiswahili and D+ in Mathematics or its equivalent	Full Time

DIPLOMA IN PURCHASING AND SUPPLIES	KCSE Mean Grade of C- (Minus) with at least C- in English or Kiswahili and D+ in Mathematics or its equivalent	Full time
DIPLOMA IN BUSINESS MANAGEMENT	KCSE Mean Grade of C- (Minus) with at least C- in English or Kiswahili and D+ in Mathematics or its equivalent	Full time
POSTGRADUATE DIPLOMA IN STRATEGIC PLANNING AND HUMAN RESOURCE MANAGEMENT	A lower second Class lower division Bachelors degree in a relevant field OR A pass Bachelors degree in any field plus two years post-graduation work experience.	Full time/Evening/ Weekend
CERTIFICATE IN HOTEL AND HOSPITALITY MANAGEMENT	KCSE Mean Grade D+ (Plus) or its equivalent	Full time/Evening/ Weekend
CERTIFICATE IN HUMAN RESOURCE MANAGEMENT	KCSE Mean Grade D+ (Plus) or its equivalent	Full time/Evening/ Weekend
CERTIFICATE IN TOURISM MANAGEMENT	KCSE Mean Grade D+ (Plus) or its equivalent	Full time/Evening/ Weekend

CERTIFICATE IN PROJECT MANAGEMENT	KCSE Mean Grade D+ (Plus) or its equivalent	Fulltime/Evening/ Weekend
CERTIFICATE IN BUSINESS MANAGEMENT	KCSE Mean Grade D+ (Plus) or its equivalent	Fulltime/Evening/ Weekend

SCHOOL OF EDUCATION (SOE)

PROGRAMME NAME

MINIMUM REQUIREMENTS

MODE OF STUDY

DOCTOR OF PHILOSOPHY IN: EDUCATIONAL LEADERSHIP AND POLICY

A Relevant Masters degree in area of specialization recognized by Rongo University Senate.

PhD applications must be submitted together with a research proposal in the area of interest.

Weekend

MASTER OF EDUCATION IN CURRICULUM DEVELOPMENT

Bachelor of Education (Arts or Science) with a Second Class Honours (Lower Division) with at least two years of working experience.

Weekend

MASTER OF EDUCATION IN EDUCATIONAL LEADERSHIP AND POLICY	Bachelor of Education (Arts or Science) with First Class (Honours) or Second Class Honours (Upper Division), or Second Class Honours (Lower Division) with two years of relevant working experience.	Weekend
BACHELOR OF EDUCATION (ARTS)	KCSE mean grade of C+ (plus) with at least C+ (plus) in two teaching subjects. In addition one must have scored grade C (Plain) in En- glish and D+ (Plus) in Mathematics OR KCSE mean grade of C (Plain) at K.C.S.E with Diploma in Education arts with two teaching subjects with C+ (Plus) will also be consid- ered for admission in year 1. In addition the applicant must have scored grade C (Plain) in English and D+ in Mathematics.	Full time
BACHELOR OF EDUCATION (SCIENCE)	K.C.S.E mean grade of C+ and with at least C+ in two teaching subjects. In addition, one must have scored at least grade C (Plain) in both Mathematics and English OR Diploma holders in Education from a recognized Insti- tution will be admitted to year 1 of study OR KCSE mean grade of C (Plain)	Full time

BACHELOR OF EDUCATION IN EARLY CHILDHOOD AND PRIMARY EDUCATION	K.C.S.E with an aggregate grade of C+ and above at K.C.S.E and grade C Plain in English and D+ in Mathematics or Diploma holders in Education (with Credit or Distinction) from a recognized Institution will be admitted to year 1 of study Or KCSE mean	Open and Distance Electronic Learning (ODEL
	grade of C (Plain), P1 certificate, Diploma in Education and at least C (Plain) in English and D+ (plus) in Mathematics
BACHELOR OF EDUCATION IN SPECIAL NEEDS EDUCATION(SNE – ARTS)	K.C.S.E with at least a minimum mean grade of C+ and above or a Diploma Qualification in Special Needs Education (with Credit or Distinction) from Institutions recognized will be admitted to year 1 of study. In addition one must have scored at least C (Plain) in English and grade D+ in Mathematics at K.C.S.E	Full time
BACHELOR OF EDUCATION IN SPECIAL NEEDS EDUCATION(SNE – SCIENCE)	K.C.S.E with at least a minimum mean grade of C+ and above or a Diploma Qualification in Special Needs Education (with Credit or Distinction) from Institutions recognized will be admitted to year 1 of study. In addition, one must have scored at least grade C (Plain) in both English and Mathematics at K.C.S.E	Full time

DIPLOMA IN SECONDARY EDUCATION (ARTS)

KCSE mean grade of C+ (plus) with at least C+ (plus) in two teaching subjects. In addition one must have scored grade C (Plain) in English and D+ (Plus) in Mathematics at K.C.S.E

(Teaching Subjects: English and Literature, Kiswahili, Religion, Geography, History)

DIPLOMA IN SPECIAL NEEDS

EDUCATION(Primary Option)

KCSE mean grade of at least C (Plain) or its equivalent qualification from recognized Institutions.

Full time & ODEL

DIPLOMA IN EARLY CHILDHOOD AND PRIMARY EDUCATION (ECPE)

KCSE mean grade of at least C (Plain) or its equivalent qualification from recognized Institutions.

An applicant with P1 certificate from recog- nized Institutions may also be considered for admission in year 1 of study.

Full time

SCHOOL OF INFORMATION, COMMUNICATION AND MEDIA STUDIES (INFOCOMS)

PROGRAMME NAME	MINIMUM REQUIREMENT	MODE OF STUDY
DOCTOR OF PHILOSOPHY IN COMMUNICATION STUDIES	Relevant Masters Degree in communication or relevant field from a recognized institution.	Weekend
DOCTOR OF PHILOSOPHY IN MEDIA & SECURITY STUDIES	Multidisciplinary Any masters degree	Weekend
MASTERS OF SCIENCE IN MEDIA AND SECURITY STUDIES	Multidisciplinary A lower Second Class Honors degree , two (2) years relevant work experience & evidence of research & publications.	Weekend
MASTERS OF SCIENCE IN COMMUNICATION STUDIES	A lower Second Class honors degree in a relevant field, two (2) years relevant work experience evidence of research & publications.	Weekend
MASTERS OF SCIENCE IN INFORMATION SCIENCE	A lower Second Class Honors degree in a relevant field, two (2) years relevant work experience & evidence of research & publications.	Weekend

MASTERS OF SCIENCE IN INFORMATION TECHNOLOGY	A lower Second Class honors degree in a relevant field and two (2) years relevant work experience	Weekend
MASTERS OF SCIENCE IN HEALTH SYSTEMS & INFORMATICS	A lower Second Class honors degree in a relevant field and two (2) years relevant work experience	Weekend
BACHELOR OF INFORMATION SCIENCE	K.C.S.E mean K.C.S.E mean grade of C+ (plus) or C (plain) with relevant diploma from recognized institution	Full time
BACHELOR OF SCIENCE IN INFORMATICS	K.C.S.E mean grade of C+ (plus) or C(plain) with relevant diploma from recognized institution.	Full time
BACHELOR OF SCIENCE IN COMMUNICATION AND MEDIA STUDIES (WITH SPECIALISATION IN JOURNALISM & PUBLIC RELATION)	K.C.S.E mean grade of C+ (plus) or C (plain) with relevant diploma from a recognized institution.	Full time
BACHELOR OF SCIENCE IN HEALTH RECORDS AND INFORMATION MANAGEMENT	K.C.S.E mean grade of C+ (plus) or C (plain) with a relevant diploma from a recognized institution	Full time

BACHELOR OF SCIENCE IN HEALTH SYSTEM MANAGEMENT	K.C.S.E mean grade of C+ (plus) or C (plain) with relevant diploma from a recognized institution.	Full time
DIPLOMA IN HEALTH RECORDS AND INFORMATION TECHNOLOGY	K.C.S.E mean grade of C (plain) with C (plain) in English or Kiswahili and C- (minus) in Mathematic, Biology/Biological Sciences and C- (minus) in any of the following: Mathematics, Physics, Physical Science, Chemistry , Computer Studies.	Full time
DIPLOMA IN PUBLIC RELATIONS	K.C.S.E mean grade of C- (Minus) or relevant certificate from recognized institution.	Fulltime
DIPLOMA IN MEDIA AND JOURNALISM WITH IT	K.C.S.E mean grade of C- (Minus) or relevant certificate from recognized institution.	Full time
DIPLOMA IN INFORMATION AND COMMUNICATION TECHNOLOGY (ICT)	K.C.S.E mean grade of C- (Minus) or relevant certificate from recognized institution	Full time
DIPLOMA IN LIBRARY AND INFORMATION STUDIES	K.C.S.E mean grade of C- (Minus) or relevant certificate from a recognized institution	Full time
CERTIFICATE IN LIBRARY & INFORMATION STUDIES	K.C.S.E mean grade of D (plain)	Full time

CERTIFICATE IN HEALTH RECORDS AND INFORMATION TECHNOLOGY

K.C.S.E mean grade of C- ( Minus) in English or Kiswahili and D+ in Biology/ Biology Science and D+ (plus) in any of the following: Mathematics, Physics, physical science, Chemistry, General Science , Com- puter Studies, Agriculture or Home Science

Full time

SCHOOL OF SCIENCE TECHNOLOGY & ENGINEERING (SSTE)

PROGRAMME NAME	REQUIREMENTS	MODE OF STUDY
BACHELOR OF SCIENCE (PHYSICS)	KCSE mean grade of C+ (Plus) with C+ in Physics and any two of the following subjects (Mathematics, Biology and Chemistry) or Relevant Diploma.	Full time
BACHELOR OF SCIENCE (CHEMISTRY)	KCSE mean grade of C+ (Plus) with C+ in Physics and any two of the following subjects (Mathematics, Biology and Chemistry) or Relevant Diploma	Full time
BACHELOR OF SCIENCE (FASHION DESIGN & TEXTILE TECHNOLOGY)	KCSE mean grade of C+ (Plus) with C+ in Physics and any two of the following subjects (Mathematics, Biology and Chemistry) or Relevant Diploma	Full time
BACHELOR OF SCIENCE (MATHEMATICS)	KCSE mean grade of C+ (Plus) with C+ in Mathematics and any two of the following subjects (Physics, Biology and Chemistry) or Relevant Diploma	Full time
BACHELOR OF SCIENCE (COMPUTER SCIENCE)	KCSE mean grade of C+ (Plus) with C+ in Mathematics and Physics or Relevant Diploma	Full time

BACHELOR OF SCIENCE (APPLIED STATISTICS WITH COMPUTING)	KCSE mean grade of C+ (Plus) with C+ in Mathematics and Physics or Relevant Diploma	Full time
BACHELOR OF SCIENCE (BIOCHEMISTRY)	KCSE mean grade of C+ with C + in Chemistry and Biology subjects or Relevant Diploma.	Full time
DIPLOMA IN SCIENCE LABORATORY TECHNOLOGY	KCSE mean grade of C – (Minus), or a relevant certificate	Full time
DIPPLOMA IN COMMUNITY HEALTH AND DEVELOPMENT	KCSE mean grade of C – (Minus), or a relevant certificate	Full time
DIPLOMA IN PUBLIC HEALTH	KCSE mean grade of C (Plain), or relevant certificate with C Plain in English/Kiswahili, Biology/Biological Sciences. In addition the applicant should have C – (Minus) in Mathematics and any of the following a) Physics b) Chemistry c) Physical Sciences.	Full time

Enrolment for Certificate, Diploma, Degree and Postgraduate Programmes

School of Agriculture, Natural Resources and Environmental Studies School of Arts & Social Sciences

School of Business and Human Resource Development School of Education

School of Information, Communication and Media Studies School of Science, Technology & Engineering

HOW TO APPLY:

Application forms may be obtained from the Office of the Registrar – Academic Affairs at the Main Campus in Kitere Hill or from our website at www.rongovarsity.ac.ke.

Duly filled application forms with copies of result slips, application fee bank deposit slip (Ksh. 500 for Diploma and Certificate, Ksh. 1000 for Undergraduate and Ksh. 2,000 for Postgraduate Programmes) and other relevant certificates / transcripts should be sent by post, courier or hand delivered to:

ffice of the Registrar – Academic Affairs

O Box 103 – 40404, Rongo Email: regaa@rongovarsity.ac.ke

Tel: + 254 770 308 256, +254 773 296 379

Teachers' Resources

Biology syllabus pdf

October 5, 2025 Maverick John Leave a comment

231 KNEC BIOLOGY SYLLABUS

FORM 1

1. Introduction to Biology

1. Definition of Biology
2. Branches of Biology
3. Importance of Biology
4. Characteristics of living organisms
5. Comparison between plants and animals

2. CLASSIFICATION 1

1. Review and use of magnifying lens
2. External features of plants and animals
3. Necessity and significance of classification
4. Major units of classification( naming)
1. Kingdoms
5. Discussion of Binomial nomenclature

3. THE CELL

1. Definition of cell
2. Structure and functions of parts of a light microscope
3. Use and care of the light microscope
4. Cell structure and functions as seen under microscope
5. Preparation of temporary slides of plant cells
6. Estimation of cell size
7. Cell specialization, tissues, organs, and organ systems.

2. Practical activities
1. Observe, identify, draw and state the functions of parts of the light microscope
2. Prepare and observe temporary slides of plant cells
3. Observe permanent slides of animal cells
4. Comparison between plant and animal cells
5. Observe, estimate size and calculate magnification of plant cells

4. CELL PHYSIOLOGY

1. Meaning of cell physiology
2. Structure and properties of cell membrane (Theories of
membrane structure not required)
3. Physiological properties- diffusion, osmosis and active
transport

4. Factors affecting diffusion, osmosis and active transport
in living organisms

5. Role of diffusion, osmosis and active transport in living
organisms

6. Water relations in plant and animal cells: turgor, plasmolysis, wilting and haemolysis
1. Diffusion as demonstrated with potassium permanganate or potassium iodide/ flower dyes/ coloured plant extracts/ smoke
2. Experiments with visking tubing and living tissues: fresh arrow roots/ cassava/ sweet potatoes/ leaf petioles/ irish potatoes
3. Demonstration of plasmolysis

5. NUTRITION IN PLANTS AND ANIMALS

1. Meaning, importance and types of nutrition
2. Nutrition in plants
1. Definition of photosynthesis and its importance in nature
2. Adaptations of leaf to photosynthesis
3. Structure and function of chloroplast
4. Process of photosynthesis- light and dark stage
5. Factors influencing photosynthesis

3. Chemical compounds which constitute living organisms
1. Chemical composition and functions of carbohydrates, proteins and lipids
2. Properties and functions of enzymes

4. Nutrition in animals

1. Modes of feeding in animals
2. Dentition of a named carnivorous, herbivorous and
omnivorous mammal
3. Adaptation of the three types of dentition of feeding.
4. Internal structure of mammalian teeth.
5. Common dental diseases, their causes and treatment

5. Digestive system and digestion in a mammal human)
1. Digestive system, regions, glands and organs
associated with digestion.
2. Ingestion, digestion, absorption, assimilation and egestion.
6. Importance of vitamins, mineral salts, roughage and water in human nutrition
7. Factors determining energy requirements in humans

FORM 2

1. TRANSPORT IN PLANTS AND ANIMALS

1. Meaning and importance of transport systems
2. Absorption of water and mineral salts
1. Internal structure of root and root hairs
2. Absorption of water
3. Active intake of mineral salts

3. Transpiration
1. Definition of transpiration
2. Review of structure of the leaf.
3. Structure and function of xylem
4. Factors affecting transpiration
5. Forces involved in water movement in plants

4. Translocation
1. Structure and function of phloem
2. Materials translocated

5. Comparison between closed and open circulatory system.
6. Mammalian circulatory system
1. Structure and function of heart, arteries, veins and capillaries
2. Diseases and defects of the heart, arteries, veins, and capillaries
3. Diseases and defects of the circulatory system.
7. The structure and functions of blood
1. Composition of blood
2. Functions of blood plasma
3. The structure and functions of red blood cells and white blood cells
4. Mechanism of blood clotting and its importance

8. Blood groups (ABO system and rhesus factor)
9. Immune responses
1. Natural and artificial immunity
2. Allergic reactions
3. Importance of vaccinations against diseases

10. Practical Activities
1. Observe permanent slides of sections of stems and roots
2. Carry out experiments to compare transpiration on lower and upper surfaces
3. Observe wall charts/models
4. Analyse data on transpiration rate under different environmental conditions in plants
5. Dissect a small mammal and observe its transport system.
6. Make longitudinal section of the mammalian heart
to display the chambers and associated blood vessels
7. Record pulse rate at the wrist before and after vigorous activities and analyse the results
8. Demonstrate the unidirectional flow of blood in the cutaneous veins of the fore arm

2. GASEOUS EXCHANGE

1. Gaseous exchange in living organisms
2. Gaseous exchange in plants
1. Mechanism of opening and closing of the stomata
2. The process of gaseous exchange in roots, stem and
leaves of both aquatic and terrestrial plants

3. Gaseous exchange in animals
1. Types and characteristics of Respiratory surfaces-
cell membrane, gills, buccal cavity, skin and lungs
2. Mechanism of gaseous exchange in protozoa,
insect (grasshopper), fish (bonnyfish),Frog and human
3. Factors affecting rate of breathing in humans

4. Respiratory diseases: Asthma, Bronchitis, Pulmonary
Tuberculosis, Pneumonia and Whooping cough

3. RESPIRATION

1. Meaning and significance of respiration
2. Tissue respiration
1. Mitochondrion- structure and functions
2. Aerobic respiration
3. Anaerobic respiration in plants and animals
4. Application of anaerobic respiration in industry
and at home
5. Compare energy output of aerobic and anaerobic
respiration

4. EXCRETION AND HOMEOSTASIS

1. Excretion in Plants
1. Methods of excretion in plants
2. Useful and harmful excretory products of plants and their economic importance e.g. coffee, caffeine

2. Excretion and homeostasis in animals
1. Difference between excretion, homeostasis and egestion
2. Excretion in a named unicellular organism (protozoa)
3. Structure and functions of skin and kidney
4. Neuro-endocrine system and homeostasis
5. Common kidney diseases, their symptoms and possible methods of prevention and control
6. The role of skin in thermoregulation, salt and water balance
7. Major functions of liver and their contributions to homeostasis
8. Common diseases of liver, their symptoms and possible methods of prevention, control

FORM 3

1. CLASSIFICATION II

1. Review of binomial nomenclature
2. General principles of classification
3. General characteristics of kingdoms
1. Monera
2. Protoctista
3. Fungi
4. Plantae
5. Animalia

4. Main characteristics of major divisions of plantae
1. Bryophyta
2. Pterodophyta
3. Spermatophyta

5. Main characteristics of phyla Arthropoda and Chordata
1. Arthropoda
2. Diplopoda
3. Chilopoda
4. insecta
5. crustacea
6. arachnida

6. Construction and use of simple dichotomous keys based on observable features of plants and animals

2. ECOLOGY

1. Concepts of ecology
1. Habitat
2. Niche
3. Population
4. Community
5. Ecosystem
6. Biomass
7. Carrying capacity

2. Factors in an ecosystem

1. Abiotic factors
2. Biotic factors
3. Inter-relationships:- Competition, predation,
saprophytism, parasitism and symbiosis

4. Nitrogen cycle

3. Energy flow in an ecosystem:- Food chains, food webs, decomposers, pyramid of numbers and pyramid of biomass
4. Population estimation methods
1. Quadrat method
2. Line transect
3. Belt transect
4. Capture-recapture method

5. Adaptations of plants to various habitats
1. Xerophytes
2. Mesophytes
3. Hydrophytes
4. Halophytes

6. Effect of pollution on human beings and other organisms
Causes, effects and control of pollutants in air, water and soil

7. Human diseases
1. Bacterial diseases- Cholera and Typhoid
2. Protozoa- malaria and amoebic dysentry
3. Ascaris lumbricodes and schistosoma

3. REPRODUCTION IN PLANTS AND ANIMALS

1. Concept of reproduction
1. Importance of reproduction
2. Chromosomes, meiosis and mitosis
3. Asexual reproduction
1. Binary fission in amoeba
2. Spore formation/ reproduction in mucor / Rhizopus
3. Budding in yeast

4. Sexual reproduction in plants
1. Structure and function of parts of named insect and
wind pollinated flowers
2. Pollination and agents of pollination
3. Features and mechanisms that hinder self
fertilization and self pollination
4. The process of fertilization
5. Fruit and seed formation and dispersal

5. Sexual reproduction in animals
1. External fertilization in amphibians
2. Structure of the reproductive system of a named mammal (human)
3. Functions of the parts of reproductive system
4. Fertilization, implantation and role of placenta.
5. Gestation period
6. Role of hormones in reproduction in humans

6. Sexually transmitted infections (S.T.Is)
1. Gonorrhea
2. Herpes Simplex
3. Syphilis, Trichomoniasis, hepatitis, Candidiasis
4. HIV/AIDS

7. Advantages and disadvantages of asexual and sexual
reproduction

4. GROWTH AND DEVELOPMENT

1. Concepts of growth and development
2. Growth and development in plants
1. Dormancy and ways of breaking it
2. Conditions necessary for germination
3. Measurement of one aspect of growth in a named
seedling e.g. region of growth
4. Primary and secondary growth
5. Role of growth hormone in plants
6. Apical dominance

3. Growth and development in animals
1. Complete and incomplete metamorphosis in insects
2. Role of growth hormones in insects

* FORM 4

1. GENETICS

1. Concept of genetics
1. Variation within plant and animal species
2. Review of chromosomes
3. Brief mention of genes and DNA (Without details of molecular structure of genes and DNA)

2. First law of heredity
1. Mendel’s experiments- monohybrid inheritance(3:1 ratio)
2. Complete and incomplete dominance, back/ testcross
3. Inheritance of ABO blood groups and Rh factor

3. Sex determination in humans
4. Linkage: sex linked genes, sex linked characteristics e.g. colour blindness, Haemophilia, Hairy ears and nose

5. Mutations
1. Types of mutations
2. Causes and consequences of chromosomal mutations
3. Gene mutations

2. EVOLUTION

1. Meaning of evolution

2. The origin of life
1. Special creation
2. Chemical evolution

3. Evidence of organic evolution
1. Fossil records
2. Geographical distribution – Continental drift
3. Comparative embryology
4. Comparative anatomy
5. Cell biology- occurrence of cell organelles and
blood pigments

4. Mechanisms of evolution
1. Lamarck’s theory (Brief mention)
2. Evolution by natural selection
3. Natural selection in action e.g. peppered moth
4. Resistance to drugs, pesticides and antibiotics

3. RECEPTION, RESPONSE AND COORDINATION IN PLANTS AND ANIMALS

1. Meaning of stimulus, response and irritability
2. Reception, response and coordination in plants

1. Response to a variety of external stimuli
2. Tropisms and tactic movements and their survival values
3. Production of auxins and their effects on plant growth

3. Reception, responses and coordination in animals
1. Components of the nervous system in a mammal
2. Structure and functions of the neurones
3. Functions of major parts of human brain
4. Simple and conditioned reflex actions

4. The role of hormones in coordination in a mammal
1. Effects of over secretion and under secretion of
adrenaline and thyroxine in humans

5. Effects of drug abuse on human health

6. Structure and functions of parts of the mammalian eye
1. Accommodation, image formation and interpretation
2. common eye defects and their corrections

7. Structure and functions of parts of the mammalian ear (human)
1. Hearing
2. Balance and posture

4. SUPPORT AND MOVEMENT IN PLANTS AND ANIMALS

1. Plants
1. Necessity for support and movement in plants
2. Review of tissue distribution in monocotyledonous
and dicotyledonous plants

2. Animals
1. Necessity for support and movement in animals
2. Types and functions of the skeleton

3. Locomotion in a finned fish

4. Identification of the bones of axial and appendicular
skeletons (names of individual bones of coccyx not required)
5. Types and functions of movable joints (ball and socket, hinge joint)
6. Structure, function and location of cardiac, smooth and skeletal muscles

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Form One Termly Exams in All Subjects Plus Marking Schemes

October 5, 2025 Maverick John Leave a comment

Form One Termly Exams in All Subjects Plus Marking Schemes

KISWA QNS.docx
AGRI MS.docx
AGRI QNS.docx
BIO MS.docx
BIO QNS.docx
BUST MS.docx
BUST QNS.docx
CHEM MS.docx
CHEM QNS.docx
COMP MS.docx
COMP QNS.docx
CRE MS.docx
CRE QNS.docx
ENG MS.docx
ENG QNS.docx
FRENCH MS.docx
FRENCH QNS.docx
GEO MS.docx
GEO QNS.docx
HIST MS.docx
HIST QNS.docx
HOMESCIENCE MS.docx
HOMESCIENCE QNS.docx
IRE F1 MS.docx
IRE F1 QNS.docx
IRE MS.docx
IRE QNS.docx
KISWA MS.docx
MATHS MS.docx
MATHS QNS.docx
PHY MS.docx
PHY QNS.docx

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Tharaka University Course List, Fees, Requirements, How to Apply

October 5, 2025 Maverick John Leave a comment

Tharaka University Course List, Fees, Requirements, How to Apply

Bachelor Courses

Business Studies Grade 7 CBC Free Schemes of Work

October 5, 2025 Maverick John Leave a comment

GRADE 7 BUSINESS STUDIES SCHEME OF WORK TERM 3

SCHOOL………………………….TEACHER’S NAME…………………. TERM THREE YEAR

Week	Lesson	Strand	Sub-strand	Specific-Learning outcomes	Learning Experience	Key Inquiry Question(S)	Learning Resources	Assessment Methods	Reflection
1	1	Government and global influence in business	Government and business	By the end of the lesson, the learner should be able to: a) Define the term investor. b) Identify the need for government involvement in business in Kenya. c) Justify the need for government involvement in business activities in Kenya. d) Appreciate the importance of government in conducting businesses.	Learners are guided to define the term investor. Learners to identify the need for government involvement in business in Kenya. In groups or in pairs, learners to justify the need for government involvement in business activities in Kenya.	Who is an investor?	Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 133-135 Photographs Pictures Video clips Charts Computing devices	Oral questions Oral Report Observation
	2	Government and global influence in business	Government and business	By the end of the lesson, the learner should be able to: a) Explain ways in which the shopkeeper in their local community may exploit his or her customer. b) Debate on the need for government involvement in business in Kenya. c) Debate on the motion: “There is need for the government of Kenya to be involved in Business” d) Have fun and enjoy the debate.	Learners to explain ways in which the shopkeeper in their local community may exploit his or her customer. In groups or in pairs, learners are guided to debate on the need for government involvement in business in Kenya, on the motion: “There is need for the government of Kenya to be involved in Business”	What is exploitation? What is foreign dominance?	Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 135-137 Photographs Pictures Video clips Charts Realia Computing devices	Oral questions Oral Report Observation
	3	Government and global influence in business	Legal requirement for starting and operating a simple business in Kenya	By the end of the lesson, the learner should be able to: a) Identify the legal requirement for starting and operating a simple business in Kenya. b) Give examples of businesses in Kenya that require a Health Certificate. c) Draw the certificate of registration in learner’s book. d) Appreciate the importance of the legal requirements for starting and operating a simple business in Kenya.	Learners to identify the legal requirement for starting and operating a simple business in Kenya. Learners to give examples of businesses in Kenya that require a Health Certificate. Individually, learners to draw the certificate of registration in learner’s book 7 pg. 139	What is considered a simple business in Kenya? What are the examples of simple business in Kenya?	Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 137-141 Photographs Pictures Video clips Charts Computing devices	Oral questions Oral Report Observation
2	1	Government and global influence in business	Extended activity	By the end of the lesson, the learner should be able to: a) Organize to visit county or sub-county office of the Ministry of Industrialization, Trade and Enterprise Development or their local county offices. b) Write a report of their findings. c) Have fun and enjoy the visit to the county or sub-county offices.	As a class, learners to organize visit county or sub-county office of the Ministry of Industrialization, Trade and Enterprise Development or their local county offices. Learners to write a report of their findings.	Which office did you visit?	Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 141 Photographs Pictures Video clips Charts Realia Computing devices	Oral questions Oral Report Observation
	2	Government and global influence in business	Role of government in business in Kenya	By the end of the lesson, the learner should be able to: a) Outline the role of government in business in Kenya. b) List any malpractices by traders against which consumers may need protection from the government. c) Recognise the role of government in business in Kenya. d) Appreciate the role of government in business in Kenya.	Learners to outline the role of government in business in Kenya. Learners to list any malpractices by traders against which consumers may need protection from the government. Learners are guided to recognise the role of government in business in Kenya.	Why does the government of Kenya train and advise people in business?	Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 141-144 Photographs Pictures	Oral questions Oral Report Observation
	3	Government and global influence in business	Taxation in Kenya	By the end of the lesson, the learner should be able to: a) Define the term tax and taxation. b) Read the case study in learner’s book. c) Have a desire to learn more about tax.	Learners to define the term tax and taxation. In groups, learners to read the case study in learner’s book 7 pg. 145	What is the meaning to tax and taxation?	Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 144-146 Pictures Video clips Computing devices	Oral questions Oral Report Observation
3	1	Government and global influence in business	Taxation in Kenya	By the end of the lesson, the learner should be able to: a) Identify values that patriotic Kenyan citizens who pays taxes possess are hidden. b) Play the game; unscramble the letters to find the hidden words. c) Have fun and enjoy playing the game.	Learners to identify values that patriotic Kenyan citizens who pays taxes possess are hidden. In groups or in pairs, learners to play the game; unscramble the letters to find the hidden words.	Who is a patriotic Kenyan citizen?	Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 146 Assessment books. Computing devices	Oral questions Oral Report Observation
	2	Government and global influence in business	Importance of paying taxes in Kenya	By the end of the lesson, the learner should be able to: a) Examine the importance of paying taxes. b) Role play the conversation in learner’s book. c) Have fun and enjoy the role playing.	Learners to examine the importance of paying taxes. In groups or in pairs, learners to role play the conversation in learner’s book 7 pg. 149	What is the importance of paying taxes?	Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 148-150 Photographs Pictures Video clips Computing devices	Oral questions Oral Report Observation
	3	Government and global influence in business	Importance of paying taxes in Kenya	By the end of the lesson, the learner should be able to: a) State the importance of paying taxes. b) Make a poster with messages on the importance of paying taxes. c) Appreciate the importance of paying taxes.	In groups or in pairs, learners to state the importance of paying taxes. In groups or in pairs, learners to make a poster with messages on the importance of paying taxes	What other ways can the government utilize taxes to promote development in the community?	Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 150-152 Pictures Video clips Computing devices	Oral questions Oral Report Observation
4	1	Government and global influence in business	The need for paying taxes in Kenya	By the end of the lesson, the learner should be able to: a) Identify the need for paying taxes in Kenya. b) Investigate the need for paying taxes in Kenya. c) Appreciate the importance of paying taxes in Kenya.	Learners to identify the need for paying taxes in Kenya. In groups or in pairs, learners to investigate the need for paying taxes in Kenya.	What services provided by the government do they think their tax money should be used for?	Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 152-154 Photographs Pictures Computing devices	Oral questions Oral Report Observation
	2	Government and global influence in business	Promoting regional unity	By the end of the lesson, the learner should be able to: a) Explain the meaning of good governance. b) Read the story in learner’s book and answer the questions that follow. c) Develop a desire to pay taxes as a Kenyan citizen.	Learners are guided to explain the meaning of good governance. In pairs, learners are guided to read the story in learner’s book 7 pg. 155 and answer the questions that follow	What is good governance?	Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 154-156 Photographs Pictures Video clips Computing devices	Oral questions Oral Report Observation
	3	Financial records in business	Business transactions	By the end of the lesson, the learner should be able to: a) Define the terms business transaction and financial documents. b) Differentiate business and non-business transactions. c) Have a desire to do business transactions.	Learners to define the terms business transaction and financial documents. Learners to differentiate business and non-business transactions.	What is business transaction? What are financial documents?	Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 157-159 Photographs Pictures Video clips Computing devices	Oral questions Oral Report Observation
5	1	Financial records in business	Extended activity	By the end of the lesson, the learner should be able to: a) Identify and visit two or three local business people in their neighbourhood. b) Talk to them about the activities they carry out in their businesses daily. c) Have fun and enjoy visiting local business people.	As a class, learners are guided to identify and visit two or three local business people in their neighbourhood. As a class, learners are guided to talk to them about the activities they carry out in their businesses daily.	What are donation?	Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 159-160 Photographs Pictures Computing devices	Oral questions Oral Report Observation
	2	Financial records in business	Cash and credit transactions in business	By the end of the lesson, the learner should be able to: a) Explain the meaning of cash and credit transactions in business. b) Match each business transaction with the correct description of how it occurs. c) Have a desire to learn more about cash and credit transactions in business.	Learners to explain the meaning of cash and credit transactions in business. In pairs, learners are guided to match each business transaction with the correct description of how it occurs.	What is cash and credit transactions?	Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 160-161 Photographs Pictures Video clips Computing devices	Oral questions Oral Report Observation
	3	Financial records in business	Extended activity	By the end of the lesson, the learner should be able to: a) Role-play the scenarios of various business transactions (people buying and selling) b) Identify whether they are cash transactions or credit transactions. c) Take turns until everyone has demonstrated business transactions and other members have identified their types. d) Enjoy demonstrating to their class members.	In groups or in pairs, learners to role-play the scenarios of various business transactions (people buying and selling) Learners to identify whether they are cash transactions or credit transactions. In groups or in pairs, learners to Take turns until everyone has demonstrated business transactions and other members have identified their types.	What have you learnt about cash and credit transactions?	Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 161-164 Photographs Pictures Video clips Charts Realia Computing devices	Oral questions Oral Report Observation
6	1	Financial records in business	Financial documents used in buying and selling	By the end of the lesson, the learner should be able to: a) Brainstorm on and list financial documents used in buying and selling. b) Share experiences on financial documents in buying and selling in everyday life. c) Appreciate the documents used in buying and selling.	In groups or in pairs, learners to brainstorm on and list financial documents used in buying and selling. In groups or in pairs, learners to share experiences on financial documents in buying and selling in everyday life.	Which other documents are used in buying and selling?	Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 164 Photographs Pictures Video clips Charts Computing devices	Oral questions Oral Report Observation
	2	Financial records in business	Financial documents used in buying and selling	By the end of the lesson, the learner should be able to: a) Identify financial documents used in buying and selling. b) Act out the conversation in learner’s book. c) Have fun and enjoy acting out the conversation.	Learners to identify financial documents used in buying and selling. In groups or in pairs, learners to act out the conversation in learner’s book 7 pg. 165	What is the conversation about?	Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 165-166 Photographs Pictures Video clips	Oral questions Oral Report Observation
	3	Financial records in business	Extended activity	By the end of the lesson, the learner should be able to: a) Search for samples of financial documents used in buying and selling. b) Draw the samples in learner’s book and answer the questions that follow.] c) Appreciate the financial documents used in buying and selling.	In groups or in pairs, learners to search for samples of financial documents used in buying and selling. In pairs or individually, learners to draw the samples in learner’s book 7 pg. 167 and answer the questions that follow	Who sends the financial document and to whom is it sent?	Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 167-168 Pictures Realia Computing devices	Oral questions Oral Report Observation
7	1	Financial records in business	Extended activity	By the end of the lesson, the learner should be able to: a) Visit a financial expert to discuss financial documents used in a business transaction. b) Identify each sample financial document that he or she will provide. c) Have fun and enjoy the visit to a financial expert.	As a class, learners are guided to visit a financial expert to discuss financial documents used in a business transaction. As a class, learners to identify each sample financial document that he or she will provide.	What information is found on the financial document?	Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 168-169 Photographs Pictures Video clips Charts Computing devices	Oral questions Oral Report Observation
	2	Financial records in business	Extended activity	By the end of the lesson, the learner should be able to: a) Organise and visit business people in the local market or trading centre neighbouring the school. b) Identify the gaps in keeping financial records. c) Enjoy visiting business men and women.	As a class, learners are guided to organise and visit business people in the local market or trading centre neighbouring the school As a class, learners are guided to identify the gaps in keeping financial records	What is the financial document used for?	Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 169-179 Photographs Pictures Video clips Charts Realia Computing devices	Oral questions Oral Report Observation
	3	Financial records in business	Methods used in making payments for goods and services	By the end of the lesson, the learner should be able to: a) Identify the methods used in making payments for goods and services. b) Analyse methods of payment for goods and services. c) Discuss how each method of payment is used. d) Appreciate the methods used in making payments for goods and services.	Learners are guided to identify the methods used in making payments for goods and services. In groups or in pairs, learners to analyse methods of payment for goods and services. In groups or in pairs, learners to discuss how each method of payment is used.	What are the methods used in making payments for goods and services?	Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 173-174 Photographs Pictures	Oral questions Oral Report Observation
8	1	Financial records in business	Methods used in making payments for goods and services	By the end of the lesson, the learner should be able to: a) State the advantages and disadvantages of each method of payment for goods and services. b) Match methods of payments for goods and services. c) Appreciate the advantages each method of payment for goods and services.	Learners are guided to state the advantages and disadvantages of each method of payment for goods and services. Learners are guided to match methods of payments for goods and services.	What are the advantages and disadvantages of each method of payment for goods and services?	Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 175-176 Photographs Video clips Charts Computing devices	Oral questions Oral Report Observation
	2	Financial records in business	Extended activity	By the end of the lesson, the learner should be able to: a) Interview the school bursar or business people in the local community. b) Discuss with them the method of payment which they preferred and why. c) Appreciate the role of financial documents in record keeping business.	As a class or in groups, learner to interview the school bursar or business people in the local community. As a class or in groups, learner to discuss with them the method of payment which they preferred and why.	What is the difference between mobile money transfer and electronic funds transfer?	Spark; StoryMoja, Business Studies Learner’s Book Grade 7 pg. 176-178 Video clips Computing devices	Oral questions Oral Report Observation
	3			REVISION
9				ASSESSEMENT

Teachers' Resources

Maths Top Student Form one to four Revision Kit

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.	Y LogT 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 = 2 3 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 – 8 dt (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 = ¼

y – 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 = 3-1

5-9

= 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 60 AE 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 9 9y – 27 = x – 2 9y – x = 25	B1 B1 M1 A1	Grad equ Grad of normal Eqn Eqn
		4
5	700 = 100 + n 2200 = 400 + n 1500 = 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 49.50 110 = 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

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HOW AGRICULTURE KCSE EXAMS ARE SET (KNEC GUIDE)

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AGRICULTURE (443)

PAPER 1 (90 MKS)

FORM 1 TOPICS

Introduction to Agriculture
Factors influencing Agriculture
Crop production I – Land preparation
Water supply, Irrigation and Drainage
Organic manure
Basic economics – Farm Records

FORM 2 TOPICS

Inorganic fertilizers
Crop production 2 – Planting
Nursery practices
Field practices
Vegetables

FORM 3 TOPICS

Land tenure and Land reforms
Soil and water conservation measures
Weeds and weed control
Crop pests and pest control
Crop diseases and disease control
Specific crops

Maize, sorghum, finger millet, bulrush millet, beans, rice
Industrial crops – cotton, sugarcane, pyrethrum

Forage crops

FORM 4 TOPICS

Production economics
Farm accounts
Agricultural marketing and organizations
Agro forestry.

PAPER 2 (90 MKS)

FORM 1 TOPICS

Farm tools and equipment
Livestock breeds

FORM 2 TOPICS

Introduction to livestock health
Parasitwes and parasite control
Livestock nutrition

FORM 3 TOPICS

Selection and breeding
Livestock rearing practices
Farm structures
Livestock diseases

FORM 4 TOPICS

Poultry production
Cattle production
Farm power and machinery

C. S. E. FORMAT

SECTION A (30 MKS)

State, list, name, give

SECTION B (20 MKS)

Photographs, diagrams,

SECTION C (40 MKS)

Explain, describe, outline, State and explain

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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

MATHEMATICS II

Using 74.5 as the Assumed mean, calculate:

MATHEMATICS II

MATHEMATICS IV

A1

A1

B1

A1

A 1

A 1

A1

A1

A1

A1

B1

B1

A1

A1

B1

A1

B1

A1

LOG E. 0.3010 0.6021 0.7782 0.9031

B1

B1