TSC score sheet 2021-2022 for p1

The Teachers Service Commission, TSC, has produced the final and newly updated Scoresheet For Primary Teachers Recruitment in 2021-2022.

Just like the case of secondary teacher recruitment, length of stay and P1 qualifications also play a vital role in employment of P1 teachers. Apart from length of stay since completion of teaching course, the age bracket of an individual has some weight in ones employment.

During the selection day one should carry the following original documents to sub-county selection panel;

  1. National Identity card.
  2. CPE/KCPE certificate (including first attempt certificate if one repeated exam).
  3. PTE certificate.
  4. School/college leaving certificates.

Sub county selection panel shall include the following;

  1. Sub-County Director – Chairperson
  2. TSC Sub-County Human Resource Officer – Secretary
  3. All zonal Curriculum Support Officer – Members

Here is the score sheet for P1 recruitment and employment 2021/2022

It is important to note that those teachers who are currently serving under internship programme will have an advantage of thirty(30) marks.

P1 Qualifications (SORING AREAS) SCORE
6-14 points 40
15-26 points 35
27-38 points 30
39-52 points 25
53-64 points 20
MAXIMUM SCORE 40
Evidence of internship service 30
LENGTH OF STAY SINCE GRADUATION FOR THOSE WHO HAVE NEVER BEEN EMPLOYED BY TSC
2015 and before 30
2016 25
2017 20
2018 15
2019 10
2020 and after 5
MAXIMUM SCORE 30
LENGTH OF STAY SINCE GRADUATION FOR THOSE WHO WERE PREVIOUSLY EMPLOYED BY TSC
2020 and after 30
2019-2018 25
2017-2016 20
2015-2014 15
2013-2012 10
2011 and before 5
MAXIMUM SCORE 30
GRAND TOTAL 100

Where applicants tie in score, the panel is advised to apply the criteria below in the order given:

  • Age of the applicant (44-45 years) with previous teaching experience of three years and had not previously been employed by the Commission.
  • Give preference to applicants with highest grade obtained in Teaching Practice;
  • Use quality of the academic KCSE certificates;
  • Give preference to applicants who graduated earlier;
  • Give preference to applicants who had not been previously employed by the Commission.

Here is the score sheet for P1 recruitment and employment for teachers with Special Needs;

P1 QUALIFICATIONS (SORING AREAS) SCORE
6-27 points 40
28-39 points 35
40-53 points 30
54-64 points 25
MAXIMUM SCORE 40
Evidence of internship service 30
LENGTH OF STAY SINCE GRADUATION
2017 and before 40
2018 35
2019 30
2020 and after 25
MAXIMUM SCORE 40
GRAND TOTAL 100

Where applicants tie in score, the panel is advised to apply the criteria below in the order given:

  • Age of the applicant (44-45 years) with previous teaching experience of three years and had not been previously employed by the Commission;
  • Give preference to applicants with highest grade obtained in Teaching Practice;
  • Use quality of the academic certificates KCSE;
  • Give preference to applicants who graduated earlier.
  • Give preference to applicants who had not been previously employed by the Commission.

Complaints (What to do in case of dissatisfaction)

Keep in mind that the process of selection is transparent and only a qualified candidate carries the day. However, if one is not satisfied with process, there is a laid down mechanism to register your dissatisfaction.

Any candidate who is dissatisfied with the whole exercise of employment should complain immediately in writing to the TSC county director and a copy sent to TSC headquarters within a period of not more than 7 days after the selection exercise. TSC county director will analyze such complaints and submit a report to TSC headquarters with necessary recommendations.

Must read;

TSC New Updated Scoresheet For Intern Recruitment 2021-2022

How to apply for TSC internship vacancies online at https://hrmis.tsc.go.ke/app/recruitP/

TSC guidelines for intern teachers recruitment, score sheet and marking scheme 2021/2022

Gikigie Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Gikigie Secondary School is a public Mixed, (Boys’ and Girls’) Sub-County Level Day  School that is located at Kigumo Subcounty in Murang’a County of the Central Region, Kenya. The School’s Official Phone Number Contact is: +254721316549

Key Details about the school.

Country where found: Kenya.

Region: Central.

County: Murang’a.

Subcounty: Kigumo.

School Type/ Ownership: A Public School.

Nature os School/ CBE Level: Senior School (SS).

Category: Regular School

School’s Official Name: Gikigie Secondary School

Sex: Mixed, (Boys’ and Girls’)  School.

School Cluster/ Level: Sub-County School whose Classification is C4.

Accomodation Type: Day  School.

Knec Code:  10227317

School’s Official Phone Number:  +254721316549 

Total Number of Subjects Combinations Offered at the School: 3

Subject Combinations Offered at Gikigie Secondary School

View all available subject combinations at this school

SOCIAL SCIENCES

1
LANGUAGES & LITERATURECode: SS1081
Fasihi ya Kiswahili,History & Citizenship,Literature in English
3 SubjectsSOCIAL SCIENCES

STEM

2
APPLIED SCIENCESCode: ST2044
Agriculture,Biology,Business Studies
3 SubjectsSTEM
PURE SCIENCESCode: ST1020
Advanced Mathematics,Chemistry,Physics
3 SubjectsSTEM

📍 How to get more Information about the School

For more information about admission requirements, facilities, and application procedures, contact the school directly. Use the official phone number indicated above to get information about the school’s fees, uniform, meals and performance.

How to Select Grade 10 Subjects and schools

To select Grade 10 schools and subjects under the Competency-Based Curriculum (CBC) in Kenya, Grade 9 learners should first choose a career pathway (STEM, Social Sciences, or Arts & Sports Science). Then, they’ll select three subject combinations within that pathway and finally, choose four schools for each combination, totaling 12 schools. To select preferred Grade 10 Schools and Subject Combinations, use the Ministry of Education portal selection.education.go.ke.

1. How you can Choose a Career Pathway:

  • Identify your interests and potential career aspirations.
  • Select one of the three pathways: STEM, Social Sciences, or Arts & Sports Science.
  • Confirm your choice to proceed with the pathway.

2. Select Subject Combinations:

  • The portal will provide you with a list of subject combinations available within your chosen pathway.
  • Choose three subject combinations that align with your interests and strengths.

3. Select Preferred Senior Schools:

  • For each subject combination, select four schools from the available clusters.
  • This ensures a diverse range of options and equal representation from different categories of schools.
  • A total of 12 schools will be selected: 4 for the first subject combination, 4 for the second, and 4 for the third.

LIST OF ALL SENIOR SCHOOLS PER COUNTY.

Senior School Subjects and Pathways selection Form.
Senior School Subjects and Pathways selection Form.

Senior School Selection Form educationnewshub.co.ke

Diploma in Medical Social Work Course, Kuccps Requirements and Campuses

Diploma in Medical Social Work Course, Kuccps Requirements and Campuses

S.N

O

PROGRAM

ME CODE

CAMPUS PROGRAMME NAME MEAN

GRADE

SUBJECT 1 SUBJECT 2 SUBJECT 3 SUBJECT 4
106. 4785K18 IMENTI DIPLOMA IN MEDICAL SOCIAL WORK C ENG/KIS:C BIO/BSC:D+ CHE/PHY/PSC/MAT A/HSC/AGR/BST/GEO/

HAG/CRE/IRE:C-

 
107. 4820K18  

KAPKATET

DIPLOMA IN MEDICAL SOCIAL WORK C ENG/KIS:C BIO/BSC:D+ CHE/PHY/PSC/MAT

A/HSC/AGR/BST/GEO/ HAG/CRE/IRE:C-

 
108. 4835K18 KARURI DIPLOMA IN MEDICAL SOCIAL WORK C ENG/KIS:C BIO/BSC:D+ CHE/PHY/PSC/MAT A/HSC/AGR/BST/GEO/

HAG/CRE/IRE:C-

 
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Important KMTC Links

KMTC Students,

KCSE MATHS TOP STUDENT REVISION RESOURCE

MATHEMATICS 1

PART I

SECTION A: 

  1. Use logarithm tables to evaluate                      (4 mks)

 

0.0368 x 43.92

361.8

 

  1. Solve for x by completing the square                           (3mks)

2x2  – 5x + 1 = 0

 

  1. 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)

 

  1. 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)

 

  1. 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)

 

  1. Find the value of θ within θ  < θ < 360if  Cos (2 θ + 120) =  γ3                                                     (3mks)

2

 

  1. 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)

 

  1. A rectangle measures 3.6 cm by 2.8 cm. Find the percentage error in calculating its perimeter.                                                                                                                                                 (3mks)

 

  1. Evaluate:          11/6   x  ¾  –  11/12                                                                                              (3mks)

½  of 5/6

 

  1. 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)

 

  1. A regular octagon has sides of 8 cm. Calculate its area to 3 s.f.             (4mks)

 

  1. Find the values of x and  y if                                                                                                       ( 2 mks)

3          x          1   =     2

2          1          -1         y

 

  1. An equation of a circle is given by x2 + y2 – 6x + 8y – 11 = 0                                           (3mks)

Find its centre and radius

 

 

 

 

 

 

  1. In the figure given AB is parallel to DE. Find the value of x and y

 

 

 

 

 

 

 

 

  1. 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.

 

  1. 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:

 

  1. Construct without using a protractor the triangle ABC so that BC=10cm, angle ABC = 600 and

BCA = 450

  1. On the diagram , measure length of AC
  2. Draw the circumference of triangle ABC
  3. Construct the locus of a set of points which are equidistant from A and B.
  4. Hence mark a point P such that APB = 450 and AP = PB
  5. Mark a point Q such that angle AQB = 450 and AB = AQ

 

  1. (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 AI B I CI DI. under a second transformation

0 – 2

which has a rotation centre (0,0) through –900 , the image AII  BII  CII  DII  of AI  BI  CI  DI  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

  1. Express AB and MC in terms of a and b
  2. By writing MN in two ways, find the ratio in which N divides
  3. AB
  4. MC

 

 

 

 

 

 

  1. In the figure below, SP = 13.2 cm, PQ = 12 cm, angle PSR = 80O 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

 

 

 

 

 

 

 

 

 

 

 

  1. 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

 

  1. 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)

 

  1. 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

  1. Find the times taken to complete the uphill and downhill sections of the course.
  2. Calculate the cyclists average speed over the 150km.

 

  1. 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
  2. Find the length of XB as a fraction
  3. Show that XAD is similar to XBC
  4. Given that the area of AXD = 6cm2, find the area of BXC
  5. Find the value of the ratio

Area of       AXB

Area of        DXC

 

 

 

 

 

 

 

  1. 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 2550 from 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)

(c)  The bearing of  C from A                                                                                 (1mk)

 

MATHEMATICS I

PART 1

MARKING SCHEME             (100MKS)

 

 

  1. 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

 

  1. 2 x2 – 5x + 1 = 0

x2 5 x + ½ = 0

2

x25 x   = ½

2

x – 5x  +     5 2    =  ½   +     5    (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

 

  1. 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

  1. 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)

 

  1. 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

25

 

  1. 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= 315o    ( for >2)

2 θ =  750 – 120   = 6300 ,

 

  1. 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

 

  1. 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 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)

 

  1. 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

                          8

Tan 67.5 =  h

4

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)

 

  1. 3   2        -1             2

=

2              1          -1           y

 

3 – x = 2       (1)       x = 1                          (2)

2 –  1 = y                 y = 1  (A)

 

  1. 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)

 

  1. (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 + 10   = 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)

 

 

  1. Acceleration = Change in velocity

Time

= (34 – 10) m/s                  = 24 m/s

60 x 10                                600

 

= 0.04m/s2-                                (2)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

17.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Triangle                                    (8)

AC = 9cm

Circumference Centre

Circle

Perpendicular bisector of AB

P

Q

 

 

 

  1. (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

= 3 – 6/5 = 9/5

s = 9/25

 

(i)    AN :     NB = 2 : 3

 

(ii)   MN :    9   :  16

 

 

 

 

 

 

 

 

20.

 

 

θ x pr2

360

 

  1. 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

 

 

  1. 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.

 

 

 

 

 

 

  1. 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

2

V2    =   – 20 – 80      X   impossible

2

speed uphill      =  30 km /hr,  T = 75  time =  2 ½ hrs

30

speed downhill =  50 km /hr  Time = 75      Time =  2 ½ hr

50

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

 

 

  1. A                 B

 

 

 

 

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

4

 

24.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. a) AD =   50km

DC   =   35km

BD  = 90km

Bearing is 020

Bearing is 134o                                                                                                       (8mks)

 

 

MATHEMATICS  I

PART II

 

SECTION (52 MARKS)

 

  1. Without using tables, simplify

1.43 x 0.091 x 5.04

2.86 x 2.8 x 11.7                                                                                             (3mks)

 

  1. Make x the subject of the formula if

y = a/x  +  bx                                                                                                    (3mks)

 

  1. Give the combined solution for the range of x values satisfying the inequality

2x + 1<  10 – x  <   6x – 1                                                                                 (3mks)

 

  1. 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)

 

  1. 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 140o.  Find

(i) The distance of A to C                                                                                        (2mks)

(ii) The bearing of A from C                                                                                          (3mks)

 

  1. Expand (x + y)6 hence evaluate (1.02) to 3d.p.                                                         (3mks)

 

  1. Rationalise the denominator in                                                                               (2mks)

 

Ö 3

1 – v3

 

 

 

  1. 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)

 

  1. Find the image of the line y = 3x = 4 under the transformation whose matrix is.

3mks

2           1

-1         2

 

  1. Three points are such that A (4 , 8), B(8,7), C (16, 5). Show that the three points are collinear                                                                                                                                          (3mks)
  2. 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)

 

  1. 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)

  1. 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 60o and 30o respectively, find the height of the hill                                                                                           (4 mks)
  2. 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)
  3. If log x = a and log y = b, express in terms of a and b

Log  x 3 

VY                                                                                                             (2mks)

 

SECTION B:

 

  1. 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

  1. The mean
  2. The standard deviation
  3. Find the median mark

 

 

 

  1. Draw the graph of y = 2x2 – x – 4 for the range of x -3  = x  =  3.  From your  graph

State the minimum co-ordinates

  1. Solve the equations
  2. 2x2 – x – 4 = 0
  3. 2x2 – 3x – 4 = 0

 

 

 

 

 

 

  1. 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

 

  1. Two unbiased dice are thrown. Find the probability of obtaining (4mks)
  2. A product of 6
  3. 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 = 60o), Find

  1. BD (Length)
  2. CD

iii.  Arc length CHG and DEF, hence find the length of the rubber.

 

  1. 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

  1. The perpendicular height of the pyramid
  2. The length of the slant edge VA
  3. The angle between edge VA and base A B C D
  4. The area of the face VAB

 

  1. Plot the graphs of y = sin xo and y = cos 2xo on the same axes for –180 £ x £180o.

Use your graphs to solve the equation 2 sin x = cos  2x

 

  1. 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)

 

 

 

  1. 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.

  1. Draw on a cartesian plane, the line y = x  and mark points B1 and D1
  2. Mark the points B and D before reflection
  3. 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.

  1. Find the image of D under a rotation, through – 900, Center the origin.

 

 

MATHEMATICS I

PART II

MARKING SCHEME.

  1. 1.43 X 0.091 X 5.04100000        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)

  1. y = a/x + bx yx = a + bx2

Either

bx2 – yx + a = 0

 

x =     y   ±   v y2 –  4ab

2b                                                         (3)

 

  1. 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

 

  1. 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

 

  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)

 

  1. (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)

 

  1.       3(1 +  3)                 =  3  +  3          3 + v3

(1-  3)(1+  3)                     1-3                          2

 

  1. 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

 

  1. 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

 

  1. 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)

 

  1. 2          -3

 

4          3          det. = 6 + 12 = 18

Inv.=     1         3          3

18

-4         2

1         3      3     2     -3   x       1           3   3       7

18                                            18

-4    2      4       2  y                     -4  2       5

x                       36

1

y          18        -18                    (3)

x = 2, y = -1      (A)

 

  1. 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)

10

= 0.1429 + 0.2419 + 67.96                                 (3)

=70.52                             (A)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. (i) ADC = 2×73

= 1460

 

(ii) OCB = x = 180 – 146 = 34

(iii) 360 – 107 – 146 – 34

= 73 0

 

  1. Tan 300 = y/x y  =  x tan  30

Tan 600  = 1000 + y       ;     y = x tan 60 – 1000

X

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)

 

 

  1. 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

2

x = -20 ±  220

2

Spd = 100 km/h

And x = 120 km/h                                 (A)

  1. Log x = a log y = b

Log  x3  = Log x3  –  log y ½

y

= 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 Ft2   v
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)

(c)    Median 45.5th value  = 39.5  + (13.5 x 10/ 26)

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)

 

  1. (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)

  1. 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)

 

  1. 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 r2 x 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)

 

  1. Use tables to evaluate

3Ö 0.09122 + Ö 3.152                                                           (5mks)

0.1279 x 25.71

  1. Simplify  (a – b)2

a2 – b2                                                             (2mks)

 

 

  1. The gradient function of a curve that passes through point: (-1, -1) is 2x + 3.

Find the equation of the curve.                                                                               (3mks)

  1. Find the value of k for which the matrix k     3

has no inverse.   (2mks)                            3     k

  1. Without using tables, evaluate       log 128 – log 18

log 16 – log 6                                                          (3mks)

  1. Find the equation of the locus of points equidistant from point L(6,0) and N(-8,4). (3mks)
  2. 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)
  3. Use reciprocal tables to evaluate to 3 d.p. 2             1   

0.321           n2.2                                          (4mks)

  1. 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)
  2. Solve the equation for 00 £ q £ 3600 and Cos2q + ½ Cosq = 0 (3mks)
  3. 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)

  1. How many multiples of 3 are there between 28 and 300 inclusive. (3mks)
  2. 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)
  3. 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)

 

 

 

 

 

 

 

 

 

 

 

  1. Make X the subject of the formula (3mks)

V = 3Ö k + x

sk – x

 

 

 

 

 

 

 

 

  1. Write down the inequalities that describe the unshaded region below. (4mks)

y

 

 

0.5         2                   x

 

-1.5

-2

 

      SECTION B (48 MARKS)

     

  1. 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)

 

  1. 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)

  1. 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

R

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)

 

  1. 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)

  1. 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)

  1. 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)

  1. 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)

(c) The bearing of port R from port P.                                                                      (3mks)

  1. 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

 

  1. Use logarithm tables to Evaluate

3Ö 36.5 x 0.02573

1.938                                                                                                              (3mks)

  1. 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)
  2. If K = ( y-c  )1/2

4p

  1. a) Make y the subject of the formula.       (2mks)
  2. b) Evaluate y, when K = 5, p = 2 and c = 2                                                                   (2mks)
  3. Factorise the equation:

x + 1/x = 10/3                                                                                                             (3mks)

  1. DA is the tangent to the circle centre O and Radius 10cm. If OD = 16cm, Calculate the area of the shaded Region.       (3mks)

 

 

 

 

 

 

 

 

 

 

 

  1. Construct the locus of points P such that the points X and Y are fixed points 6cm apart and

ÐXPY =     600.                                                                                                            (2mks)

  1. In the figure below, ABCD is cyclic quadrilateral and BD is diagonal. EADF is a straight line,

CDF = 680, BDC = 450 and BAE = 980.

 

 

 

 

 

 

 

 

Calculate the size of:                                                                                               (2mks)

  1. a) ÐABD                                       b) ÐCBD
  2. 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)
  3. Calculate the distance:
  4. i) In nautical miles (nm)
  5. ii) In kilometres (km)

Between the two places along the circle of Latitude:

  1. a) A(300N, 200E) and B(300N, 800E) (Take Radius of Earth = 6371Km).                (2mks)
  2. b) X(500S, 600W) and Y(500S, 200E) (Take Radius of Earth = 6371Km).                  (2mks)
  3. 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)
  4. 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)  
  5. 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)
  6. 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)
  7. Find the following indefinite integral ò 8x5 – 3x dx                                                                  (4mks)

x3

  1. 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)

 

 

 

 

 

 

 

 

  1. Use the data below to find the standard deviation of the marks.

 

Marks (x ) Frequency (f)
5

6

7

8

9

3

8

9

6

4

(4mks)

 

SECTION II (48MKS)

 

  1. The figure below shows a cube of side 5cm.

 

 

 

 

 

 

 

 

 

 

 

 

 

Calculate:

  1. a) Length FC                                                                                                      (1mk)
  2. b) Length HB                                                                                                        (1mk)
  3. c) Angle between GB and the plane ABCD. (1mk)
  4. d) Angle between AG and the Base.       (1mk)
  5. e) Angle between planes AFC and ABCD. (2mks)
  6. f) If X is mid-point of the face ABCD, Find angle AGX. (2mks)
  7. Draw on the same axes the graphs of y = Sin x0 and y = 2Sin (x0 + 100) in the domain 00 £ x0 £ 1800
  8. i) Use the graph to find amplitudes of the functions.
  9. ii) What transformation maps the graph of y = Sin x0 onto the graph of : y = 2Sin (x0 +100).
  10. 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.

  1. 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.

 

 

 

 

  1. 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:
  2. a) Using a suitable scale draw a diagram to show the positions of Aeroplanes after 2hrs.
  3. b) Use your graph to determine:
  4. i) The actual distance between the two aeroplanes.
  5. ii) The bearing of B from A.

iii) The bearing of A from B.

  1. 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.
  2. a) it will either rain and not.
  3. b) Neither will not take place.
  4. c) One Event will take place.
  5. Calculate the Area of each of the two segments of y = x(x+1)(x-2) cut off by the x axis. (8mks)
  6. 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:

 

  1. 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)

 

  1. (a – b)(a – b) a – b

(a – b)(a + b)       a + b

 

  1. dy = 2x + 3

dx

y = x2 + 3x + c

-1 = 1 – 3 + c

c = 1     ;     E.g  y = x2 + 3x + 1

 

  1. K2 – 9 = 0

K = ± 3

 

  1. log 128    =  log       64

18                    9

 

log   16        log     8 

6                    3

2 log (8/3)

log (8/3)

= 2

 

  1. 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

 

  1. 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

 

  1. 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.

 

  1. Cos q (cos q + ½ ) = 0

cos q = 0        cos q = -0.5

q = 900, 2700    q = 1200, 2400              

\ q = 900, 1200, 2400, 2700

 

  1. MP = 4 MK MK =      -9

9                                   -18

MP = 4 ( -9  ) = ( -4 )

9  -18          8

\ P is ( -1,0 )

 

  1. a = 30 d = 3   l = 300

300 = 30 + 3 (n – 1 )

300 = 30 + 3n – 3

300 – 27 = 3n

273 = 3n

91 = n  

 

 

 

 

  1. y = mx – 1

1 = 3m – 1

m = 2/3 = 0.6667

tan q = 0.6667  ;     q = 33.690    

 

  1. FK x FC = FA2

FC = 25/3 = 8 1/3 cm

CX = 81/3 – 9/2 = 23/6 = 35/6 cm

CX x XK = XA x XN

33/6 x 3/2 = 3 x XN

\ XN = 111/12 cm

 

  1. V3 = k + x

k – x

V3k – V3x = k + x

V3k – k = x + V3x

V3k – k = x( 1 + v3)

V3k – k  = x

1 + V3

 

  1. (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

 

  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

60

= 74.5 – 28  =    46.5

(ii) Standard deviation = Ö 77600 – ( –1680 )2

60            60

= Ö 1283.3 – 784

= Ö 499.3 = 22.35

 

  1. 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

 

  1. 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

4

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

3

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

 

 

  1. 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

X

Time taken by B = 300 hrs

X + 10

300300  =  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

 

 

 

 

 

 

 

 

 

  1. (a) Distance = 15 x 40 = 300km

2

(b)

 

 

 

 

 

 

 

 

 

 

 

PR2 = 2002 + 3002 –2x 200 x 300 cos700

= 130,000 – 41040   =   88,960

PR = 298.3 km

 

(c) 298.3  = 300

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

 

  1. (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

 

  1. 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)

 

  1. 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

  1. = 250

Shirt = sh 200  ;   Blouse = sh 250

     

  1. (a) K2 = y – c

4p

y – c = 4pK2

y = 4pK2 + c

(b)    y = 4 x 2 x 25 + 2   ;      y = 202

 

  1. x2 + 1 – 10x = 0

3

3x2 – 10x + 3 = 0

3x (x – 3) – 1(x – 3) = 0

(3x – 1) (x – 3 ) = 0

x = 1/3  or x = 3                                                                                                             

 

  1. 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

 

 

 

 

 

 

 

 

  1. ÐXPY = 600

\ÐXC1Y = 1200

              B1             \ÐC1XY = ÐC1YX

= 1800 – 1200  = 300

2

 

 

 

 

Construct 300  angles

at XY to get centres

B1           C1 and C2  mojar arcs drawn

2            on both sides with C1X and C2X

as centres.

 

 

 

 

 

 

 

 

 

 

  1. 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 = 980                                                                                   82 + 98 = 1800

        1800  – (980 – 450) =

ÐCBD = 370                                                                                  180 – (98 + 45)

= 370

  1. 10 x 320

100     Discount = sh 32

Sold at      sh 288

If no Discount = ( 320 x 20 ) % = 22.7%

288

 

  1. (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

 

  1. 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.

 

  1. 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)+ 5 (-0.2)4

1 – 1.0 + 0.4 – 0.08 + 0.008  =   0.3277 (4d.p)                                                                                                                     

 

  1. Area of metal : Material – Cross section.

p(R2 – r2)

3.14 (21 –19)

Vol  6.28cm2 x 3400cm

= 215.52m3        

                                       

  1. 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

 

  1. ò( 8x5/x3 – 3x/x3) d4

ò( 8x2 – 3x-2) d4                                                                

16x3/3 + 6x-3/-3  + C                                                 

16x3/3 – 2/x+ C

 

  1. 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)

30            30

=  Ö 50.4 – 49

=   Ö 1.4  = 1,183                                                       

 

       SECTION II                                               .

 

  1. (a) FC = Ö 52 + 7.072 = Ö 50 = 7.071

(b) HB = Ö 52 + 7.072    = Ö 75 = 8.660

(c) q = Tan-1 5/5 = Tan-1   = 450                                                         

(d)  b = Tan-1 5/7.071 = Tan-1 0.7071  =  35.30                                                        

(e)  y = Tan-1 5/3.535   = Tan-1    = 54.70                                                        

(f) ÐAGX = 19.40

 

 

  1. 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

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. 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

 

  1. (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

  1. 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         

                            

  1. y = x3 – 3x2

dy  = 3x2 – 6x

At stationary

Points      dy = 0

dx

i.e   3x2 – 6x = 0

3x(x – 2) = 0

x = 0 or 2

Distinguish

dy = 3x2 – 6x

dx

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)

  1. Without using tables evaluate:

 

Ö7.5625 x 3Ö3.375

15                                                                                                        (5 mks)

 

  1. Make k the subject of the formula.

y = 1  Ök + y                                                                            

T2      k                                                                                                       (3 mks)

 

  1. If A = (x, 2) and xB     =     x     and if AB = (8), find the possible values of x.

-2                                                                                 (3 mks)

  1. Simplify completely. (3 mks)

rx4 – r

2xr – 2r

 

  1. Solve the equation. (3 mks)

Log 3 (8-x)  –  log 3 (1+x) = 1

 

  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)

 

  1. Find the equation of the line perpendicular to the line 4x-y = -5 and passing through the point (-3,-2).       (2 mks)
  2. Find the standard deviation of the data below:

3,5,2,1,2,4,6,5                                                                                                   (4 mks)

 

  1. What is the sum of all multiples of 7 between 200 and 300? (4 mks)

 

  1. Solve the equation.

½ tan x  =  sin x for -1800  £  x  £  3600.                                                            (3 mks).

 

  1. Expand (1-2x)4. Hence evaluate (0.82)4 correct to 5d.p. (4 mks)

 

  1. The line y = mx – 3 passes through point (5,2). Find the angle that the line makes with the x-axis. (2 mrks)
  2. 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)

 

 

 

 

 

 

  1. 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)

 

 

 

 

 

 

 

 

 

 

 

 

  1. 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)

 

  1. Find the inequalities that describe the unshaded region R below. (4 mks)

y

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

SECTION  2 (48 Marks)

 

  1. 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)

 

  1. 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)

 

  1. 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)

 

  1. 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)

 

  1. 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)

 

  1. 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)

 

  1. 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)

 

 

 

 

 

 

 

 

  1. Two variables T and X are connected by the equation T = abx where 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)

 

  1. Use mathematical tables to evaluate:

 

8.67                                                                                                                        (3 mks)

Ö 0.786 x (21.72)3

 

  1. Simplify completely. (3 mks)

4      –    1

x2 – 4        x-2

 

  1. 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)

 

  1. The last of three consecutive odd numbers is (2x+3). If their sum is 105, find the value of x. (4 mks)

 

  1. a S  b is defined by:           a S b  =  (a + b)

ab

If B S   (2  S   3)  =  4  S   1, Find B.                                                                                   (3 mks)

  1. Find the value of M. (3 mks)

 

 

M

 

850

 

1600

 

 

  1. (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)

 

 

  1. Show that x is the inverse of : Y =    3          -3      1           X =       2      1                       (3 mks)

-5        2                     5      3

 

 

 

 

 

 

  1. 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)

 

  1. 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

  1. 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)

 

  1. Evaluate 0 (2 mks)

( 3x2 –  1 ) dx

4 x 2

1

  1. 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)
  2. Write down the three inequalities which define the shaded region. (3 mks)

 

 

 

(3,2)

 

 

 

 

 

 

(2,1)                                   (4,1)

 

 

 

 

  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)

  1. 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)

 

  1. 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)

 

  1. 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)

 

  1. 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)

 

  1. 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)
  2. 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)

 

  1. (a) On the same axes, draw the graphs of : y  =  sin x  ;  y  =  cos x

y  =  cosx  +  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.

(c) Use your graph to solve:

Cos x  = – sin x for 00 Ð X Ð 3600 .

 

  1. 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.

B

 

 

A                                                                              E

C

 

 

D

(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)

(c)  Calculate the value of angle BED.                                                                     (2 mks)

(d)  Calculate the value of angle DCB.                                                                     (2 mks)

 

  1. 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

 

3 Ö33 x 53 x 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  =  log3 3

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

 

 

1st equation

2nd 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 21

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           33t  =  1k

4   4       2

33t = 1 t

4    4     2

5t  =   3

4       4

t  =  3 + 4

4    5

= 3

5

\   k = 3

5

\   PM  =  3r  +  3h

5       10

 

 

 

1

 

1

 

 

1

 

 

1

 

1

 

 

1

 

1

1

 

 

 

 

 

PM

 

PM

 

 

PM simplified

 

 

 

 

 

 

 

Both equations

 

method

 

 

 

 

Value of k

 

Value k

PM

 

8
 

 

 

24.

YLogT

 

 

 

Log T  =  log a + x log b

Log T  Þ  0.82, 1.25, 1.68, 2.11, 2.54, 2.97, 3.40, 3.84

 

y – intercept = log a = 0

a = 1

gradient  =  3.84 – 0.82  =   3.02

9 – 2                  7

= 0.4315

 

log b = 0.4315   =  0.4315

b = antilog 0.4315

b  =  2.7

 

1
1

 

 

 

 

 

 

 

1

2

 

1

 

1

 

 

 

 

1

8

Plotting
Labeling of axis

 

 

 

 

Linear

All correct logs

 

Value of a

Method of gradient

 

Value of  b

MATHEMATICS III

PART II

MARKING SCHEME

 

  1. SOLUTION MARKS    AWARDING
1.    No                                      log

 

8.69                                   0.9390

0.786                                 1.8954

21.72                                 1.3369

1.2323

1.7067 – 2

 

21.7067

2           2

– 1  +  0.8533

0.7134 x 10 -1     =  0.07134

 

 

 

 

 

M1

 

 

M1

 

 

 

A1

 

 

ü reading to 4 s.f

 

 

 

 

 

 

Rearranging

3
2.  

 4                   –         1

(x-2)(x+2)                  (x-2)

 

 – x+2

(x-2(x+2)

– (x-2)

(x-2(x+2)

 

-1

x+2

 

 

 

M1

 

 

M1

 

 

A1

 

 

3
3.  

Re6000  =  Ksh. 75000

Spent 5000 Rem 2500

Rem    2500

1.25

Re 2000

M1 

 

M1

 

A1

 

3
4. 2x – 1  ,  2z + 1  ,  2x + 3

6x +  3  =  105

6x  =  102

x  =  17

M1

M1

A1

A1

 

Allow M1 for us of different variable.
4
5.  

4 * 1  =  5

4

2 * 3  =  5

6

A * 5  =  5

6      4

A + 5  =  5  x  5A

6      4       6

A +  5  =  25 A

6       24

A   =  20

 

 

M1

 

 

 

 

M1

 

 

A1

3

6.  

 

 

 

 

180 – M + 20 + 95  =  180

295  –  M  =  180

– M  =  – 115

M  =  115

 

 

 

B1

 

 

B1

 

 

A1

 

3
 

7.

 

1 + 2x + 60x2 + 160x3 +

1 + 0.2 + 0.006 + 0.00016

=  1.20616

=  1.206

 

M1

M1

M1

A1

4

 

Only upto term in x3.

Correct substitution

 

Only 4 s.f.

 

8.  

3   -1      2    1    =    I

-5   2       5    3

 

6   -5             3    -3

-10 +10         -5 + 6

 

1      0

0       1

 

 

M1

 

M1

 

 

A1

 

 

Matrix multiplication gives :

 

I       1   0

0   1

3
9. (a)   2  x  3  x  4      =  23      4      5           5

(b)

2  x  3  x 1     +     2  x  1  x  4     +     1  x  3  x  4
3      4     5            3      2      5             3      4      5

 

1  +  4  +  1

10     15     5

 

=     17

10

M1

 

 

M1

 

 

 

 

A1

 

 

3
10. ÐQCB  =  300

180 – (27 + 30)  =  1230

\     BAC  =  570.

 

 

 

 

OBA  =  370

OAB  =  370

 

 

AOB  =  1060

\ ACB  =  530

 

 

 

M1

 

 

 

 

 

M1

 

 

A1

 

 

 

 

 

 

 

 

 

Isosceles triangle.

 

Angle at centre is twice angle at circumference.

3
11. V  =  1  x  3.14  x  r 2  x 10  =  270

L.S.F.      20   =  2

30       3

V.S.F  =    2   3        =     8

3                   27

Vol. of cone  =  8  x  270

27               =      80cm3

\ Vol. Of frusturm  = (270 – 80)  =  190cm3

 

 

 

M1

 

M1

 

 

A1

 

2
12.

 

 

 

 

 

 

 

 

3x 3  –  x  -1          2

3       -1         1

 

x 3  +  1     2

x     1

 

8  +  1     –   ( 1  –  1)

2

8 1  –  2     =         6  1

2                           2

 

 

 

 

 

 

M1

 

 

 

A1

2

13. (2 x 3 x 1.5)  volume

9 m3

1L  º  1000 cm3

1000 L  =  1 m3

9000 L  =  9 m3

1000 L  =  1 tonne

9000 L  =  9 tonnes.

 

 

M1

 

 

 

A1

 

2
14.      y   ³ 1            (i)

y   <  x – 1     (ii)

y   <  5 – x     (iii)

 

B1

B1

 

3
15. M1  =  5.1  +  4.9  +  4.7  =  4.9

3

M2  =  4.9 + 4.7 + 4.5  =  4.7

3

M3  =  4.7 + 4.5 + 4.0  =  4.4

3

M1

M1

M1

 

 

3
16. Original contribution per woman  =  6300

N

Contribution when 7 withdraw  =  6300

(n-7)

Increase   –  Diff.

6300   –   6300

n-7          n

6300n  –  6300(n-7)

n(n-7)

6300n – 6300 + 44100

n(n-7)

44100

n(n-7)

 

 

 

M1

 

 

M1

 

1

3

SECTION II (48 Marks)

 

17. (i)

1150

 

A                                B

 

Centre of circles of latitude 500 S.  R Cos 500

AB  =  115  x  2p R Cos 50o

115  x  40192  x  0.6428

360

=  8252.98  km

 

(ii)   Arc AB 60 x 115  Cos 50 nm

60 x 115 x 0.6428 nm

4435 nm

 

 

 

 

 

M1

M1

 

 

M1

 

A1

 

M1

M1

M1

A1

 

 

 

 

 

 

 

No.                     log

60                      1.7782

1+5                    2.0607

0.6428               1.8080

4435nm             3.6469

8
18. (a)  V  =  ds  =  6t2 + 8t – 8dt

(i)  t  =  2

V  =  6×4 + 8×2 – 8

= 32 ms-1

(ii)  t  =  3

V =  6×9 + 8×3 – 8

= 70ms-1

 

(b)  Particle is at rest when V = 0

6t2 + 8t – 8 = 0

2(3t – 2) (t+2) = 0

t  =  2                   t  =  -2

3

particle is at rest at t = 2 seconds

3

   

 

 

 

 

 

 

 

 

 

 

 

Do not accept t = -2. Must be stated.

8
19. Area under velocity – time.

graph  gives distance.

 

A  = { h ½  (y1 + y6 ) + y2 + y3 + y4 + y5 )}

 

= 1 { ½ ( 12+47) + 24 + 35 + 41 + 45)}

=  29.5 + 14.5

=  174.5m

 

 

B1

B1

M1

M1

B1

B1

A1

 

Trapezium rule only accepted.

Formula.

 

Substitution into formular.

8
20.                  Drawing actual

Scale 1cm  =  2cm

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Radius      1cm

=  2cm

 

M1

 

M1

 

M1

 

M1

 

M1

M1

 

M1

M1

 

 

Bisect ÐA

 

Bisect Ð B

 

Intersection at centre of inscribed circle.

Draw circle.

 

Measure radius.

Arcs must be clearly shown.

8
 

 

 

21.

 

 

 

 

mean = 44.5 +  130

100

=  44.5  +  1.3

=  45.8

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(b)  Variance  S (x – A) 2  =  2800

Sf               100

= 28

S.D.  =  Ö 28  =  5.292

 

 

 

 

M1

 

 

 

 

 

A1

 

M1

 

A1

M1

A1

 

8
 

 

 

 

 

22.

y = sin x

x    0        60        120        180     240      30      360

sin x 0    0.866     0.866      0     -0.866   -0.866    0

y = cos x

x     q        60        120        180     240    300  360

cos x 1     0.5       -0.5       -1.0     -0.5     0.5   1.0

y = cosx + sinx

x            q        60       120        180     240      30     360

cosx + sinx 1  1.366   0.366       -1   -1.366  -0.366 1.0

(c)      Cos x = – sin x

x  =  450 , 2250

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

23.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(i)  amplitude   =  1.366

(ii)  Period  =  3000

 

 

 

(a)  (x+1) 2  +  (y-6)2  =  32

x2 + 2x + 1 + y2 – 12y + 36  =  9

x2 + 2x + y2 – 12y + 28  =  0

 

(b)  cos 10  =  OD             DE  =  3

DE                   0.9848

DE  =  3.046

 

(c)  Twice ÐOED

100 x 2  =  200

 

(d)  DAB  =  800

\ DCB  =  1000

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

M1

 

A1

 

M1

A1

 

 

M1

A1

 

M1

A1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Formular

(x-a)2 + (y-b)2 = r2

 

 

 

 

 

 

 

 

 

Cyclic quad.

 

 

8
24. Let number of trips by 12 tonne lorry be x.

Let number of trips by 7 tonne lorry be y.

 

(a)   x > 0  ;  y > 0

24x + 21y  £  150

 

12 x 25 x X + 15 x 7 x y £ 1200

300x + 105y  £  1200

x > 2y

 

(b)  Ref. Graph paper.

Minimising:

3 – 12 tonne lorry and 2 – 7 tonne lorries should be deployed.

 

 

 

B1

 

 

 

B1

B1

 

 

 

 

 

 

MATHEMATICS IV

PART I

 

SECTION 1 (52MKS)

 

  1. Evaluate using logarithms 3Ö7.673 – 15.612

12.3                                                              (4mks)

 

  1. Solve x   –  3x  –  7    =  x – 2                                                                                   (3mks)

3            5             5

 

  1. In the given figure CD is parallel to BAC, calculate the values of x and y. (3mks)

 

 

C                                       D

 

 

 

 

 

 

B                                                A

 

  1. 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)

 

  1. A line perpendicular to y = 3-4x passes through (5,2) and intercepts y axis at (0,k)

Find the value of K.                                                                                              (3mks)

 

  1. 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)

 

 

  1. 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)

 

  1. 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)

  1. 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

 

 

  1. Express 3.023 as a fraction                                                                                      (2mks)
  2. Point A (1,9), Point B(3,5) and C (7,-3). Prove vectorically that A,B and C are collinear.       (4mks)
  3. 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)

  1. 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)
  2. Rationalize the denominator in Ö3

Ö 7 – 2

Leaving your answer in the form Öa + Öb

C

Where a ,b, and c are integers                                                                              (3mks)

  1. For positive values of x, write the integral solutions of 3£ x2  £  35                 (4mks)
  2. 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)

 

  1. 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)

 

  1. The sum of the 2nd and third terms of a G.P is 9/4 If 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)

  1. E and F are quantities related by a law of the form E = KFn Where 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)

 

  1. 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)
  2. 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)

 

  1. 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

 

  1. 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)

  1. 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)

 

  1. Without using mathematical tables, evaluate                                                                    (3mks)

 

Ö 0.0784 x 0.27                                              (leave your answer in standard form)

0.1875

 

  1. 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)

 

  1. 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

  1. Given that a = -2, b = 3 and c = -1, Find the value of   a3 – b – 2c2                                    (2mks)

2b2 – 3a2c

 

  1. 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)

 

  1. 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)

 

  1. 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)

  1. 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)

  1. Under an enlargement of linear scale factor 7, the area of a circle becomes 441.p

Determine the radius of the original circle.                                                              (3mks)

  1. A circle has radius 14cm to the nearest cm . Determine the limits of its area.                     ( 3mks)
  2. Expand (1 + 2x)5 up to the term with x3. Hence evaluate 2.045 to the nearest 3 s.f. (4mks)
  3. The nth 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)

  1. 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)
  2. On a map scale 1:20,000 a rectangular piece of land measures 5cm by 8cm. Calculate its actual area in hectares.                                                                                                                                      (3mks)
  3. 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)

 

  1. 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)

 

  1. 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)
  2. 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)

(c) Two are white and one is blue                                                                                         (3mks)

 

  1. 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)

(c)  Each carton costs shs. 800. He spends shs. 100 to transport  each 5 cartons. How  much would one

sell each carton to make 20% profit ?                                                                             (4mks)

  1. 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)

  1. In the given figure, OA = a , OB =b,  OP: PA =3:2,  OQ:QB = 3:2

Q

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)

(c) PR  =  K PQ Where K is a scalar, write OR in terms  of k, a and b                           (1mk)

(d) Calculate the value of k and h                                                                                               (3mks)

 

  1. 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 A1B1C1 on the given grid.

Transformation Q maps A1B1Conto A11 (-6,2) B11(-2,3) and C11(-6,6). Find the matrix Q and plot

A11B11C11on the same grid. Describe Q fully.                                                           (8mks)

 

  1. 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)

  1. 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

 

1.  

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

B1

 

 

 

 

B

 

M1

 

A1

4

 

 Subtraction

 

 

 

 

Logs

 

Divide by 3

 

Ans

2. 5x – 3 (3x –7 )    =  3(x – 2 )

5x – 9x + 21    =   3x – 6

-7x             = -27

x              =  36/7

 

M1

M1

 

A1

3

Multiplication

Removal ( )

 

Ans

3. 3x +5y + x =  180

9x   =  180

x    =   20

y   =    60

M1

A1

B1

3

Eqn

X

B

 

 

4.  

.                               r   =       3v      1/3

4P

 

.                              r   =        S       ½

4P

 

\ 3V      1/3              =            ½

4P                                 4P

 

3V                         =       S       3/2

4P                                 4P

 

V             =       4P      S     3/2

3            4P

 

 

 

B1

 

 

 

 

 

 

 

M1

 

 

 

A1

3

 

 

 

Value r

 

 

 

 

 

 

 

Equation

 

 

 

Expression

5. 

 

 

 

6.

Grad  line          = ¼y – 2        = ¼

x – 5

y            =  ¼ x + ¾

k             =      ¾

P in Alloy         = 4/10  x 800

= 320g

100 x 320

20

=  3.2 kg

 

M 1 

A1

A 1

3

 

B1

 

M1

 

A 1

 

Equation 

Equation

K

 

 

P in alloy

 

Expression

 

Ans

 

 

 

 

7.

 

 

 

 

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)

 

 

 

 

 

B1

 

M1

 

 

A1

3

 

 

 

 

B conduct

 

Formular

 

 

C

8.  

 

 

 

 

 

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

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B1

M1

 

 

 

M1

 

A1

4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Expression for  h both

Equation

 

 

 

Expression for h

 

Ans

9.                 2pr    =  90  x 2p x 50

360

r    =  12.5

h     =  Ö2500 –  156.25

=   Ö2343.75

=   48.41 cm

 

M1

P

A1

M1

 

A1

4

Equation

 

.r

expression for h

 

ans

 

 

10.

 

100 n      =   302.323

     n      =      3.023   

99n       =   299.3

n      =    2993

990

=    323/990

 

M1

 

 

A1

4

 

 

Equation

 

 

Ans

 

11. AB        =     3-15-9

=     2

-4

BC         =     4

-8

AB         = ½   BC

\ AB // BC

But B is common

\ A,B,C are collinear.

 

 

 

 

B1

 

 

 

 

 

B1

 

 

B1

3

 

 

A B &  BC

 

 

 

 

 

Both

 

 

Both

 

12.       4% of 200,000  = 8000/=

balance                   = 4200/=

6% of  x                 = 4200/=

x                 = 4200 x 100

6

=  70,000

sales                 =  sh. 270,000

B1

 

 

M1

A1

B1

4

 

 

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

 

 

 

 

 

M1

M1

 

M1

 

A1

4

 

 

 

 

 

 

Vol tank

Vol tank

 

Div x 3600

 

Tank

 

 

14.

 

 

 

 

 

 

 

    Ö3                      =     Ö3           Ö7 + Ö2

Ö7Ö2                         Ö2Ö2         Ö7+ Ö2

 

= Ö3 Ö7 + Ö2

5

 

= Ö21 + Ö6

5

M1

 

M1

 

A1

3

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

B1

B1

B1

4

Lower limit

Upper limit

Range

Integral values

 

16.  No of days   =  8/6 x 5/8  x 12

=   10 days

M1

A1

2

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

B1

B1

B1

 

B1

 

B1

B1

 

B1

8

 

 

 

 

 

 

 

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)

½

= 6 ( 31/32)

= 6 31/32

 

B1

B1

 

B1

 

M1

A1

 

M1

 

 

 

M1

 

 

A1

8

 

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

B1

 

 

S1

 

 

P1

 

 

L1

 

 

M1

A1

 

B1

8

 

Log E

Log F

 

 

Scale

 

 

Plotting

 

 

Line

 

 

Gradient

 

 

K

 

 

 

20  

.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

 

 

 

 

 

 

 

B2

 

B1

 

B1

 

S1

P1

C1

 

L1

 

B1

 

8

 

 

 

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

M1

M1

M1

M1

 

A1

 

M1

M1

A1

8

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

 

 

M1

 

 

A1

 

M1

A1

M1

M1

M1

 

A1

8

Method

 

 

 

 

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

2

.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

 

 

 

 

 

 

 

B1

 

 

 

M1

 

 

 

 

M1

 

 

 

 

A1

 

M1

A1

 

M1

 

A1

8

 

 

 

 

 

 

 

Ans .

 

 

 

Expression both correct

 

 

 

Equation

 

 

 

 

Ans

 

 

 

 

Expression

 

Equation

 

MATHEMATICS IV

PART II

MARKING SCHEME

 

 

1.                784 X 27        =

187500

Ö 784 x 9           =    4 x 7x 3

62500                      250

=       42

125

=       0.336

 

 

 

M1

 

M1

 

 

A1

 

 

 

Factors for

Fraction or equivalent

 

C.A.O

3
2.      Father 3x ,  r son  = x

2(x +10)        = 3x + 10

2x +20       =  3x + 10

x        = 10

father            = 30

M1

 

 

A1

B1

 

Expression

 

 

 

 

 

 

3
3. 3   = sin   60AE

AE  = 3

Sin 60

= 3.464

perimeter  = 5×2 + 3.464 x 3

= 10+10.393

= 20.39

M1 

 

 

A1

 

 

B1

Side of a triangle 

 

 

 

 

 

Perimeter

3
4.    .a3 – b-2c2  =  (-2)3 – 3 –2(-1)2

2b2 – 3a2c      2(3)2 –3(-2)2(-1)

= -8 –3-2

18 + 12

= -13

30

M1

 

 

M1

 

A1

Substitution

 

 

Signs

 

C.A.O

3
5.        Ksh  189,000          =   $ 189,000

75.6

= $ 2500

balance                    = $ 2500

=  Kshs. 189,000

Kshs. 189,000          =             189,000

115.8

Uk    ₤1632

M1

 

A1

 

M1

A1

 

A1

4

 

Conversion

 

 

 

Conversion

 

6. Area of 2 triangles  =   2 (½ x 8x 5 sin 60)

=   40 sin 60

=   40x 0.8660

= 34.64 cm2

Area of rectangle    = 300 x 8 + 300 x 5 +300 x BC

BC              = Ö64 +25 – 2 x 40cos 60

= Ö89 – 80 x 0.5

= Ö89 – 40

= Ö49

= 7

Total   S.A.              = 300 (8+5+7) + 34.64 cm2

= 6000 + 34.64

= 6034.64 cm2

M1

 

 

 

 

M1

 

 

 

 

M1

 

A1

Areas of D

 

 

 

 

B.C. expression

 

 

 

 

Area

 

4
7.    AF2    = 32+42+-2+12x cos 50

= 25 – 24 x 0.6428

= 25-15.43

= 9.57

AF      =  3.094 x 50

AF      =  154.7m

Sin Q  =  200 sin 50o

154.7

= 0.9904

Q   = 82.040

Bearing = 117.96

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   + 918

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 2n –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   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  57  +  587

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

 

 

 

 

 

 

 

 

 

 

1st slab

2nd slab

8
21.  (a)            PQ                 =  –3/5 a   +  3/1b

=  31/23/5 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)

=  (3/53/5k)a +3k b

(d)                      1 – h     =  3/53/5k    (i)

3k    =  h                   (ii)

Sub (i)              1 – 3k    =  3/53/5k

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

 

 

 

 

All BII CII

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)

 

 

 

  1. Use logarithms to evaluate 6 Cos 40   0.25
    63.4                                                                                                                                                                                                       (4mks)
  2. Solve for x in the equation (x + 3) 2 – 5 (x + 3) = 0 (2mks)
  3. In the triangle ABC, AB = C cm. AC = bcm. ÐBAD = 30o and ÐACD = 25o. Express BC in terms of b and c.                                                                                                     (3mks)
  4. Find the equation of the normal to the curve y = 5 + 3x – x3 when x = 2 in the form
    ay + bx = c                                                                                                             (4mks)
  5. 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)
  6. 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)
  7. Express as single factor 2     –     x + 2         +       1
    x + 2    x2 + 3x + 2         x + 1                                       (3mks)
  8. 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)
  9. 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)
  10. 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)
  11. Make T the subject of the formulae 1       –  3m   +  2
    T2         R         N                        (3mks)
  12. Vectors, a =   2     b =   2   and   c –   6
    2              0                   4
  13. 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)
  14. A piece of wood of volume 90cm3 weighs 54g. Calculate the mass in kilograms of 1.2 m3 of the wood.      (2mks)
  15. 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)
  16. 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)

 

  1. 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)

  1. 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)
  2. 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(50oN, 10oW) 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 = 37.5o.                                (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 = 90o.              (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)

 

 

 

  1. 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)

 

  1. Evaluate without using mathematical tables (2.744 x 15 5/8)1/3                              (3mks)
  2. If 4 £ x £ 10 and 6 £ y £5, calculate the difference between highest and least
    (i) xy                                                                                                                    (2mks)
    (ii)  y/x                                                                                                                     (2mks)
  3. 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)
  4. Matrix 1        2x   has on inverse, determine x                                                     (3mks)
    x +3      x2
  5. The school globe has radius of 28cm. An insect crawls along a latitude towards the east from A(50o, 155oE) to a point B 8cm away. Determine the position of B to the nearest degree.                                                                                                                                                 (4mks)
  6. The diagonals of triangle ABCD intersect at M. AM = BM and CM = DM. Prove that triangles ABM and CDM are Similar.       (3mks)
  7. Given that tan x = 5/12, find the value of 1  –   sinx
                                                                         Sin x + 2Cos x,   for 0 £ x £ 90           (3mks)

 

  1. 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)
  2. MTX is tangent to the circle at T. AT is parallel to BC. Ð MTC = 55o and Ð XTA = 62o. Calculate Ð (3mks)
  3. 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)

  1. A2 digit 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)
  2. Without using logarithm tables, evaluate log5 (2x-1) –2 + log5 4 = log5 20             (3mks)
  3. 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)
  4. The interior angles of a regular polygon are each 172o. Find the number of sides y lie polygon.                                                                                                                            (2mks)
  5. Evaluate 2x   =       2    +        3
    341       9.222                                                                           (2mks)
  6. A water current of 20 knots is flowing towards 060o. 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 350o.

Find the bearing of A from B.                                                                                (3mks)

SECTION II  (48 MARKS)

 

  1. 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)
  2. 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)
  3. 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)

 

 

 

 

 

 

 

 

 

 

  1. 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 = 30o 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)

  1. 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)

(c)     Hire charge for type A is Sh.1000 while hire for type B is Sh.1200 per vehicle. Find the cheapest

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)

 

  1. Airport A is 600km away form airport B and on a bearing of 330o. Wind is blowing at a speed of

40km/h from 200o. A pilot navigates his plane at an air speed of 200km/h from B to A.
(a)     Calculate the actual speed of the plane.                                                                (3mks)
(b)     What course does the pilot take to reach B?                                                          (3mks)
(c)     How long does the whole journey take?                                                                (2mks)

 

MATHEMATICS V

PART I

MARKING SCHEME

 

1 SOLUTION MKS AWARDING
No         Log

13.6        1.1335   +

Cos 40    1.8842

1.0177   –

63.4       1.8021

1. 2156

(4 + 3.2156) 1/4

1.8039

Antilog    0.6366

 

B1

 

M1

 

 

M1

 

A1

 

Log

 

+

 

 

divide by 4

 

C.A.O

4
2. (x + 3) (x + 3 – 5) = 0

(x +3)b (x – 2) = 0

x = -3 or x = 2

M1

 

A1

 

Factors

 

Both answers

3 BD = C Sin 30  = 0.05

CD = b Cos 25

= 0.9063b

‹ BC = 0.9063b + 0.5 C

B1

 

B1

B1

 

BD in ratio from

 

CD in ratio form

Addition

3
4  Dy  = 3 – 3x2
dx
x = 2, grad = 1
9
Point (2,3)
y – 3  = 1
x – 2     99y – 27  = x – 2
9y – x   =  25
B1

 

B1

 

M1

 

 

A1

 

Grad equ

 

Grad of normal

 

Eqn

 

 

Eqn

 

4
5   700 = 100 + n
2200 = 400 + n1500 = 300m

m = 5

n = 200

P = 5 + 200
q2
When q = 5 P = 13

M1

 

 

A1

 

 

B1

B2

Equan

 

 

Both ans

 

 

Eqn (law)

Ans (P)

4
 

6

 

4 Sin x + 2 cos y = 6

3 Sin x – 2 Cos y = 1
7 sin x                  = 7

Sin x            = 1

X                = 90

Cos y          = 1

Y        = 0o

 

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

= 47o

Q (50oN, 37oE)

 

(b) Time diff = 47 x 4
60

= 3.08

Time at P = 9.12am

(c) QR = 2700 nm

x o   = 2700

60

= 45o

R (85oN, 133oW)

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   53     1/3

1000            23

 

23 x 73  1/3  x   5

103                         2

 

2 x 75   = 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

= 36.03o

 

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 60o

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

= 32.73o

= 33o

 

 

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 +……..yn)

= 1(2.225 + 4.25 + 8.25 +14.25 + 22.25)

= 51.25 sq units

 

 

 

B1

 

 

M1

 

A1

 

 

 

Ordinals

3
 

9

ÐCBA = 117o

Ð ACD = 55

Ð BAC = 180 – (117 + 55) = 8o

 

 

 

 

 

 

 

 

 

 

 

 

 

B1

B1

B1

 

3

10  

 

 

 

B1

B1

B1

B1

1994

1996

1997

1998

4
11. Xy = 35

y = 35/x

9x – 9y = -18

Sub x2 + 2x – 35 = 0

x2 + 7x – 5x – 35 = 0

x (x + 7) – 5(x + 7) = 0

(x – 5) (x + 7) = 0

x  = -7

x = +5

y = 7

Smaller No.

= 57

= 75

B1

 

M1

 

 

 

 

A1

 

 

 

B1

 

3
12 Log5 (2x – 1 )4  = log552

20

4(2x – 1)  = 52

20

2x – 1 = 25

5

2x – 1 = 125

2x = 126

x = 63

M1

 

M1

 

 

 

 

 

A1

3
13 C.P = 100 x 49.50110

= 45/-

52x + 40y = 45

x + y

45x + 45y  = 52x + 40

-7x  = -54

x/y  = 5/7

x : y = 5 : 7

 

 

B1

M1

 

 

M1

 

A1

4
14  

2n – 4 it angle = 172

n

(2n – 4) x 90 = 172n

n

90 (2n – 4) x 90 = 172

n

180 n – 360 = 172n

180n – 172n = 360

8n = 360

n = 45

 

M1

 

A1

 

M1

  2
15 2 x = 2.    1    +    3.    1

6.341                  9.22

2x = 2 x 0. 1578 + 3 x 0.1085

= 0.3154 + 0.3254

= 0.6408

x = 0.3204

 

 

B1

 

 

A1

 

 

Tables

2
16 Bearing 140o

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  – 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    = 15.12o

 

 

 

 

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         = 8.81o

Ð ACB = 180 – (50 + 8.81)o

= 121.19o

    x             =   200
Sin 121.19     Sin 50

 

= 200 x Sin 121.19
Sin 50

= 200 x 0.855645
0.7660

= 223.36Km/h

 

(b)  Course = 330o – 8.81o

= 321.19o

 

(c) Time  =    600
321.19o

 

= 2.686 hrs

 

 

 

 

 

 

 

 

 

 

 

 

 

 

M1

 

 

 

 

 

 

A1

 

 

 

M1

 

 

M1

 

 

A1

 

B1

 

 

 

M1

 

A1

 

 

8

 

 

 

MATHEMATICS VI

PART I

 

SECTION I (52 MARKS)

 

 

  1. Evaluate without mathematical tables leaving your answer in standard form

0.01712 X 3

855 X 0.531                                                                                                                  (2 Mks)

  1. 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)

 

 

 

 

 

  1. 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

 

  1. By matrix method, solve the equations

5x + 5y = 1

4y + 3x = 5                                                                                                                         (3 Mks)

 

 

  1. In the given circle centre O, ÐABC = 1260.

Calculate ÐOAC                                           (3 Mks)

 

A                                     C

 

 

 

B

 

  1. Solve the equation

2(3x – 1)2 9 (3x – 1) + 7 = 0                                                                                               (4 Mks)

  1. 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)

  1. Expand (2 + 1/2x)6 to the third term. Use your expression to evaluate 2.46 correct to 3 s.f     (3 Mks)
  2. 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)

  1. Line y = mx + c makes an angle of 1350 with the x axis and cuts the y axis at y = 5. Calculate the

equation of the line                                                                                                             (2 Mks)

  1. During a rainfall of 25mm, how many litres collect on 2 hectares? (3 Mks)
  2. Solve the equation a 3a – 7 = a – 2 (3 Mks)

3       5          6

  1. 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)

  1. The curved surface of a core is made from the shaded sector on the circle. Calculate the height of

the cone.                                                                                                                            (4 Mks)

 

 

 

 

 

O

20cm      1250                   20 cm

 

 

 

 

 

 

  1. Simplify (wx – xy – wz + yz) (w + z) (3 Mks)

z2 – w2

  1. 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)

  1. In the given circle centre O, ÐQTP = 460, ÐRQT = 740 and ÐURT = 390

 

 

U                                                   T                                P

 

 

Q

S          390

      Calculate                                                                                    R

(a)  ÐRST                                                        (1 Mk)

(b)  ÐSUT                                                       (3 Mks)

(c)  Obtuse angle ROT                                    (2 Mks)

(d)  ÐPST                                                        (2 Mks)

  1. 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)

(c) When he flies back to Kenya, the exchange rate for 1 Fr = Kshs.12.80. How much more in

Kshs. does he receive for his balance than he would have got the day he left?                 (3 Mks)

  1. 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)

  1. (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)

(c)  Calculate the difference between area of the circle formed and triangle PQR                   (4 Mks)

  1. The given Region below (unshaded R) is defined by a set of inequalities. Determine the inequalities (8 Mks)

Y

 

4

 

 

 

2                   R              (3,3)

  

 

X

-3                           5

 

 

 

 

 

 

 

  1. 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)

(c)   Draw a histogram for the data                                                                                       (2 Mks)

  1. 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)

Z

 

 

C

 

 

 

 

..                    B

X

 

A                    Y

 

 

(a)  Calculate, length XA                                                                                                    (2 Mks)

(b)  The shaded area                                                                                                                  (6 Mks)

  1. 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)

 

 

  1. Simplify 32a10   -2/5 ÷  9b4      11/2

b15             4a6                                                                                                 (2 Mks)

 

  1. Use logarithm tables to evaluate

Ö0.375 cos 75

tan 85.6                                                                                                       (4 Mks)

  1. 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)
  2. 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)
  3. 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

 

B

 

 

  1. Simplify     2       –        1                                                                                             (3 mks)

5 – 2Ö3     5 + 2Ö3

 

 

  1. 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)
  2. 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                      120o C                     E

in the tent.                                                                                                            8m                   (4 mks)

B                              D

10m

  1. 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)
  2. 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

  1. Expand (1 + 2x)6 to the 3rd term hence evaluate (1.04)6 (4 mks)
  2. 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)

  1. 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)

  1. 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)

  1. 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)

  1. 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)

 

  1. 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)

  1. 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)

(c)  At least two are correct                                                                                     (3 Mks)

  1. 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)

 

 

 

 

  1. 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)

(c)  QN                                                                                                                   (2 Mks)

(d)  PN                                                                                                                   (2 mks)

 

  1. 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)

  1. 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)

  1. 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)

  1. The vertices of triangle P,Q,R are P(-3,1), Q (-1,-2), R (-2,-4)

(a)  Draw triangle PQR and its image PIQIRI of PQR under translation T =    3    on the provided grid                                                                                                                4                        (2 Mks)

(b)  Under transformation matrix m =    4  3  , PIQIRI is mapped on to PIIQIIRII. Find the

co-ordinates of PIIQIIRII and plot it   1  2    on the given grid                                          (4 Mks)

(c)  If area of D PIQIRI is 3.5 cm2, find area of the images PIIQIIRII                                        (2 Mks)

 

MATHEMATICS VI

PART 1

MARKING SCHEME

 

  1. 171 X 171 X 3 X 10-5 M1

                                  855 X 531

= 2 X 10-6                                                                                     A1

  2

 

  1. No. of men = 6 X 14 X 8 X 2500 M1

                                  2 X 5 X 2240

= 75                                                                            A1

Extra men        = 75 – 6 = 69                                                                B1

 3

  1. 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

  3

 

 

 

 

 

  1. 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

 3

 

  1. Reflex ÐAOC = 126 x 2 = 2520 B1

Obtuse ÐAOC = 360 – 252 = 1080                                                               B1

= 1/2 (180 – 108)0

= 360                                                                                B1

 3

  1. 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

4

 

  1. 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

3

  1. (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

 4

  1. P (FF) = 7/20 X 7/20

= 49/100                                                                                                            B1

P (at least one fail) = 1 – P (FI FI FI)

= 1- 13/20   3                                                      M1

= 1 – 2197                                                       M1

8000

= 5803

                                                     8000                                                                        A1

 4

 

  1. grad = term 135

= -1                                                                                                            B1

y  = mx + c

y  = -x + 5                                                                                          B1

 2

 

  1. Volume = 2 x 10,000 x 10,000 x 25 M1Ö x section area

1000                 10                                                            M1Ö conv. to litres

= 500,000 Lts                                                               A1

 3

 

  1. 10a – 6(3a – 7) = 5(a -2) M1

10a – 18a + 42 = 5a – 10

– 13a    = -52                                                                                        M1

a        = 4                                                                                           A1

 3

  1. 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

 5

 

  1. 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

 4

  1. = (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

 3

 

R

250                                                                                B1Ö sketch

  1. 550

Q  125                                            PR = 4 sin 125                                              M1

Sin 25

A1

30

P                                                                                                          3

  1. (a) <RST = 1800 – 740 = 1060                                                              B1

(b) < RTQ = 900– 740           = 160                                                                B1

< PTR = 460 + 160         = 620                                                                B1

< SUT = 620 – 390         = 230                                                                B1

(c)  Reflex ÐRQT = 180 – 2 x 16

= 180 – 32 = 1480                                                                      B1

Obtuse ROT = 360 – 148 = 2120                                                                   B1

(d)  < PTS = 46 + 180 – 129 = 970                                                                      B1

< PST = 180 – (97 + 39) = 440                                                                      B1

8

(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

(c) Value on arrival = Kshs.7580 X 12.80

= 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

 8

X -2 -1 0 1 2 3
Y -11 0 5 4 -3 -16

B1Övalues

 

y

S1Ö scale

8 —                                                                P1Ö plotting

6 —                                                                C1 Ö curve

4 —

2

 

-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

8

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

 8

 

 

 

 

 

 

 

 

 

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

 8

  1. 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

  8

CLASS

F x Fx Cf
60 – 64

65 – 69

70 – 79

75 – 79

80 – 84

85 – 89

8

14

18

15

3

2

62

67

72

77

82

87

 496

938

1296

1155

246

174

8

28

40

55

58

60

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.

 8

(c)

 

Histogram

 

 

 

20  —

 

 

15  —

 

 

10 – –

 

 

5  —

 

 

 

 

55    60        65        70        75        80        85        90

 

  1. (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 = 1/2 X 10 X 12 sin 41.41 – 22/7 X 2.6452                                    M1 Ö D & circle

= 39.69 – 21.99

= 17.7 cm2                                                                    A1Ö

 8

 

 

 

 

 

 

 

  1. (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

 8

MATHEMATICS VI

PART II

MARKING SCHEME

 

SECTION I (52 MARKS)

 

 

  1. = 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

 

  1. S. Price =  80   X 600

100

= Shs.480                                                                         B1

Cost Price = x

96x       = 480                                                                            M1

100

x  =   Shs.500                                                                      A1

 3

  1. r2 + hr = A/2p 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

 

  1. (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

 3

 

  1. = 2 (5 + 2Ö3) – 1 (5 – 2Ö3) M1

(5 – 2Ö3) (5 + 2Ö3)

= 10 + 4Ö3 – 5 +2Ö3                                                                                                M1

13

= 5 + 6Ö3                                                                                                    A1

13                                                                                                      3

  1. 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

 4

  1. Volume = 1/2 X 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

 3

 

  1. Max. error = 2(8.5 + 5.5) – 2(7.5 + 4.5)

2

= 2                                                                                           B1

% error = 2/26 X 100                                                                                 M1

= 7.692%                                                                                  A1

G                                                                          3

 

 

  1. 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

E

F

  1. (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

 4

 

 

 

 

  1. 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

  4

 

 

  1. 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

  3

 

  1. 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

  3

  1. 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

  3

 

  1. 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

  2

 

  1. 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

  8

 

 

 

 

 

(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

(c) P (at least 2 correct)

= P(2 correct or 3 correct)

= 46/125 + 12/125                                                                           M1

= 46 + 12                                                                                  M1

125

=  58

                                         125                                                                          A1

  8

  1. Old price/pen = 440

x

New price/pen = 494                                                                 B1Öboth expressions

x + 3

440494   = 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

  8

  1. (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

(c)  QN = QO + ON

= 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

  8

  1. (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

  8

  1. (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

  8

  1. 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

  1. (a) T 3 PQR ® PIQIRI

4    PI (0,5), QI (2,2) RI (1,0)

PI QI RI       PII  QII  RII

(b)  4  3    0   2   1   =   15    14   4                                                              M1 Ö

1  2     5   2   0        10     6    1                                                             A1 Ö

 

PII (15,10), QII (14,6), RII (4,1)                                                               B1Ö

(c)  Area s.f = det M

= 5

 

area of PII QII RII = 5 (area PIQIRI)

= 5 X 3.5                                                          M1Ö

= 16.5 cm2                                                        A1

  8

 

Teachers’ new salary and allowances- What TSC wants to award teachers in new salaries

The Teachers Service Commission, TSC, has presented new salary increments to the Salaries and Remuneration Commission, SRC. The salaries increment are to be approved so as to be included in the next next Collective Bargaining Agreement, CBA; cycle. In the proposals, the Commission wants a basic salary increment of between 16 percent and 32 percent; with classroom teachers getting the higher perks.

The 16 percent rise in basic pay should be for teachers in administrative grades (C4 to D5) who reaped big from the 2016-2021 CBA. Classroom teachers in lower grades (B5 to C3) are to be awarded an increment of 30 percent.

But, the teachers’ unions ,Kenya National Union of Teachers (KNUT) and Kenya Union of Post Primary Education Teachers (KUPPET), are vehemently opposed to the proposals terming them a drop in the ocean. The unions claim there were no talks between them and TSC to come up with the new salary scales; as should be the case during the CBA negotiations.

KNUT is proposing a basic pay rise of between 120 and 200 percent, while KUPPET wants 30 to 70 percent increment.

For allowances, TSC is proposing a 20 percent increment in commuter and leave allowances. On its part, house allowance is to be increased by 10 percent. These proposals are by far much lower than what the unions are proposing.

DEMANDS BY KUPPET.

KUPPET is rooting for the expansion of teachers’ allowances so as to include post graduate and township allowances (meant for teachers staying in towns). The union also wants the Commission to harmonize house allowances for all teachers. Currently, tutors plying their trade in towns and former municipalities earn higher house allowances as opposed to their counterparts in rural areas. Harmonization of house allowances should be based on job group as opposed to regions.

Another allowance being fronted by KUPPET is special school allowance to be paid at a rate of Sh15,000 per month.

They also want the readers’ facilitation allowance to be reviewed by 30 percent so as the teachers will get Sh.19,500 per month. This allowance is paid to a visually impaired teacher who has engaged a reader whose minimum qualification is not below KCSE D+/KCE Division III. The allowance is paid at a fixed rate determined from time to time by the commission.

Also called facilitation or aid allowance, reader’s allowance is currently paid at a rate of Sh15,000 per month to the blind teachers and those confined to wheel chairs by virtue of their disability.

The current CBA cycle has expired, paving the way for implementation of the new one with effect from July 2021.

Still on allowances, KUPPET demands that leave allowance be paid based on one’s basic pay i.e. an equivalent of one month’s basic pay for all cadres.

KUPPET says the Commission has no scheme of service for teachers who have attained a Masters and Doctorate degrees. Instead, the employer awards three increments to the holders of such qualifications. The proposed scheme seeks to allow TSC recruit teachers possessing post-graduate qualifications at entry level.

The Union is demanding that the Post graduate scheme of service be developed and be eligible to all teachers holding a Master’s and Doctorate degree. The said teachers shall be paid an allowance equivalent to 40% of the basic salary. Finer details show that the union is seeking to have a holders of Masters Degree getting an increment of 20 percent that is pegged on their basic salaries and 40 percent for PHD degree holders.

See also;

KNUT DEMANDS ON ALLOWANCES.

The Sossion led outfit is agitating for a 50 percent increment in both house and Commuter allowances. Just like KUPPET, KNUT also wants an upward adjustment on leave allowance equivalent to a teacher’s basic pay for one month.

For hardship allowance, KNUT wants the Commission to award a rise of 50 percent of basic salary.

The medical scheme currently being offered by Minet will still be in place.

TSC OFFER ON ALLOWANCES

The Commission is willing to offer an increase of 20 percent for both commuter and leave allowances. It is also keen to retain the current clusters used to determine the teachers’ house allowances but review it upwards by 10 percent.

Also to be retained is hardship allowance at the current rates.

Proposals by KUPPET to have the TSC introduce new allowances have fell on deaf ears. The Commission has declined the introduction of special school, township and post graduate allowances.

Related TSC news

TSC releases circular on teachers’ promotions

The Teachers Srvice Commission, TSC, has come out clean on the state of promotions for teachers. According to the commission a number of school administrators who are due for promotion have not been moved. Instead, some classroom teachers are enjoying higher perks due to an error during data collection by TSC county and sub county directors.

The teachers’ employer  collected data for institutional administrators in in June, 2017. This was supposed to validate and move the administrators to higher grades once the current Collective Bargaining Agreement (CBA) was to be implemented.  But, TSC boss Dr Nancy Macharia now says the commission’s field officers provided inaccurate data leading to disparities in promotions for institutional heads, their deputies, senior teachers and senior masters.

“It has been established that some of you (TSC field officers) provided inaccurate and misleading data leading to errenous conversions (promotions to the new grades after implementing the SRC salary guidelines in 2017),” says Dr Macharia.

“Classroom teachers were converted to grades exclusively reserved for institutional administrators, teachers serving as deputy or senior teachers were wrongly captured as heads while in other instances, staffing levels in terms of required administrators in a school exceeded the optimum establishments contrary to the established staffing norms,” she further explains.

Related content;
COLLECT DATA AFRESH

A number of school deputy heads and principals have been crying foul over the move by the TSC to snub them.

But, the commission now seeks to recollect the data and clean its payroll. Dr Macharia explains that having unreliable data in the TSC data base has ‘negatively affected the operations of the commission in terms of payroll management and the process of selection, appointment and deployment of new administrators’.

The commission has now directed the field officers to recollect the data and submit the same to the headquarter by June 30, 2020.

“You are hereby directed to critically analyze the data and reconfirm all institutional administrators,” says the TSC boss in her latest circular to the field officers.

Also required is data for schools with extra administrators (teachers with minimum of grade C4 in secondary schools and their counterparts in primary schools having C2) so that they can be appointed to serve in their respective positions.

The issue of teachers’ promotions has been thorny. Most affected are tutors serving in grade C3 (Secondary Teacher) with some having stayed in the job group for over 15 years. and, this has been worsened by the commission’s failure to advertise promotion vacancies for these and other classroom teachers; citing budgetary constraints.

TSC is set to implement the fourth phase of the 2017-2021 Collective Bargaining Agreement (CBA) that was agreed between the Commission and teachers’ unions. The Commission has already implemented the first, second and phases; that were effected in 1st July, 2017, 1st July, 2018 and July 1, 2019 respectively. Read full article here; what teachers will earn per job group after TSC implements the fourth phase of CBA in July, 2020.

Form 2 Physics Exams and Marking Schemes Free

NAME……………………………………………………………..ADM/NO……………………

 

DATE…………………………. ………………… ………….CLASS……………………………

 

JOINT EXAMINATIONS

FORM 2

TERM THREE

PHYSICS 232

TIME: 2HRS

 

INSTRUCTIONS TO STUDENTS

  • Write your name and admission number in the spaces provided above.
  • The paper contains two sections, section A and B
  • Answer all the questions in the spaces provided
  • All working must be clearly shown
  • Candidates should check the question paper to ascertain that all the 12 pages are printed as indicted and that no questions are missing.
  • Candidates should answer the questions in English

 

 

SECTION QUESTION MAXIMUM SCORE CANDIDATE’S

SCORE

 

A 1-12 25  
B 13-18 55  
                                                                                        TOTAL  

 

 

SECTION A (25 MARKS)

Answer all questions in the spaces provided

 

  1. The micrometer screw gauge below has a zero error of – 0.19mm.

Determine the actual thickness of the object.                                                     (2mks)

 

 

  1. Two mirrored walls stand at an angle to each other. A student standing in the room counts nine images of himself in the mirrors. Determine the angle between the walls.

(3mks)

 

 

 

 

 

  1. a) What is meant by the term anomalous expansion of water? (1mks)

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

 

  1. b) Explain any two applications of contraction and expansion in solids. (2mks)

………………………………………………………………………………………………………

………………………………………………………………………………………………………

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………………………………………………………………………………………………………

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………………………………………………………………………………………………………

………………………………………………………………………………………………………

  1. A body weighs 600N on the surface of the earth and 450N on the surface of another planet. Calculate the value of g in that planet (g on the earth = 10N/Kg)               (3mks)

 

 

 

 

 

 

  1. State two applications of electrostatic charges (2mks)

 

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

 

  1. 200 coulombs of charge passes through a point in a circuit for 0.6 minutes. What is the magnitude of the current flowing? (3mks)

 

 

 

 

 

 

  1. While heating water in a beaker, a wire gauze is placed below the beaker explain.(2mks)

 

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

 

  1. What is the relationship between physics and technology (2mks)

 

………………………………………………………………………………………………………

………………………………………………………………………………………………………

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………………………………………………………………………………………………………

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  1. Using the domain theory distinguish between magnetic material and a magnet (2mks)

 

………………………………………………………………………………………………………

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………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

  1. State two application of convection in fluids (2mks)

 

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

 

  1. Convert a temperature of 234K to degree celsious                                (1mk)

 

 

 

 

SECTION B (55 MARKS)

  1. (a) State Hooke’s law                                                                                         (1mk)

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

(b) In an experiment to verify Hooke’s law, a piece of rubber was fixed to a rigid support and the other end pulled with a force of ranging magnitude.  The values of force and the extension were recorded as in the table below:-

Force (N) 0 0.20 0.55 0.75 1.00 1.30 1.40
Extension(cm) 0 1.5 2.5 3.5 4.5 6.0 7.0

 

  • Plot a graph of force ( Y axis) against extension (X-axis) on the gird provided

(5mks)

  • From the graph, determine the spring constant of the rubber within elastic limit (3mks)

 

 

\

 

 

 

 

 

 

 

  • What is the size of force at the elastic limit             (1mks)

 

 

 

 

 

 

 

 

  1. (a) State three characteristics of a brake fluid. (1mk)

 

………………………………………………………………………………………………………

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……………………………………………………………………………………………………

(b) The height of a mercury barometer at a particular place is 70cm. given that the density of mercury is 13600kgm3, determine;

(i) The atmospheric pressure at the place.                                                                             (3mks)

 

 

 

 

 

 

 

(ii) The height of a water barometer at the same place. (Density of water=1g/cm3)                       (2mks)

 

 

 

 

 

 

(iii) Give a reason why mercury is preferred as a barometric liquid.                                   (1mks)

 

 

 

 

(c) Calculate the minimum pressure a block of dimensions 3cm by 10cm by 15cm and mass 12kg could exert on a horizontal surface.                                                                                          (3mks)

 

 

 

 

 

 

 

 

  • a) Differentiate between transverse waves and longitudinal waves.(2mks)

 

………………………………………………………………………………………………………

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  1. The figure below shows a wave form in a string.

 

 

 

 

 

 

 

Given that the speed of the wave is 10m/s. With reference to this wave motion, determine;

  1.           (1mk)

 

 

  1.           (1mk)

 

 

  • (2mks)

 

 

 

  1. Period                                                   (2mks)

 

 

 

 

  1. A person standing 49.5m from the foot of a cliff claps his hands and hears an echo 0.3 seconds later. Calculate the velocity of the sound in air.           (3mks)

 

 

 

 

 

 

 

  1. (a) What property of light is suggested by the formation of shadows?       (1mk)

…………………………………………………………………………………………………

(b) A building standing 200m from a pinhole camera produces on the screen of the camera an image 2.5cm and high 5.0cm behind the pinhole.

Determine the actual height of the building                                                                          (3mks)

 

 

 

 

 

 

(c) An object of height 2.0cm is placed 5.0cm in front of a convex mirror of focal length 10.0cm

 

(i) On the grid provided, draw to scale a ray diagram to locate the position of the image.(4mks)

 

(ii) State two applications of concave mirrors                                                          (2mks)

………………………………………………………………………………………………………

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  1. (a) When is an object said to be in stable equilibrium?       (1mk)

 

………………………………………………………………………………………………………

………………………………………………………………………………………………………

 

 

 

 

(b)  A uniform metal rod of length 80cm and mass 3.2kg is supported horizontally by two vertical spring’s balances C and D. Balance C is a 20cm from one end while balance D is 30cm from the other end. Find the reading on each balance.                                                   (4mks)

 

 

 

 

 

  1. (a) The figure below shows an electric bell. Briefly explain how it works.  (4mks)

 

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

  • State the right hand grip rule for straight conductor carrying current (1mks)

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

………………………………………………………………………………………………………

  1. a) State the equation of continuity (1mk)

…………………………………………………………………………………………………….

 

 

  1. b) The velocity of glycerin in a 5cm internal diameter pipe is 1.00m/s. Find the velocity in a 3cm internal diameter pipe that connects with it, both pipes flowing full. (3mks)

 

____________________________________________________________________

NAME……………………………………………………………..ADM/NO……………………

 

DATE…………………………. ………………… ………….CLASS……………………………

 

JOINT EXAMINATIONS

FORM 2

TERM THREE

PHYSICS 232

TIME: 2HRS

MARKING SCHEME

 

 

 

SECTION A (25 MARKS)

Answer all questions in the spaces provided

 

  1. The micrometer screw gauge below has a zero error of – 0.19mm.

 

 

 

 

Determine the actual thickness of the object.                                                           (2mks)

 

Reading shown is                  3.50

                                         +    0.01

                                       3.51mm

 

Actual thickness        =          3.51

                                                      +    0.19

                                                3.70mm

 

 

 

  1. Two mirrored walls stand at an angle to each other. A student standing in the room counts nine images of himself in the mirrors. Determine the angle between the walls.

(3mks)

n =– 1         9=– 1

 

Ѳ=

 

 =360

 

  1. a) What is meant by the term anomalous expansion of water? (1mks)

Anomalous expansion of water is defined as the unusual behavior of water in which it contracts when heated and expands when cooled between 0 and 40C

 

  1. b) Explain any two applications of contraction and expansion in solids. (2mks)
  2. Expansion joints in Steam Pipes

Pipes carrying steam are fitted with loops or expansion joints to allow for expansion when steam is passing through them and contraction when they are cooledFixing of Railway Line

  1. Railway lines are constructed in sections with expansion gaps and the sections held together by fishplates. The bolt holes in the rails are oval to allow free expansion and contraction of rails as the bolts move freely in the holes.

A modern method of allowing for expansion and contraction in railways is to plane slant the rails so that they overlap

  1. Installation of Telephone/ Electric Wires

They are loosely fixed to allow for contraction. Telephone or electric wires appear to be shorter and taut in the morning. When it is hot, the wires appear longer and slackened

  1. Fixing of Steel Bridges

In bridges made of steel girders, one end is fixed and the other end placed on rollers to allow for expansion and contraction.

1

  1. A body weighs 600N on the surface of the earth and 450N on the surface of another planet. Calculate the value of g in that planet (g on the earth = 10N/Kg)               (3mks)

Mass  on the earth’s surface      =          600

                                                                   10

                                                      =          60Kg

 

      Value of g in the other planet           =          450

                                                                  60

 

                                                      =          7.5N/Kg

 

  1. State two applications of electrostatic charges (2mks)

 

  • Electrostatics precipitators
  • Finger printing
  • Spray painting

Photocopying

 

 

  1. 200 coulombs of charge passes through a point in a circuit for 0.6 minutes. What is the magnitude of the current flowing? (3mks)

 

Q=It

             200=I×0.6×60

 

I=

 

=5.556A

  1. While heating water in a beaker, a wire gauze is placed below the beaker explain.(2mks)

 

The vessel is placed on a wire gauZe because the gauZe is a good conductor of heat it therefore spreads the heat to a large area of the vessel.Without gauze heat will concentrate at one point which may lead to uneven expansion of glass hence break.

  1. What is the relationship between physics and technology (2mks)

 

  • Machines used in the field of medicine such as x-rays, body scanners and lasers are all applications of physics.
  • Manufacture and use of satellites and microwave dishes used in information technology to relay information is based on physics knowledge
  • Physics knowledge is also used in defense industry in the manufacture and useof most modern and complex machines.

 

  1. Using the domain theory distinguish between magnetic material and a magnet (2mks)

 

  • For a magnet the dipoles in all domains align towards a common direction while in magnetic magnetic material dipoles in every domain are aligned in different direction.
  1. State two application of convection in fluids (2mks)
  • Domestic hot water system
  • Cooling system of car engine
  • Ventilation
  • Land and sea breeze

 

  1. Convert a temperature of 234K to degree celsious                                (1mk)

 

T=Ѳ +273

234=   Ѳ+273

Ѳ= 234-273= -39c

 

 

 

 

 

 

SECTION B (55 MARKS)

 

  1. (a) State Hooke’s law                                                                                         (1mk)

 

For a helical spring or other elastic material, the extension is directly proportional to thestretching force, provided elastic limit is not exceeded.

 

(b) In an experiment to verify Hooke’s law, a piece of rubber was fixed to a rigid support and the other end pulled with a force of ranging magnitude.  The values of force and the extension were recorded as in the table below:-

 

Force (N) 0 0.20 0.55 0.75 1.00 1.30 1.40
Extension(cm) 0 1.5 2.5 3.5 4.5 6.0 7.0

 

  • Plot a graph of force ( Y axis) against extension (X-axis) on the gird provided

(5mks)

     Plotting=2mk

Scale=1mk

Line=1mk

Axes= 1mk

  • From the graph, determine the spring constant of the rubber within elastic limit

(3mks)

 

                                   K  =         F/E

                                               

Gradient = 0.75  –  0.2

3.5  – 1.0

 

                =  0.5

                    2.5

 

              =  0.2N/cm

 

 

  • What is the size of force at the elastic limit             (1mks)

 

1.3 N  -check on student graph where the straight line starts a bend

 

 

 

 

 

 

 

  1. (a) State one characteristic of a brake fluid. (1mk)

-must be incompressible

-should not corrode the parts of break system

-have high boiling point

-have  low freezing point

(b) The height of a mercury barometer at a particular place is 70cm. given that the density of mercury is 13600kgm3, determine;

(i) The atmospheric pressure at the place.                                                                             (3mks)

 

 

Pa = 0.7×13600×10

 

=95200N/m2

 

(ii) The height of a water barometer at the same place. (Density of water=1g/cm3)                       (2mks)

 

95200=h×1000×10

                        h =

                    =9.52m

 

(iii) Give a reason why mercury is preferred as a barometric liquid.                                   (1mks)

 

Mercury is much denser than water. It also supports a small and measurable column

 

 

(c) Calculate the minimum pressure a block of dimensions 3cm by 10cm by 15cm and mass 12kg could exert on a horizontal surface.                                                                                          (3mks)

 

Pmin=

 

=8000N/m2

 

  • a) Differentiate between transverse waves and longitudinal waves.(2mks)

 

Transverse waves, the vibration of the particles is a right angles to the direction of wave travel while longitudinal waves, the vibration of the particles is in a direction parallel to the direction of the wave travel.

 

b)The figure below shows a wave form in a string.

 

 

 

 

 

 

 

 

Given that the speed of the wave is 10m/s. With reference to this wave motion, determine;

  1.           (1mk)

                             50-10= 40m

 

  1.           (1mk)

 

5m

  • (2mks)

 

  f= velocity/wavelength

   =10/40 = 0.25Hz

 

  1. Period                                                   (2mks)

 

t= 1/f     =1/0.25  = 4s

 

 

 

 

 

 

  1. A person standing 49.5m from the foot of a cliff claps his hands and hears an echo 0.3 seconds later. Calculate the velocity of the sound in air.           (3mks)

 

 

  1. (a) What property of light is suggested by the formation of shadows?       (1mk)

 

Rectilinear propagation of light/light travels in a straight line

 

(b) A building standing 200m from a pinhole camera produces on the screen of the camera an image 2.5cm high 5.0cm behind the pinhole.

Determine the actual height of the building                                                                          (3mks)

 

                        hi  =  v

                        ho      u

 

                        ho  =2.5  x  20000

                                         5

                        =  10000cm  or 100m

 

 

 

 

 

 

(c) An object of height 2.0cm is placed 5.0cm in front of a convex mirror of focal length 10.0cm

(i) On the grid provided,draw to scale a ray diagram to locate the position of the image.(4mks)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. ii) State two applications of concave mirrors (2mks)

 -shaving mirrors

-By dentist to examine teeth

 

  1. (a) When is an object said to be in stable equilibrium?       (1mk)

 

A body is said to be in a stable equilibrium if it returns to the original position after being displaced slightly.

 

(b)  A uniform metal rod of length 80cm and mass 3.2kg is supported horizontally by two vertical spring’s balances C and D balance C is also from one end while balance D is 30cm from the other end. Find the reading on each balance                                                    (4mks)

 

 

 

 

When pivot at C:

Then C.M = (30 x 32) Ncm

A.C.M = 50D

C.M = A.C.M =>

D = 19.2N

 

C + D = 32N

C = 32 – 19.2

= 12.8N

Reading on C = 12.8N

Reading on D = 19.2N

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. (a) The figure below shows an electric bell. Briefly explain how it works.  (4mks)

 

 

When circuit is closed, soft iron coregets magnetized and attracts the soft iron ammarture.The hammer being attached to the armature thus strikes the gong. the contacts breaks, core looses magnetism and armature returns to normal position.The process repeats.

  • State the right hand grip rule for straight conductor carrying current (1mks)

 

When a straight conductor is held with right hand,with thumb showing the direction of current then the fingers show direction of magnetic field

  1. a) State the equation of continuity (1mk)

 

AV= constant   / A1V1=A2V2

a-area v- velocity

 

  1. b) The velocity of glycerin in a 5cm internal diameter pipe is 1.00m/s. Find the velocity in a 3cm internal diameter pipe that connects with it, both pipes flowing full. (3mks)

 

A1=   22/7*2.5*10-4

A2= 22/7* 1.5*10-4

 

A1V1=A2V2

V2=  A1V1/A2

 

22/7*2. 52*10-4 *1)/22/7* 1.52*10-4

=2.778M/S

St Peter’s Nyakemincha Secondary School KCSE 2020 results analysis, grade count and ranking

St Peter’s Nyakemincha Secondary KCSE 2020/2021 RESULTS ANALYSIS (SCHOOL MEAN, INDIVIDUAL CANDIDATES’ RESULTS AND MEAN GRADE SUMMARY)- St Peter’s Nyakemincha Secondary school has been posting impressive performances in KCSE over the years and 2020 is no exception. In the 2020 KCSE examinations the school posted a mean score of 8.026 which is a B- (minus).

St Peter’s Nyakemincha Secondary SCHOOL KCSE 2020 RESULTS ANALYSIS MEAN GRADE SUMMARY

Looking for KCSE 2020/2021 results for all schools and candidates? Visit this portal; KCSE RESULTS PORTAL.

The school registered a total of 340 candidates in the KCSE 2020 exam. In the just released KCSE 2020 results, the school posted an impressive mean score of 8.026 which is a B- (minus). The good news is that 308 candidates managed to score above C+ (plus), which is the minimum university entry grade. This translates to a percentage of 90.59% securing direct entry to university under the placement body, KUCCPS.

Download KCSE 2020/2021 results for this school here; Official Knec KCSE Results Portal

Here is a complete distribution of grades for the school in KCSE 2020 results;

KCSE RESULTS 2020
Grade Entry
A 0
A- 0
B+ 34
B 96
B- 101
C+ 77
C 28
C- 2
D+ 0
D 0

All KCSE results are available here; KCSE 2019, 2020-2021 Results analysis and ranking for all schools and candidates.

Related news; How to get the KCSE 2020/2021 results via Knec SMS Code and online portal

SCHOOL’S KCSE 2019 RESULTS

Read more details here; KCSE 2019 list of top 200 schools nationally; Full list.

You may also like; KCSE 2019 national results and ranking per subject- Physics

Complete information on Strathmore School, Nairobi; KCSE Performance, Location, History, Fees, Contacts, Portal Login, Postal Address, KNEC Code, Photos and Admissions

Strathmore School is a a prestigious private, independent, boys’ day school located in Nairobi; Westlands Constituency, Kenya that follows the 8-4-4 curriculum. The school was started as a Sixth Form college in March 1961. In 1977 the first Form One students were admitted, and in 1987 the first Standard One pupils.

Today, there are around 630 students in the school with a significant number of them receiving financial aid. Strathmore began as a Sixth Form College offering a full range of Arts and Science subjects. The first Form Five students enrolled in March, 1961, and sat the Higher School Certificate Examination in November 1962.

Accountancy studies were added to the College Curriculum in January, 1966, when the first students registered to prepare for the Association of Certified Accountants examinations. The first Form One students enrolled in January, 1977; during the years 1977-1980 the Secondary School was expanded to include Forms One to Four.

The College is managed by Study Centres of Kenya. The land and buildings of the College are held in Trust by Strathmore Registered Trustees which is incorporated as a charitable Trust under the Land (Perpetual Succession) Ordinance.

With the starting of the Primary School, the Trustees have organized Strathmore College into two levels: ‘Strathmore College School of Accountancy’ for post-secondary studies and ‘Strathmore School’ for primary and secondary Studies.

Here are links to the most important news portals:

Co-Curricular Activities

The school participates in the following Extra Curricular activities:

For complete information on all schools in Kenya, including best private and public schools, please visit this link: Schools Portal; Complete guide to all schools in Kenya

Mission

The School’s chief purpose is to develop the minds and consciences of its students so that they can become self-disciplined and resourceful, and therefore better able to serve society in a useful way. Above all, the School emphasizes academic excellence and moral integrity. The School aims at creating an environment that is conducive to learning; both experienced teachers and well-motivated students strive to create this environment.

Through daily close contact with the students in the classroom, in extra-curricular activities and on the sports fields, the School Staff strive to develop in the students such qualities, such as honesty, responsibility, good use of freedom, tolerance, and hard work. The School aims at providing an education based on Christian principles and accepts worthwhile students regardless of creed.

Srathmore Court of Arms

The lion is the symbol of strength, power, courage of determined fight for excellence and justice. It also represents Kenya, our country, which has all the qualities mentioned above.

The Rose in full blossom represents, love, the source of all good desires and actions, even if at times it requires loving sacrifice, represented by the thorns the rose shows in its stem. The rose has some supernatural meaning. Love, in capital letters, is love of God. And also it has some historical meaning associated with the life of Blessed Josemaria who inspired and encouraged those who started Strathmore: He received a sign by means of a wooden gilded carved rose in a very trying moment of his life.

The Three hearts represent the three races which, in 1961 when the School started, were segregated in the colonial system of education. The heart represents the person, since it is taken as the source of all our actions, and the source of love. Being the three hearts of the same colour it shows the radical equality of all people. At the beginning it clearly meant the unity sought of the three major races. These days it means the co-operation of parents, teachers and students towards the same aim.

The motto “Ut ommnes unum sint” is in Latin. It is a quotation from a passage of the Gospel, and translated means “That all may be one”. It expresses the desire of working together towards the same aim, in spite of individual legitimate opinions, preferences and tastes.

The colours in heraldry (the science of making coats of arms) have associated meanings as follows:

  • Blue (Azure) Sky blue means high ideals, high aims;
  • Red (Gules) Blood red means sacrifice, love, fortitude;
  • Yellow (Gold ) Gold means eternity, perfection.

More articles on Education News;

STRATHMORE SCHOOL KCSE RESULTS ANALYSIS

The school has posted impressive performance at the Kenya Certificate of Secondary Education, KCSE, exams over time.

2017 KCSE RESULTS ANALYSIS

Strathmore School presented 88 candidates. The school posted a good performance, with an average of 9.190.  3 students scored straight A, 15 scored A-, 23 scored B+, 24 scored B, 9 scored B-, 6 scored C+, 6 scored C, and 2 scored C-. This, therefore, gives a transition rate of over 90% to university.

The subjects that scored the best means were Computer Studies with a mean of 12,  Mathematics with a mean of 10.557 and French with a mean of 10.133. Improvement was noted in Computers Studies, Chemistry and English.

The table below is a summary of the performance.

2018 KCSE Results Analysis

In 2018, Strathmore School presented 84 candidates who registered 9.988 points (a performance index of 72.28) to emerge the best school nationally.

Strathmore School 2018 KCSE Results

In 2019, the school registered a total of 84 candidates.

Strathmore school Contacts

At one time if you wish to contact the school, use any of the contacts below:

  • School Name: Strathmore school
  • Physical Location: Nairobi
  • Address: P. O. Box: 25095, NAIROBI 00603
  • City/Town: Nairobi
  • School Code: 20406020
  • Phone Number: +254722221221, +254718222222, +254733937945, +2540202398488, +2540202398311
  • Website: http://www.strathmore.ac.ke/
  • County: Nairobi
  • School Category: International School

Best Study Guide to a Doll’s House By Henrik Ibsen Free

Best Study Guide to a Doll’s House By Henrik Ibsen Free

 

A STUDY GUIDE TO

 

A DOLL’S HOUSE

 

BY HENRIK IBSEN

 

 

TABLE OF CONTENTS

  1. About the author………………………………………..5
  2. Genre………………………………………………………6
  3. About the title…………………………………………..7
  4. Tone………………………….…………………………8
  5. Setting…………………………………………….…….9
  6. Structure………………………..………………………10
  7. Character list………………………………….……….12
  8. Synopsis……………………………………………….16
  9. Plot summary and analysis…………………………….17
  10. Character, characterization and role……………………27
  11. Themes………………………………………………….61
  12. Stylistic/literary devices…………………………………83
  13. Revision questions……………………………………..108

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ABOUT THE AUTHOR

Henrik Ibsen was born on March 20, 1828, in Skien, Norway. He was the second son in a wealthy family that included five other siblings. When he was about 8 years old, his family was thrown into poverty due to complications with his father’s business. It was after this when Ibsen started to invest his time reading, writing, painting, and doing magic tricks.

Ibsen wrote his first play, Catiline, in 1850 which generated little interest. His second play, The Burial Mound, however, was performed at the Christiania Theatre on September 26, 1850.

Later, he wrote a series of plays which included Lady Inger (1855), The Feast at Solhoug (1856), Olaf Liljekrans (1857), The Vikings at Helgeland (1858), The Pretenders (1863), Peter Gynt (1867), The League of Youth (1869), Emperor and Galilean (1873), Pillars of Society (1877), A Doll’s House (1879), Ghosts (1881), An Enemy of the People (1882), The Wild Duck (1884), Hedda Gabler (1890), The Master Builder (1892), Little Eyolf (1894), John Gabriel Borkman (1896) and When We Dead Awaken (1899). He also wrote a dramatic epic poem, “Brand” (1866).

He married Suzannah Thoresen in 1858 and their only child, Sigurd, was born the following year. In 1900, Ibsen suffered his first of several strokes and poor health ended his writing career. He died on May 23, 1906.

GENRE

A Doll’s House is a family drama for the obvious reason that it concerns a family. It is a “drama” because it is a play—a piece of literature that is never fully realized until it is put on stage in front of an audience.

It is also a modern tragedy because it focuses on the trials and tribulations that face women in a patriarchal society.  The play explores not only the status of women, but how they are victims of social forces to the extent that they are left with the role of a “doll-wife.” In this tragedy, we don’t get blood and death at the end; we get the death of a marriage and of the characters’ old selves. Ibsen shows Nora, and maybe all the other characters, trapped in a society defined by restrictive gender roles. In order to become more than a doll, Nora must shatter the cornerstone that her entire society is based on: marriage.

The play can also be categorized as a realist drama. In a realist drama, the characters talk in a close approximation of everyday speech. The speeches are straightforward, conversational and concerned with normal, everyday things; which makes the play really easy for a modern audience to associate with.

Example

MRS. LINDE: “You must not forget that I had a helpless mother and two little brothers. We couldn’t wait for you, Nils; your prospects seemed hopeless then.” (Pg 86)

The vast majority of modern plays, TV shows, and movies are written in a similar style.

ABOUT THE TITLE

Just before Nora leaves her husband and children at the end of the play, she has the following to say to her husband, Torvald: “Our home has been nothing but a playroom. I have been your doll-wife.”(Pg 111-112)

It is therefore not too hard to guess where the play’s title might have come from. Torvald has never treated Nora as anything more than a plaything. He admires her beauty. He gets her to dance for him. He even dresses her up in costumes. In effect, she is his doll. The home they live in seems perfect and picturesque, but in reality it is just like the Helmers’ marriage: all for show.

Nora adds, “at home I was papa’s doll-child.”(Pg 112) She has never been anything but a man’s plaything. Every house she’s ever lived in has been just as artificial; first her father’s house, and now her husband’s house.

No wonder the play is titled A Doll’s House!

TONE

In the beginning, the play seems to be biased toward Nora. We are definitely inclined to sympathize with her. It is very hard to be on Torvald’s side. From his reaction toward Nora for eating macaroons, we know that he is overbearing. His demeaning little pet names for Nora further confirm this.

Torvald, however, redeems himself in the end with the last line, “The most wonderful thing of all?”(Pg 120) The line seems to indicate that he is heading toward the same spiritual awakening as Nora.

This makes us move from seeing Nora as Torvald’s prisoner to seeing that all the characters, Torvald included, have been prisoners in some way.

In the end, the tone of the play becomes more objective. Sympathy can be found for all its characters. Hence the play can be said to end with a serious, intense and somber tone.

SETTING

Setting can be discussed from three dimensions: Geographical, historical and social setting.

Geographical setting

This refers to the place or location where the events in the play are taking place. In the play A Doll’s House, the events take place in The Helmers’ Living Room. The dwelling contains comfortable and stylish furniture and such items as a china cabinet, a bookcase with well-bound books, and a piano on carpeted floor—all of which demonstrate a stable financial situation. On a broader level, it is assumed that the events take place in Norway in Europe; however there are no references to anything specially Norwegian. This assumption is made because that is where Ibsen was born and raised.

Historical setting

This refers to the time in history when the events in the play took place. The events in A Doll’s House took place in The Victorian Era, presumably around the late 1870s. During this time, gender roles were very stiff and clearly outlined. Women were expected to be submissive to their husbands; husbands were expected to dominate. Women raised the children; men went out to work. Anyone who challenged these deeply entrenched values faced some serious consequences. This charged atmosphere of gender division was the reason that the play became such a phenomenon.

Social setting

Social setting refers to the kind of a society in which the events in the play are taking place. The play involves a middle-class society of family and friends who are reeling under the pressure of strict Victorian values which eventually result to conflicts.

STRUCTURE

Henrik Ibsen’s A Doll’s House is divided into three Acts. Ibsen followed the form of a well-made play. Features of a well-made play include increasing suspense by methodical plotting, introducing past events early on and unraveling a secret, which leads to the climax of the play.

The play circumvents through four major stages:

  1. Major conflict – This comes in the form of Nora’s struggle with Krogstad, who threatens to tell her husband about her past crime, incites Nora’s journey of self-discovery and provides much of the play’s dramatic suspense. Nora’s primary struggle, however, is against the selfish, stifling, and oppressive attitudes of her husband, Torvald, and of the society that he represents.
  2. Rising action – This comes in Nora’s first conversation with Mrs. Linde; Krogstad’s visit and blackmailing of Nora and Krogstad’s delivery of the letter that later exposes Nora.
  3. Climax – This is reached when Torvald reads Krogstad’s letter and erupts angrily.
  4. Falling action – This finally comes in Nora’s realization that Torvald is devoted not to her but to the idea of her as someone who depends on him and her decision to abandon him to find independence.

 

 

 

 

 

 

CHARACTER LIST

Nora Helmer

Nora is the play’s protagonist and the wife of Torvald Helmer. She is the central character, who is a “doll” for her husband to dress up, show off, and give direction to. She is childlike and romps easily with her three children. She has never lived alone, going immediately from the care of her father to that of her husband. Inexperienced in the ways of the world as a result of this sheltering, Nora is impulsive and materialistic. She takes a loan from Krogstad to make her husband’s holiday possible. Later, she emerges as a fully independent woman who rejects both the false union of her marriage and the burden of motherhood.

Torvald Helmer

Torvald Helmer is Nora’s husband of eight years. At the beginning of the play, he has been promoted to manager of the bank. He was once gravely ill and needed to go to a southern climate to improve his health. He has built his own legacy through his own work and not from family money. He lives his life according to society’s norms – both professionally and personally. He spends a great deal of his time at home in his study, avoiding general visitors and interacting very little with his children. In fact, he sees himself primarily as responsible for the financial welfare of his family and as a guardian for his wife. Torvald is particularly concerned with morality. He also can come across as stiff and unsympathetic. Still, the last Act of the play makes it very clear that he dearly loves his wife.

Dr. Rank

Dr Rank is a friend of the family of Torvald as well as his physician. He is sick from consumption of the spine (tuberculosis of the spine) as a result of a venereal disease contracted by his father. He confesses his desire for Nora in the second Act and dies in the third Act.

Mrs. Christine Linde

Mrs Linde is an old schoolmate of Nora’s. She is a widow. She comes back into Nora’s life after losing her husband and mother. She successfully asks Nora to help her secure a job at Torvald’s bank. Ultimately, she gets married to Krogstad.

Nils Krogstad

Nils Krogstad is a man from whom Nora borrows money to pay for her family’s trip to Italy. He is an acquaintance of Torvald’s and an employee at the bank which Torvald has just taken over. He is also a lawyer and moneylender. Krogstad was involved in a work scandal many years previously; as a result, his reputation is tainted because he once committed a forgery. When his job at the bank is threatened by Torvald, he blackmails Nora to ensure that he does not lose it. Dr. Rank calls Krogstad “morally diseased.”(Pg 25)

Ivar, Bob, and Emmy

These are Nora’s young children. They spend little time with their mother or father: they are mostly with their nurse, Anne. In the play, the children speak no individualized lines; they are “Three Children.” Ibsen facilitates their dialogue through Nora’s mouth.

Anne

Anne is the family nurse. She raised Nora too after she (Nora) lost her mother to death. She stayed on to raise Nora’s children. Nora is confident that she can leave her children in Anne’s care. She gave up her own daughter to “strangers.”

Helen

Helen is a housemaid employed by the Helmers.

 

 

Porter

The porter brings in the Christmas tree at the very beginning.

Nora’s father

Although he never makes a physical presence during the play, Nora’s father’s influence is felt throughout its course. Torvald repeatedly brings up his loose morals and past scandals to compare them to Nora.

 

 

 

 

 

 

 

 

 

 

 

 

 

SYNOPSIS

Nora Helmer once secretly borrowed a large sum of money so that her husband, Torvald Helmer, could recuperate from a serious illness in Italy. She never told him of this loan and has been secretly paying it back in small installments by saving from her household allowance. Her husband thinks her careless and childlike, and often calls her his doll.

When he is appointed bank director, his first act is to relieve a man who was once disgraced for having forged his signature on a document. This man, Nils Krogstad, is the person from whom Nora has borrowed her money! It is then revealed that she forged her father’s signature in order to get the money.

Krogstad threatens to reveal Nora’s crime and thus disgrace her and her husband unless Nora can convince her husband not to fire him. Nora tries to influence her husband, but he thinks of Nora as a simple child who cannot understand the value of money or business. Thus, when Torvald discovers that Nora has forged her father’s name, he is ready to disclaim his wife even though she had done it for him.

Later when all is solved, Nora sees that her husband is not worth her love and she leaves him.

PLOT SUMMARY AND ANALYSIS

ACT I

SUMMARY

The play opens on the day before Christmas. Nora returns home from shopping; although her husband is expecting a promotion and payrise, he still criticizes her excessive spending. In response, Nora plays around with her husband as a child might, and, indeed, Torvald addresses her as he might a child. He hands her more money but only after having criticized her spending. Their relationship compares with that of a daughter and father and, indeed, is exactly like the relationship Nora had with her father. Early in this act the audience is aware that the relationship between the Helmers is based on dishonesty when Nora denies that she has eaten macaroons, knowing that her husband has forbidden her to do so.

Nora is visited by an old friend, Christine Linde. Mrs. Linde tells Nora that she has had some difficult problems and is looking for employment. Nora confesses to Mrs. Linde that she, too, has been desperate and reveals that she had been forced to borrow money several years earlier when her husband was ill. The money was necessary to finance a trip that saved her husband’s life, but Nora forged her father’s signature to secure the loan and lied to Torvald that her father had given them the money. Thus, she has been deceiving her husband for years as she worked to repay the loan. She tells this story to Mrs. Linde to demonstrate that she is an adult who is capable of both caring for her family and conducting business. Unfortunately, Nora’s secret is known by Krogstad, an employee at Torvald’s bank. After a confrontation with Krogstad, Torvald decides to fire Krogstad and hire Mrs. Linde in his place.

Krogstad threatens Nora, telling her that if he loses his job he will expose her earlier dishonesty. For her part, Nora cannot believe that forging her father’s signature – an act that saved her husband’s life – could lead to a serious punishment. Still, she is concerned enough to plead with Torvald on behalf of Krogstad. Torvald refuses to reconsider firing Krogstad and forbids Nora to even mention his name.

ANALYSIS

The Helmers’ house is decorated tastefully, showing they are relatively well-off. Nora’s happiness as she returns with the Christmas shopping reveals that she enjoys both spending money and doing nice things for her husband and children. At the same time, it will soon become clear that eating the macaroons is an act of deceit and disobedience, as she has been forbidden by Torvald.

Torvald’s nicknames for Nora suggest that he thinks of her almost as a child or a pet. This impression is emphasized when Nora hides the macaroons, like a mischievous child afraid of getting caught. Torvald’s parent-like attitude is highlighted by the way he talks to Nora about money, implying that he thinks she’s not intelligent enough to be financially responsible.

Nora’s happiness shows she enjoys performing the role of a wife and mother. At the same time, her request for money to buy something for herself suggests she wants to be allowed to make decisions for herself. But Torvald clearly doesn’t trust Nora with the money.

Even though Torvald and Nora appear to be in love, Torvald does not trust her, and Nora on her part doesn’t hesitate to lie to him; she was eating macaroons earlier.

Money is central to Torvald and Nora’s happiness. Torvald’s emphasis on their new prosperity suggests how important money is to him as well.

Mrs. Linde has been visibly changed by her life experiences. Nora’s happiness in the last eight years has left her remaining girlishly innocent and naive, whereas Mrs. Linde seems much older. Mrs. Linde’s decision to travel alone was unusual for women at the time, and Nora’s admiration of her “courage” suggests a desire for independence. Mrs. Linde’s status as a widow adds to the impression that she is much older than Nora.

In this part of the play Nora is childishly impolite. Mrs. Linde is obviously in a bad situation following the death of her husband, yet instead of listening to her Nora begins to insensitively boast about her and Torvald’s good fortune. Her speech also shows that she believes money leads to freedom and happiness.

Mrs. Linde’s story shows how difficult it was for women to survive without the financial support of men. The need for money effectively forced her to marry her husband, and after his death her struggle to support her family highlights the obstacles women faced in earning a reasonable income.

Both Mrs. Linde and Nora have strange and suspicious reactions to Krogstad’s arrival. Thus when Krogstad claims he is here on “routine” business matters, we are tempted to believe there is more to the story.

Here, Krogstad reveals more about Nora’s deceitful nature; not only did she lie to Torvald (and everyone else) about where the money for the trip to Italy came from, but she also committed forgery, an illegal act. He threatens to reveal the secret unless she does him a particular favour. Nora is terrified to the point that she even seems to be going mad.

ACT II

SUMMARY

Mrs. Linde stops by to help Nora prepare for a costume ball. Nora explains to Mrs. Linde that Krogstad is blackmailing her about the earlier loan. After Nora again begs Torvald not to fire Krogstad, her husband sends Krogstad an immediate notice of his dismissal. Nora is desperate and decides to ask help from Dr. Rank, a family friend, for a loan, to clear Krogstad. Before she can ask him for his help, Dr. Rank makes it obvious that he is in love with her and Nora decides that because of this it would be unwise to ask his help. Krogstad visits Nora once again and this time leaves a letter for Torvald in which Nora’s dishonesty is revealed. To divert Torvald’s attention from the Krogstad’s letter in the mailbox, Nora engages him to help with her practice of the dance she is to perform, the tarantella. Finally, Nora asks Torvald to promise that he will not read the mail until after the party.

ANALYSIS

In the opening of the second act, the stripped Christmas tree not only shows that time has passed, but also symbolizes a negative shift from the  joy of Christmas to a sense of ruin and chaos. Nora’s obsession in checking to see if any person or letter has arrived and assurances that no one will come for two days gives a sense of time running out and impending disaster.

Nora cannot think of anything else but her secret and the possibility of someone finding out. She tries to occupy herself with the clothes but is unable to.

As the play progresses, it becomes more and more clear how possessive Torvald is. Nora’s pride at saying Dr. Rank is “her” friend suggests she doesn’t really have many friends now that she is married. Nora believes that the reason that Torvald is so controlling is because he is so in love with her.

Nora seems increasingly desperate and crazed. Her mutterings to herself when she is alone show the effect that concealing her secret in front of others is having on her. She lies easily to Dr. Rank, showing how natural lying has become to her.

Nora flirts with Dr. Rank in a very provocative manner. When she teases him with the stockings, this is a very explicit sexual gesture. Her promise to dance for him likewise betrays a disregard for the boundaries of her marriage and a delight in displaying her femininity and sexuality.

Nora is almost asking Dr. Rank to help with keeping the secret of the debt from Torvald, but she is stopped by his confession of love. The confession changes her view of Dr. Rank completely. Where before she perhaps thought flirtation was harmless, the fact that Dr. Rank seems to genuinely love her becomes too much to handle, and she retreats in a rather childlike way.

Krogstad is determined to keep his position at the bank, to the extent of lacking etiquette for Nora, which shows he is desperate. Meanwhile, Nora must cover her tracts in front of everyone—even the maid—hence increasing her isolation.

 

 

 

ACT III

SUMMARY

In this act, it is revealed that Krogstad had years earlier been in love with Mrs. Linde. At the beginning of this act they agree to marry, and Krogstad offers to retrieve his letter from Torvald. However, Mrs. Linde disagrees and thinks that it is time that Nora is forced to confront the dishonesty in her marriage. After the party, the Helmers return home and Torvald opens the letter from Krogstad. While Torvald reads it in his study, Nora pictures herself as dead, having committed suicide by drowning in the icy river. Torvald interrupts her fantasy by demanding that she explains her deception.

However, he refuses to listen and is only concerned with the damage to his own reputation. Torvald’s focus on his own life and his lack of appreciation for the suffering undergone by Nora serve to open her eyes to her husband’s selfishness. She had been expecting Torvald to rescue her and protect her, and instead he only condemns her and insists that she is not fit to be a mother to their children.

At that moment another letter arrives from Krogstad telling the Helmers that he will not take legal action against Nora. Torvald is immediately excited and is willing to forget the entire episode. But having seen her husband revealed as self-centered, egoistic and hypocritical, Nora tells him that she can no longer live as a doll and expresses her intention to leave the house immediately. Torvald begs her to stay, but the play ends with Nora leaving the house, her husband, and her children.

ANALYSIS

Here, Mrs. Linde radically disrupts the course of events in the play. While it would have been easier for her to ask Krogstad to get his letter back, thereby ensuring that life between the Helmers went on as normal, Mrs. Linde’s belief in honesty triumphs over her promise to Nora. This finally benefits Nora, as Torvald’s behaviour when he reads the letter allows her to see the reality of her situation and that she no longer wants to remain in her marriage.

In this act it is clear that Torvald is thinking of Nora far more as a possession that he can display in order to impress other people than a real person with her own thoughts and feelings. To him, Nora was at the party merely to perform for the enjoyment of him and others, not to have a good time herself.

Nora’s bitterness toward Mrs. Linde because she did not get Krogstad to retrieve the letter shows that she has cut herself off even from her close friends in her obsession with the secret of the debt. All the hope and innocence seems to have drained out of her, and she has become a much more serious, grave person.

In his speech we see that Torvald’s love and desire for Nora is revealed to be cosmetic, rather than an appreciation for whom she truly is as a person. He talks about his sexual desire for her with no consideration of whether she is feeling the same way at the moment; indeed, when she tells him that she doesn’t want to be with him that night, he dismisses her feelings by saying she must be playing a game. In reminding her that he is her husband, Torvald is suggesting that their marriage means Nora does not have the right to refuse sex with him, a commonly held belief at the time.

Nora is preparing to kill herself, perhaps the ultimate symbol of self-sacrifice. Her whispering murmurs on the stage suggest that she is becoming mad.

Throughout this whole section of the play Torvald only thinks of himself and doesn’t pause to consider the way Nora has been and will be affected by Krogstad’s threats, or that Nora did what she did purely out of love for him.

Nora has evidently undergone a transformation both visually and in the way she speaks to Torvald. For the first time, she is addressing him as an equal and demanding that he treats her with respect by listening and not interrupting.

Finally, Nora conducts what can be considered an unofficial divorce ceremony. Although Torvald doesn’t want her to go, the fact that he agrees to give her his ring and not to write or try to help her shows that he finally respects her wishes and ability to make decisions for herself.

 

 

 

 

 

 

 

 

 

 

CHARACTERS, CHARACTERIZATION AND

ROLE

NORA HELMER

Nora Helmer is the protagonist or the main character or the heroine in the play.  Still a young woman, she is married to Torvald Helmer and has three children. Nora is by far the most interesting character in the play. Her whole life is a construct of societal norms and the expectations of others.

CHARACTER TRAITS

  1. Impulsive and a spendthrift

-In her first moments onstage, we see her give the porter an overly generous tip.

-She comes in with tonnes of Christmas presents, and shrugs at the idea of incurring debt.

-Unbeknown to Torvald, Nora borrowed money so that they could afford a year-long trip to Italy.

  1. Loving

-Nora borrowed money so that they could afford a year-long trip to Italy. Doctors said that Torvald would die without it—but that he shouldn’t know how bad his condition was.

-Nora brings home lots of Christmas presents for everybody in her house.

-She plays hide-and-go-seek with her kids.

  1. Independent and farsighted

-In the past, Nora was always a passive child-like possession who followed Torvald’s orders, but towards the end of the play, she is an independent adult and is able to dominate Torvald.

  1. Wise and intelligent

-Nora uses wisdom and intelligence to confront an emergency. She forges her father’s signature in order to secure a loan from Krogstad so as to save her husband’s life.

-Nora realizes that her understanding of herself, her husband, her marriage, and even her society was all wrong. She decides that she can no longer be happy in her life and marriage, and resolves to leave Torvald and her home in order to find a sense of self and learn about the world, a newly empowered woman boldly escaping the oppressive clutches of her old life.

-Nora has been leading a double life. She has not been thoughtlessly spending their money. Rather, she has been saving to pay off a secret debt.

 

  1. Childlike, immature, ignorant and whimsical

-She happily accepts the pet names “little lark”, “little squirrel”, and “Little Miss Extravagant” that her husband calls her without any opposition. In fact she seems to enjoy and even play into it.

-The maturity level Nora exhibits demonstrates that the relationship between Torvald and Nora is more like father and daughter than husband and wife.

  1. Irresponsible and reckless

-Her first act on stage is paying the porter. Though his service only costs sixpence, she gives him a shilling. (Pg 1) The casual way in which she gives it to him is indicative of her irresponsibility. She hands him the shilling and before he can thank her, she decides in the middle of the transaction that she is not patient enough to wait for change.

-She forges loan documents to raise money for an expensive trip to Italy. Even if the documents were not forged, Nora did not have any means to repay the loan anyway.

-She has never spent serious time with her husband of nearly a decade, and is always dumping her children on the nurse rather than bonding with them herself.

 

  1. Dishonest and deceitful

-She falsely blames Mrs. Linde for smuggling forbidden macaroons into the house.

-She has been eating macaroons, something she has been forbidden by her husband, despite her promises of total obedience to him.

– At the beginning of her marriage, she secretly borrowed money from Nils Krogstad and forged her father’s signature in order to finance a trip to Italy that was necessary to save Torvald’s life.

  1. Unfeeling

– She blames Mrs. Linde for smuggling forbidden macaroons into the house. Though she is just trying to hide her indiscretions, she does not care whom she hurts in the process.

  1. Materialistic

-She is always trying to make herself happy by buying things: dresses, toys, candy etc., rather than doing anything meaningful with her life.

– She is infatuated with the acquisition of possessions.

  1. Decisive, resolute and independent-minded

– At the end of the play, it becomes clear to Nora that “[she] had been living all these years with a strange man, and [she] had born him three children.”(Pg 117-118) This realization forces her into the real world and she ceases to be a doll. At the end of the above statement, she adds “Oh, I cannot bear to think of it!”

-She tells her husband, “Our home has been nothing but a playroom. I have been your doll-wife, just as at home I was papa’s doll-child.”(Pg 118) In the end, Nora has a sort of spiritual awakening. She walks out into the night alone but, for perhaps the first time in her life, she’s on the path to becoming a fully realized, fully independent human being.

– She decides to desert her family to go on a quest of personal enlightenment.

-The act of concealing the ill-gotten loan signifies Nora’s independent streak. She is proud of the sacrifice she has made. Although she says nothing to Torvald, she brags about her actions with her old friend, Mrs. Linde.

– Nora is independent enough to negotiate the loan to make her family’s holiday possible, and over the course of the play, Nora emerges as a fully independent woman who rejects both the false union of her marriage and the burden of motherhood.

 

 

  1. Manipulative

-At the end of the play, Nora seats Torvald at the table and explains her situation to him. She does not let him speak until she has finished what she wants to say.

– Other examples of manipulation are having a nanny take care of her children, having Mrs. Linde repair her dress, behaving seductively around Dr. Rank, influencing Torvald to give her money, and most importantly convincing Krogstad to overlook the similarity between her penmanship and her “father’s.”

  1. Selfish

– She does not want to forgive Torvald. She would rather start another life than try to fix her existing one.

  1. A dreamer

– Until she comes to the realization that her life is a sham, she spends her whole life in a dream world in which she does not take anything seriously.

– In her dream world, Nora takes a back seat approach to life and becomes like an object, reacting to other’s expectations rather than advancing herself.

 

 

  1. Trusting and naïve

-She trusted that Krogstad would not blackmail her and it therefore comes as a rude shock when he does so.

-Until she comes to the realization that her life is a waste, she spends her whole life in a dream world of naivety. In this dream world, Nora does not take life seriously, an attitude that led to many of the plot’s complications.

-She believes that Torvald loves her enough to take all blame upon himself, but she is mistaken. When she realizes that he is more concerned with appearances and respectability than with her happiness, she decides to leave him and find her own way in life.

-She naively thought that Torvald would selflessly give up everything for her. When he fails to do this, she accepts the fact that their marriage has been an illusion. Their false devotion has been merely play acting. She has been his “child-wife” and his “doll.”

  1. Determined

-Whenever Nora would get money from Torvald, she would reserve half of it to repay the debt, determined to clear it all one fine day.

-She was so determined to save her husband that she committed fraud to do so.

  1. Hardworking

-She has been secretly working odd jobs to pay back the debt.

  1. Courageous, bold, daring and tenacious

-To save her husband from poor health, she committed fraud. She valued love over the law.

-She courageously confronts Torvald about the demeaning way he treats her at the end of the play.

-She slammed the door on her husband as she left.

-Although she has been forbidden from eating macaroons by her husband, she still does it anyway despite her promises of total obedience to him.

  1. Calculating

-She is blackmailed by Krogstad, so she begs Torvald to let Krogstad keep his job.

-She flirts with Dr. Rank in the hope of borrowing money from him.

– She gets Christine to go and talk to Krogstad on her behalf regarding the withdrawal of the letter.

-She dances the tarantella to distract Torvald from the mail.

  1. Principled and firm

-She decides against borrowing from Rank when he reveals that he is dying and is in love with her.

-She rejects Torvald’s drunken advances after the party.

  1. Secretive

-She has never told Torvald where the money for their trip to Italy came from, as his pride would suffer.

-She also hides her thoughts and actions from her husband even when there is no real benefit in doing so.

  1. Suicidal

-She contemplates committing suicide in order to eliminate the dishonour she has brought upon her family.

  1. Rebellious

-She eats macaroons even though she has been forbidden by her husband.

ROLE

  1. Advancement of the plot – Nora is the central character in the play around whom the play circulates. As a result of her borrowing money from Krogstad, the conflict in the play is born as Krogstad tries to blackmail her with that secret, in order to make Nora’s husband, Torvald, to allow him keep his former job at the bank. When Torvald refuses, Krogstad sends him a letter to inform him about Nora’s indebtedness. Torvald’s reaction to this letter not only betrays his true nature, but also leads to Nora leaving him and her children to seek independence and freedom.
  2. Development of themes – Nora helps in developing themes such as love and marriage, women and feminity, lies and deceit, money and materialism, the sacrificial role of women, parental and filial obligations, the unreliability of appearances, gender roles, individual vs. society, growth and development and betrayal
  3. Revealing character traits of other characters – Nora helps in revealing the character traits of other characters. For example, she helps in revealing Torvald as selfish and egoistic, naïve, strict, loving, hypocritical and hardworking. Through her, we learn that Krogstad was on one hand, vicious and ruthless, but on the other hand, merciful and forgiving. Dr Rank’s friendly but immoral nature is revealed through Nora. Mrs Linde traitorous nature is also revealed through her.
  4. Enhancing style – Nora helps in enhancing the style of symbolism (through the symbols like the Christmas tree, tarantella, the dolls, macaroons), dramatic irony, foreshadow, etc.

TORVALD HELMER

Torvald Helmer is a lawyer who at the start of the play has recently been promoted to Bank Manager. He is married to Nora Helmer, with whom he has three children. He does not seem particularly fond of his children; even once saying that their presence makes the house “will only be bearable for a mother now!”(Pg 30) His best friend is Dr. Rank, who visits him every day.

  1. Loving and affectionate

-He loves and is very affectionate towards Nora. That is why he showers her with endearments like “My little skylark”, “My little squirrel”, “My little singing bird,” “My pretty little pet,” “My little sweet-tooth,” and “My poor little Nora.” and “Little Miss Extravagant.” With every term of endearment, the word “little” is always included to show affection.

– His despair as Nora exits at the very end of the play suggests that, despite his patronizing and unjust treatment of her, Torvald really does love Nora.

  1. Generous

-He treats Nora generously, giving her extra money when she asks for it.

  1. Proud

-Typical of many contemporary heads-of-the-family, he is a proud specimen of a middle-class husband.

  1. Morally upright

– He sees Krogstad as irredeemably morally tainted, and hence decides to give his job to Mrs Linde.

– He is keenly concerned with his place and status in society and wouldn’t allow anybody to threaten his reputation, including his own wife.

  1. Selfish

-He considers Nora merely as an ornamental sex object instead of an equal partner in their marriage and the mother of his children.

-He maintains amorous fantasies toward his wife: he dresses her as a Capri fisher girl and encourages her to dance in order to arouse his desires.

-At the end of the play, Nora imagines that Torvald will defend her honour and not allow Krogstad to blackmail the Helmers. Nora imagines that Torvald would sacrifice his own reputation and future to save her, but Torvald tells her that he would not make the sacrifice, shattering Nora’s dream world. At this point it becomes clear to her that she had been living all these years with a strange man, and she had born him three children.

-He planned to cope with the scandal resulting from blackmail by stripping Nora of her spousal and motherly duties, but would keep her in the house for appearance sake.

-He is overly concerned with his place and status in society, and he allows his emotions to be swayed heavily by the prospect of society’s respect and the fear of society’s scorn.

  1. Hardworking and diligent

-He spends a great deal of his time at home in his study working, avoiding general visitors and interacting very little with his children. In fact, he sees himself primarily as responsible for the financial welfare of his family and as a guardian for his wife.

  1. Dictatorial, authoritarian and patronizing

-He restrains Nora with rules, much as a father would have to inhibit a child, forbidding her from eating macaroons and other temporal pleasures.

  1. Manipulative

-He insists on Nora wearing the fish girl costume for the tarantella. The costume and dance are part of Torvald’s fantasy of gazing upon Nora from across the room at a party and pretending that she is something exotic. Torvald made Nora take on a foreign identity; he used her as a doll.

  1. Unforgiving

-When he finds out about the debt, he fails to forgive her until he is sure that his reputation is safe.

  1. Heartless and unfeeling

-At the end of the play, Torvald seems untroubled and even a little relieved at the thought of Dr. Rank’s death.

-When he finds out about Nora’s secret debt, he instantly turns on her until he confirms that his reputation is safe.

-His heartless and unfeeling nature makes Nora not to tell him the truth about her loan, and Dr Rank not to tell him about his imminent death.

  1. Conservative and traditional

-Torvald’s focus on status and being treated as superior by people like Nils Krogstad points at his obsession with reputation and appearances.

-He has straightforward and traditional beliefs about marriage and society.

-When Nora tells him she is leaving him, Torvald at first reacts by calling her mad and saying she is acting like a stupid child.

-He is unable to cope with the disagreeable truths of life.

However, he can be said to be flexible because when he realizes how resolute Nora is in her decision, he offers to change and desperately searches for a way to make her change her mind.

  1. Shallow and vain

-He is incapable of understanding his wife or of properly returning her love.

  1. Hypocritical and self-righteous

-Though he regards her as his wife, he never considers her an equal partner in the relationship.

-Many times throughout the play, Torvald criticizes the morality of other characters. He trashes the reputation of Krogstad, one of his lesser employees. He speculates that Krogstad’s corruption probably started in the home. Torvald believes that if the mother of a household is dishonest, then surely the children will become morally infected. He also complains about Nora’s late father. When Torvald learns that Nora has committed forgery, he blames her crime on her father’s weak morals.

-In the beginning of Act Three, after dancing and having a merry time at a holiday party, Torvald tells Nora how much he cares for her. He claims to be absolutely devoted to her. He even wishes that some calamity would befall them so that he could demonstrate his steadfast, heroic nature.

Of course, a moment later, such an opportunity arises. Torvald finds the letter revealing how Nora has brought scandal and blackmail into his household. Nora is in trouble, but Torvald fails to come to her rescue as he had promised.

  1. Naïve

-Throughout the play, Torvald is oblivious to his wife’s craftiness. When he discovers the truth at the end, he is outraged.

ROLE

  1. Advancement of the plot – Torvald is also a major character in the play who plays an instrumental role in the development of the plot. It is as a result of his illness and the subsequent one-year stay in Italy that caused Nora to get into a debt trap with Krogstad. When he refuses to reinstate Krogstad to his former job at the bank, he intensifies the conflict because this makes Krogstad to send him a letter exposing his wife’s secret. The fact that he at first refuses to forgive her leads to Nora’s sudden discovery that he was a selfish, egoistic man. She has no alternative but leave him and her children to seek independence and freedom.
  2. Development of themes – Torvald also helps in developing themes such as love and marriage, pride, honour, respect and reputation, money and materialism, parental and filial obligations, the unreliability of appearances, gender roles, individual vs. society, and betrayal.
  3. Revealing character traits of other characters – Torvald helps in revealing the character traits of other characters. For example, he helps in revealing Nora as impulsive and a spendthrift, childlike and immature, irresponsible and reckless, dishonest and deceitful, manipulative, calculating and traitorous. Through him also, we learn that Krogstad was unscrupulous, vicious and ruthless but merciful and forgiving.
  4. Enhancing style – Torvald helps in enhancing the style of imagery through his pet names for Nora such as “My little skylark”, “My little squirrel”, “My little singing bird,” “My pretty little pet,” “My little sweet-tooth,” and “My poor little Nora.” He also enhances the style of symbolism like his insistence that Nora should wear the dancing costume, similes like when he says he will protect her “like a hunted dove,” dramatic irony, verbal irony, hyperbole, etc.

 

MRS CHRISTINE LINDE

Mrs. Linde is an old schoolfriend of Nora’s. She is a woman whose marriage was loveless, and based on a need for financial security, and who doesn’t have any children. She and Krogstad had been in love before, but he was too poor to support her family. She arrives in town in search of a job in order to earn money and survive independently.

CHARACTER TRAITS

  1. Honest and Truthful

-She tells Krogstad that Nora must eventually conclude, through her own sufferings, that the only way of life which can survive crises is one based on truthful relationships.

-She believes very deeply in honesty and stops Krogstad from taking the letter he wrote to Torvald back, thereby ensuring that Torvald finds out about Nora’s secret.

-She insists that, “Helmer must know everything. This unhappy secret must come out!”(Pg 90) Even though she has the power to change Krogstad’s mind, she uses her influence to make certain that Nora’s secret is discovered.

 

 

  1. Hardworking

-Towards the end of the play, she explains to Krogstad that she finds joy and meaning in work.

-She worked hard to support her helpless mother and two younger brothers since the death of her husband.

  1. Independent-minded

-She arrives in town in search of a job in order to earn money and survive independently. In this way, she is a fairly modern woman.

  1. Traitorous

-She stops Krogstad from taking the letter he wrote to Torvald back, thereby ensuring that Torvald finds out about Nora’s secret, which seems like betrayal to her friend Nora.

  1. Conservative and traditional

-She tells both Krogstad and Nora that she is miserable without other people to take care of, thereby fitting into the traditional role of women as caretakers and nurturers. It is this conviction that causes her to marry Krogstad towards the end of the play.

  1. Selfish and materialistic

-She ended up marrying another man in order to have enough money to support her dying mother and young brothers. Apparently, Krogstad was too poor at this time to marry her. This left Krogstad lost and embittered, unhappy in his own marriage, and is presented as the reason behind his moral corruption.

  1. Resilient

-She has lived an independent life as a single working woman. She has struggled financially and now that she has no one to look after, she feels empty.

ROLE

  1. Plot development– she lets the audience know the inner thoughts of the protagonist. She has a major effect on events that happen in the play.
  2. Reveals character traits of Nora and Krogstad – Mrs. Linde functions as a convenient device for exposition. She enters Act One as an almost forgotten friend, a lonely widow seeking a job from Nora’s husband. However, Nora does not spend much time listening to Mrs. Linde’s troubles. Rather selfishly, Nora discusses how excited she is about Torvald Helmer’s recent success. Through Mrs Linde, Nora launches into a dramatic explanation of all her secret activities (obtaining a loan, saving Torvald’s life, paying off her debt). Mrs Linde therefore functions as the primary means by which the audience learns of Nora’s secret and her character traits. She is the first character to see that Nora is not a child.
  3. Develops themes– she introduces the theme of deception. Through Mrs. Linde, Nora reveals that she has lied to save Helmer’s life and therefore deceived him with her cleverness.
  4. Enhances style– she foreshadows how Nora will confront a bitter future after learning that her marriage is based on deception by recounting how she herself sacrificed her rights to love and self-determination by marrying for financial security.

DR RANK

Dr. Rank is a medical doctor who is best friends with Torvald and Nora, who he visits every day. He suffers from spinal tuberculosis; a condition he believes was caused by his father’s vices, which included having extramarital affairs and consuming too much luxurious food and drink. He is unmarried and lonely, and over the course of the play it is revealed that he is in love with Nora.

CHARACTER TRAITS

  1. Friendly and loyal

-Nora explains how she always feels at ease around Dr. Rank because he does not have any expectations or demands of her.

-He visits the Helmers every day.

  1. Immoral

– It is revealed that he is in love with Nora, his best friend’s wife.

  1. Courageous

-He is unconcerned with what others think of him.

-He has accepted his fate and his impending death.

  1. Cynical

-He rejoices when he finds out that his illness is terminal, and insists that neither Torvald nor Nora visits him in his dying days. As he predicted, he is not particularly missed by the other characters.

  1. Trusting

– He trusts Nora completely. He refrains from telling Torvald of his imminent death because it is too “ugly” an idea for him to tolerate, but he does tell Nora, an indication of the bond between them. He talks with her about his coming death in a code that excludes Torvald and protects him from the harsh reality.

  1. Hypocritical

-Dr Rank is not as straightforward as he appears. His real motive for visiting the Helmers is that he is in love with Nora.

 

  1. Realistic

– On the subject of the costume party, Dr. Rank suggested that Nora should go as herself and that Torvald should be invisible. Under the surface, Rank is suggesting that Nora should not be a doll. With an invisible chaperon, Nora would not be dominated by a figure placing an identity over her.

ROLE

-To provide amusement for Nora as a change from the tiresome rules of Torvald, just as she used to seek the conversation of the maids as a refreshing change from the strictness of her father.

– Dr. Rank adds to the somber mood of the play; he is not essentially useful to the conflict, climax, or resolution.

NILS KROGSTAD

Nils Krogstad Krogstad is an employee at the bank at which Torvald is made manager. He leant Nora the money to take Torvald to Italy to recuperate. He is, at least at the beginning, the main antagonist: Everything is going well for the Helmers until Krogstad enters the story. Known to the other characters as unscrupulous and dishonest, he blackmails Nora, who borrowed money from him with a forged signature, after learning that he is being fired from his job at the bank. In the past, he too committed the crime of forgery, an act that he did not go to prison for but that nonetheless ruined his reputation and made it extremely difficult to find a respectable job.

CHARACTER TRAITS

  1. Morally Corrupt

-In the past, he too committed the crime of forgery, an act that ruined his reputation, though he did not go to prison. But it made it extremely difficult for him to find a respectable job.

-He was once in love with Kristine Linde, who ended up marrying another man in order to have enough money to support her dying mother and young brothers. This left Krogstad lost and embittered, unhappy in his own marriage, and is presented as the reason behind his moral corruption.

-Torvald, who sees Krogstad as irredeemably morally tainted, decides to give his job to Mrs Linde.

  1. Merciless, vicious and callous

-At first he treats Nora without mercy when demanding his money. He argues that no mercy has been shown to him in life.

 

 

 

  1. Unscrupulous and dishonest

-He blackmails Nora with the threat of exposing her indebtedness unless she talked her husband Torvald into giving him back his job at the bank.

-Unless Nora persuades Torvald to keep Krogstad in his job (he later extends this to a promotion), he will tell Torvald about her loan and her forgery of her father’s signature.

  1. Forgiving and remorseful

-At first he treats Nora without mercy on the basis that no mercy has been shown to him in life; however, after Mrs. Linde and he decide to marry, he becomes happier and rescinds his threats to Nora, saying he regrets his behaviour.

-He removes his threats to the Helmers and sends Nora’s bond back to her, relinquishing his power over her.

-After engaging in a conversation with his lost love, the widow Mrs. Linde, they reconcile, and once again their romance is reignited, Krogstad no longer wants to deal with blackmail and extortion. He is a changed man!

-Although Mrs. Linde suggests that he should leave the first letter in the mailbox so that Nora and Torvald can finally have an honest discussion about things, he later drops off a second letter explaining that their secret is safe and that the IOU is theirs to dispose of.

-He has been trying to remake his life after having made earlier mistakes.

– Although he has been labeled as corrupt and “morally sick,” Krogstad has been trying to lead a legitimate life. He complains, “For the last year and a half I have not had a hand in anything dishonourable, amid all that time I have been struggling in most restricted circumstances. I was content to work my way up, step by step.”(Pg 75) Then he angrily explains to Nora, “And be sure you remember that it is your husband himself who has forced me into such ways as this again. I will never forgive him for that.”(Pg 76)

– He is one of several examples in the play of a person being forced into morally questionable action as a result of the rigid and unmerciful forces of society.

  1. Loving and responsible

-As soon as Mrs Linde tells him that she has always loved him and asks him to resume their relationship, he reveals himself as a more loving, joyful and merciful character.

-In matters concerning his children, we find the true measure of the man. Nils Krogstad is a good father and is not afraid of doing whatever it takes to make sure his family is secure.

– Despite the financial strain that he found himself under as a single parent Nils Krogstad still brought up his two boys as best he could.

– Although at times Krogstad is vicious, his motivation is for his motherless children, thus casting a slightly sympathetic light on his otherwise cruel character.

  1. Hardworking and resilient

– Here is a man who is a single parent of two boys who is totally unsupported by the society he exists in. He has been dealt many blows by life, first by Mrs. Linde who rejected him on financial grounds many years ago, and then by the death of his wife. He has to work multiple jobs to support himself and his family; by day he is a lowly bank clerk and by night he is a moneylender and he even finds time to write for a paper.

  1. Frank and sincere

-When Mrs. Linde proposes they resume their old relationship, Krogstad remains truthful and makes sure she is aware of his past deeds as well as what people think of him. He even makes sure she knows about his current dealings with the Helmers.

ROLE

  1. He advances the plot – Krogstad initiates the conflict by attempting to blackmail Nora Helmer. He serves as a catalyst. Basically, he initiates the action of the play. He sparks the flames of conflict, and with each unpleasant visit to the Helmer residence, Nora’s troubles increase. In fact, she even contemplates suicide as a means of escaping his torments.
  2. Develops themes – Krogstad helps to develop the themes of love and marriage, pride, honour, respect and reputation, money and materialism, parental obligations and individual vs. society.
  3. Reveals character traits of other characters– Through him we are able to know that Nora is secretive and deceitful, and Mrs Linde is traitorous and materialistic for leaving him when he was poor.

NURSE

She is a nurse to both Nora and Nora’s children. Her name is Anne Marie. The nursemaid is an example of a woman in bad circumstances forced to do anything in order to survive.

 

CHARACTER TRAITS

  1. Kind

-She was forced to give up her own child, who it is suggested was born out of wedlock.

  1. Reliable and responsible

-When Nora first thinks of leaving, she considers the fact that her children will be raised by the nursemaid and, remembering what a good mother the nursemaid had been to her, decides that she would also raise Nora’s children well.

  1. Immoral

-She gave birth to a baby out of wedlock,

  1. Irresponsible

-She had to give up her own child in order to take up her position as the nursemaid at the Helmers.

  1. Self-sacrificial

-She had to give up her own child in order to take up her position as the nursemaid at the Helmers. Nora finally leaves her children in her care, believing that they will be better off than they would be with her.

 

 

ROLE

  1. Developing themes – She helps to develop the theme of the sacrificial role of women by giving away her child to strangers so that she could concentrate on her job.

IVAR, BOB, AND EMMY

These are Torvald and Nora’s young children. Raised primarily by Anne, the Nurse (and Nora’s old nurse), the children spend little time with their mother or father. The time they do spend with Nora consists of Nora playing with them as if she were just another playmate. The children speak no individualized lines in the play; they are “Three Children.” Their dialogue is facilitated through Nora’s mouth, and they are often cut entirely in performance.

CHARACTER TRAITS

  1. Playful

-They asked their mother to play child games with them and they played hide-and-seek.

  1. Insistent

When their mother showed reluctance to participate in the children’s game, they insisted until she gave in.

 

  1. Loving

-They loved their mother dearly and would have wanted to spend more time with her and to continue playing children’s games with them.

ROLE

  1. 1. To bring out the character of Nora as a loving mother. She showers them with Christmas gifts and even plays children games with them. When Nora later refuses to spend time with them because she fears she may morally corrupt them, she acts on her belief that the quality of parenting strongly influences a child’s development.

HELEN

She is a housemaid employed by the Helmers.

CHARACTER TRAITS

  1. Humble

-She answers Nora with a lot of humility.

  1. Hardworking

-She does her work diligently.

PORTER

This is the porter who brings the Christmas tree to the Helmers house at the very beginning of the play.

CHARACTER TRAITS

  1. Obedient

-He obediently delivers the Christmas tree to the Helmers house.

  1. Grateful

-He thanks Nora for the tip that she gives him. (Pg 1)

  1. Honest

-He honestly states his charges without exaggerating and was already giving her back the change when she told him to keep it. (Pg 1)

NORA’S FATHER

Though Nora’s father is dead before the action of the play begins, the characters refer to him throughout the play. Though she clearly loves and admires her father, Nora also comes to blame him for contributing to her subservient position in life.

CHARACTER TRAITS

  1. Manipulative

-He manipulated Nora to do according to his wishes and whims. She complains that her father and her husband both treated her like a doll.

 

 

  1. Immoral

-Torvald criticizes him as having been a morally crooked man who engaged in corrupt deals.

  1. Insensitive

-The way he treated Nora was too insensitive for a father to treat his daughter.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

THEMES

MARRIAGE AND LOVE

At the beginning of the play, Nora and Torvald appear to be very happily married, even to themselves. Nora talks joyfully about her love for Torvald, and Torvald refers to Nora using affectionate pet names. At first the Helmers seem happy, but over the course of the play, the imbalance between them becomes more and more apparent.

Their loving marriage stands in contrast with the lives of the other characters. For example, the marriage of Krogstad and Mrs. Linde was based on necessity rather than love, and were unhappy. Dr. Rank, on the other hand, was never married, and it is later revealed that he has silently loved Nora for years.

Yet although Nora and Torvald’s marriage is based on love (as opposed to necessity, as was the case with Krogstad and Mrs. Linde), it is still governed by the strict rules of society that dictated the roles of husband and wife. It is clear that Nora is expected to obey Torvald and allow him to make decisions for her.

At first it seems that Nora and Torvald both enjoy playing the roles of husband and wife in a way that is considered respectable by society. However, Nora soon reveals to Mrs. Linde that she secretly borrowed the money from Krogstad behind Torvald’s back, and therefore has already broken both the law and the rules of marriage at the time. This creates a dilemma: Nora broke the rules of marriage, yet did so in order to save her husband’s life – a true act of love.

By the end, the marriage breaks apart due to a complete lack of understanding. Nora Helmer, the “doll” wife, realizes after eight years of marriage that she has never been a partner in her marriage. At the play’s conclusion, she leaves her husband in order to establish an identity for herself that is separate from her identity as a wife and mother.

The main message of A Doll’s House seems to be that a true marriage is a joining of equals. The play centres on the dissolution of a marriage that doesn’t meet these standards.

There is a lot of talk about love in A Doll’s House. Throughout the play we hear of and see many different forms of love: familial, maternal, paternal, and fraternal. Romantic love even blossoms for two of the secondary characters, namely Krostad and Mrs Linde. However, for the main characters, the Helmers, true romantic love is elusive.  They finally discovered that true love never existed between them.

WOMEN AND FEMININITY

Nora has often been painted as one of modern feminist heroines. Over the course of the play, she breaks away from the domination of her dictatorial husband, Torvald. Also throughout this play, there is constant talk of women, their traditional roles, and the price they pay when they break with tradition.

When A Dollʼs House was written in 1879, a wife was not legally permitted to borrow money without her husbandʼs permission. On her wedding day, a woman transferred from living under the authority of her father to under that of her husband.

Poverty had already forced women into the workplace early in the nineteenth century, and the Norwegian government passed laws protecting and governing women’s employment. By the middle of the 19th century, Norwegian women were permitted inheritance rights and the right to an education. But many of the rights provided to women favoured the lower economic classes. Employment opportunities for women were limited to low paying domestic jobs, teaching, or clerical work. Middle class women, such as Nora, noticed few of these new advantages. It was the institution of marriage itself that restricted the freedom of middle class women. Universal women rights were eventually achieved in 1913, making Norway the first country in Europe to have equal voting rights for men and women.

PRIDE, HONOUR, RESPECT AND REPUTATION

The men characters in A Doll’s House are obsessed with their reputation. Some have good names in their communities and will do anything to protect it; others have lost their good names and will do anything to get them back.

Honour

Honour is extremely important to Torvald; it is what motivates his behaviour. Early in the play, his value for honour is the reason he gives for sacking Krogstad, claiming that because he once displayed a lack of honour, it means that Krogstad is forever dishonoured. When he learns of his wife’s mistake, Torvald’s first and foremost concern is for his honour. He cannot appreciate the sacrifice that Nora has made for him; he is only concerned with how society will react to his family’s shame. For Torvald, honour is more important than family and far more important than love; he simply cannot imagine anyone placing love before honour. This issue brings out the glaring difference between Nora and Torvald.

Pride

Like honour, pride is another quality that Torvald upholds. He is proud of Nora in the same way one is proud of an expensive or rare item or possession. When her scandal threatens to be exposed, Torvald is very fearful of losing his public pride. Instead of accepting Nora with her misperfections, Torvald instead rejects her when she is most in need of his support. His pride in himself and in his possessions blinds him to Nora’s worth and value. Nora is left with no choice but to leave him. Only when she has made the decision to leave Torvald does she begin to develop pride in herself.

LIES AND DECEIT

The tension that runs throughout A Doll’s House comes from Nora’s fear of her secret being discovered. Her great terror being exposed leads her to tell a lie after a lie. When her web of lies finally reaches a climax, her marriage proves too weak to bear the strain.

At the beginning of the play, Nora appears to be a dutifully obedient and honest wife, however it is quickly revealed that she is hiding a serious secret from her husband—the fact that she borrowed money from Krogstad to finance a trip to Italy that she claims saved Torvald’s life. This confirms that all her statements about never disobeying or hiding anything from him were nothing but deceitful. When she reveals her dishonesty to Mrs. Linde, Mrs. Linde insists that she ought to confess to Torvald immediately, insisting that a marriage cannot succeed when husband and wife are not completely honest with each other.

But Nora is not alone in telling lies and being deceitful. Krogstad is also revealed to have committed a forgery. The fruits of their acts of deception are devastating: Krogstad’s reputation is ruined, and Nora is forced to leave her husband and family at the end of the play.

It should however be noted that the motivation behind Nora’s dishonesty was love – she lied in order to save her husband’s life. Furthermore, she wouldn’t have been deceitful if it weren’t for societal law dictating that women were not allowed to handle financial matters independently. Therefore Nora’s deceit was not the result of a personal flaw, but rather an attempt to commit a noble act of saving her husband’s life that went awry.

Dr. Rank also comes out as deceitful and dishonest. He has been deceiving both Nora and Torvald for years about the depth of his feelings for Nora. Only when she attempts to seek his financial help does Nora finally see beneath the surface to the doctor’s real feelings. He has been lusting for his best friend’s wife all those years. Nora is so shocked to discover this that she automatically decides not to ask Dr. Rank for financial assistance.

Torvald, who has been deceived throughout most of the play, is finally revealed in the final act to be the one most guilty of deception. He has deceived Nora into believing that he loved and cherished her, while all the while he had regarded her as little more than his property.

MONEY AND MATERIALISM

Throughout the entire play everyone is talking about money, as if it was a god. As the entire issue starts over a debt, the play revolves around money and who has it as well as who does not have it. It is a prevailing theme due to that.

In the very first scene, Nora gives the porter one shilling, telling him to “keep the change”, thus indicating her relaxed attitude to money and spending. The next scene with Torvald almost entirely revolves around the subjects of money, spending and borrowing, with Nora portrayed as a spendthrift. Torvald has very strong views on borrowing and debt. He says to her, “That is like a woman! But seriously, Nora, you know what I think about that. No debt, no borrowing. There can be no freedom or beauty about a home that depends on borrowing and debt.”(Pg 3)

A need for money affects all the major characters in A Doll’s House. In the beginning of the play it is revealed that Torvald was recently promoted and will receive “a big salary.”(Pg2) However, he still criticizes Nora for overspending, arguing that they need to be cautious financially. Mrs. Linde is in desperate need of a job following the death of her husband. Krogstad’s replacement at the bank leaves him threatening to reveal Nora’s secret in order to get his job back because he fears he will lose his source of income. Indeed, the bank works as a symbol for the pervasive presence of money in the characters’ lives.

Throughout the play A Doll’s House, the characters spend a good deal of time talking about their finances. Some are said to be doing quite well financially, and some have the promise of their finances improving in the future. Others are struggling to make ends meet. Either way, each character’s financial status seems to be a defining feature.

In the play, money symbolizes the power that the characters have over one another. In the first Act, Torvald’s ability to dictate how much Nora spends on Christmas presents shows his power over her. On the other hand, the debt that Nora owes Krogstad allows him to have power over her and Torvald. Both Nora and Mrs. Linde cannot earn large incomes because they are women; their inability to access significant amounts of money shows the power that men have over the women in this society.

It is also clear that, while earning money leads to power, it can also be dangerous. For instance, even if money actualized Nora and her family’s trip to Italy, the debt she owed Krogstad soon became a source of terror, dread, and shame. The thrill of obtaining money soon became a nightmare for her.

Krogstad is a moneylender, and money (or lack of it) has had a major effect on his life. We learn that Mrs Linde ended her relationship with him many years ago because of his lack of financial security, choosing to marry a richer man instead. Throughout his life Krogstad has been poor, struggling to support his family, and it is this dependency on financial income that leads him to blackmail Nora in an attempt to keep his job at the bank. Mrs Lindeʼs life has also been directly affected by money, or lack of it. Her late husbandʼs business collapsed, leaving her with nothing to live on, and since then she has had to work hard to survive.

Dr Rank is the only main character who appears to be comfortable financially, having inherited money from his late father. However, although he is financially comfortable he is terminally ill, referring to his body as being “bankrupt.”

Torvald in particular focuses on money and material goods rather than people. His sense of manhood depends on his financial independence. He was an unsuccessful lawyer because he refused to take “unsavory cases.” As a result, he switched jobs to the bank, where he will primarily be dealing with money.

THE SACRIFICIAL ROLE OF WOMEN

In A Doll’s House, Ibsen paints a bleak picture of the sacrificial role held by women of all economic classes in his society.

In order to support her mother and two brothers, Mrs. Linde found it necessary to abandon Krogstad, her true but poor lover, and marry a richer man.

The nanny had to abandon her own child to support herself by working as Nora’s and later as Nora’s children’s caretaker. As she tells Nora, the nanny considers herself lucky to have found the job, since she was “a poor girl who has got into trouble…” (Pg 50)

Though Nora is economically advantaged in comparison to the play’s other female characters, she nevertheless leads a difficult life because society dictates that Torvald be the marriage’s dominant partner. Torvald issues rules and looks down on Nora, and Nora must hide her debt from him because she knows Torvald would never accept the idea that his wife had helped save his life. Furthermore, she must work in secret to pay off her loan because it is illegal for a woman to obtain a loan without her husband’s permission.

Nora’s abandonment of her children can also be interpreted as an act of self-sacrifice. Despite Nora’s great love for her children, as seen in her interaction with them and her great fear of corrupting them, she chooses to leave them. Nora truly believes that the nanny will be a better mother and that leaving her children is in their best interest.

All the three women in the play have made some kind of personal sacrifice in their lives in order to fulfill the roles which society expects of them. Nora, besides risking her dignity by borrowing money on behalf of her family, she also has sacrificed all her own opinions, thoughts and ideas and adopted Torvaldʼs views as her own. Besides that, she has been saving every bit of money she had and working odd hours of the night to repay Krogstad. And at the end of the play she sacrifices her home, family and children for the sake of her own self-discovery.

Mrs Linde, after her husbandʼs death, continued to make personal sacrifices for the sake of her family, taking on any work she could to support them financially.

Anne-Marie, on the other hand, sacrificed motherhood for a respectable job, which was all too common for young unmarried mothers in the 19th century.

PARENTAL AND FILIAL OBLIGATIONS

There is a strong emphasis throughout the play on the importance of parental and filial responsibility, and of the effect that the actions of parents have upon their children.

Parental obligations

Nora, Torvald, and Dr. Rank believe that a parent is obligated to be honest and morally-upright, because a parent’s immorality is passed on to his or her children like a disease.

For instance, Dr. Rank has a disease that is the result of his father’s wickedness. Dr. Rank implies that his father’s immorality, which included affairs with many women, led him to contract a venereal disease that he passed on to his son, causing Dr. Rank to suffer for his father’s misdeeds. He talks about the unfairness of this, of the sins of the father being passed on to the son.

Torvald, on the other hand, talks about a parentʼs immorality being passed on to the children like a disease. He voices the idea that one’s parents determine one’s moral character when he tells Nora, “Almost everyone who has gone to the bad early has had a deceitful mother” (Pg 30) He speaks about Krogstad poisoning his own children with lies and immorality. He also refuses to allow Nora to interact with their children after he learns of her deceit; for fear that she will corrupt them.

Nora is referred to as being like her father, having inherited a lot of his qualities. It is also important to note that she never had a mother, with Anne-Marie fulfilling the maternal role in her life.

Anne-Marie was forced to give away her own child to take on the role of Noraʼs maid; in contrast Nora chooses to leave her own children at the end of the play.

Filial obligations

Filial means the duties, feelings or relationships which exist between a son or daughter and his or her parents.

The play suggests that children too have an obligation to protect their parents. Nora recognized this obligation, but she ignored it, choosing to be with, and sacrifice herself for, her sick husband instead of her sick father.

Mrs. Linde, on the other hand, abandoned her hopes of being with Krogstad and undertook years of labour in order to tend to her sick mother. Mrs Linde has fulfilled her filial responsibility by dedicating her life to care for her mother, at the expense of her own personal happiness. Her motherʼs illness has directly affected the life she has led and the personal decisions she has made.

Ibsen does not however pass judgment on either woman’s decision, but uses the idea of a child’s debt to her parent to demonstrate that familial obligation is not one way – it is reciprocal.

THE UNRELIABILITY OF APPEARANCES

Over the course of A Doll’s House, appearances prove to be quite misleading and hide the true reality of the play’s characters and situations. Our first impressions of Nora, Torvald, and Krogstad are all later proved quite wrong.

Nora, at first, seems a silly, childish woman, but as the play progresses, we see that she is intelligent, motivated, and, by the end of the play, a strong-willed, independent thinker.

Torvald, on the other hand, though he appears as the strong, benevolent husband, reveals himself to be cowardly, petty, and selfish when he fears that Krogstad may expose him to scandal.

Krogstad, who initially appears to be a vicious, ruthless blackmailer, later reveals himself to be a much more sympathetic and merciful character. He also turns out as an earnest lover. Indeed, the play’s climax is largely a matter of resolving identity confusion – we see Krogstad as a loving merciful man, Nora as an intelligent, brave woman, and Torvald as a helpless, sad man.

Situations too are misinterpreted both by the audience and by the characters. The seeming hatred between Mrs. Linde and Krogstad turns out to be love. Nora’s creditor turns out to be Krogstad and not Dr. Rank, as the audience and Mrs. Linde had thought. Dr. Rank confesses that he is not just a friend to Nora but instead he is in love with her, to Nora’s and the audience’s surprise. The seemingly ruthless Krogstad repents and returns Nora’s contract to her, while the seemingly kindhearted Mrs. Linde fails to help Nora, leading to Torvald’s discovery of Nora’s secret.

GENDER ROLES

A Doll’s House exposes the restricted roles of women during the time of its writing and the problems that arise from a drastic imbalance of power between men and women.

Throughout the play, Nora is treated like a child by the other characters. Torvald calls her his “pet” and his “property,” and suggests that she is not smart or responsible enough to be trusted with money. Neither Krogstad nor Dr. Rank take her seriously, and even Mrs. Linde calls her a “child.” Nora seems unperturbed by the views of others about her; even calling herself “little Nora” and promising that she would never dream of disobeying her husband.

However, there are clues that she is not entirely happy with the limited position she has as a woman. For example, when revealing the secret of how she borrowed money to finance the trip to Italy, she refers to it as her “pride” and says it was fun to be in control of money, explaining that it was “like being a man.” (Pg 21) Nora seems to wish to enjoy the privileges and power enjoyed by males in her society. She seems to understand the confinement she faces simply by virtue of her sex.

Nora’s dissatisfaction with her status as a woman intensifies over the course of the play. In the final scene she tells Torvald that she is not being treated as an independent person with a mind of her own. According to her, the bitter solution to this issue is to leave married life behind, despite Torvald’s begging that he will change. Nora’s problems arise because as a woman she cannot conduct business without the authority of either her father or her husband. When her father is dying, she must forge his signature to secure a loan to save her husband’s life. That she is a responsible person is demonstrated when she repays the loan at great personal sacrifice.

The men in this play have a very conservative view of the roles of women, especially in marriage and motherhood. Torvald, in particular, believes that it is the sacred duty of a woman to be a good wife and mother. Moreover, he tells Nora that women are responsible for the morality of their children. In essence, he sees women as childlike, helpless creatures detached from reality on the one hand, but on the other hand as influential moral forces responsible for the purity of the world through their influence in the home.

The men of A Doll’s House are in many ways just as trapped by traditional gender roles as the women. The men must be providers. They must bear the burden of supporting the entire household. They must be the undoubted kings of their respective castles. Besides providing for their families, the men are obsessed by a desire to achieve higher status. Respectability is of great concern to both Torvald and Krogstad. When Nora’s borrowing is revealed, Torvald’s first thoughts are for his reputation. On the other hand, Krogstad is obsessed with achieving success now that he has changed his character. He intends to one day take over Torvald’s job and run the bank.

By the end of the play, these traditional ideas are truly put to the test.

INDIVIDUAL VS. SOCIETY

Nora, a dutiful mother and wife, spends most of the play putting others before herself. She thinks little about herself to the extent of engaging in an act of forgery and taking a debt for the sake of her husband’s health. She doesn’t stop to worry about how these actions might impact the lives of her husband and children. Even when she plans to kill herself near the end of the play, it is not to hide her shame but rather because she thinks that if she is alive then Torvald will ruin himself in trying to protect her.

Similarly, Mrs. Linde admits that, without a husband or any family members to care for, she feels that her life is pointless. Therefore both women find a sense of meaning in their lives through serving others and performing the caring, obedient role that society requires of them.

However, Nora later learns that prioritizing her duty as a wife and mother cannot lead to real happiness. She realizes that while she thought she was sacrificing herself to protect her love, in fact no such love existed. It becomes clear that Torvald would never have sacrificed his reputation to protect her. She therefore decides to leave him in order to develop a sense of her own identity. The play ends with Nora choosing to put herself as an individual before society’s expectations of her.

Some characters, however, are more concerned about themselves as individuals rather than the society. A good example is Krogstad. Throughout most of the play, it seems that he cares more about his reputation than anything else. Punished by society for his act of forgery, he is desperate to reclaim respectability in the eyes of others. However, he realizes that he will only achieve happiness through truly reforming himself and regaining the personal integrity that he lost, rather than the outward respectability.

In a similar way to Nora, Krogstad learns that society’s view of him is meaningless if he doesn’t respect himself as an individual.

BETRAYAL

Betrayal is a theme of this play in several ways. Nora has betrayed her husband’s trust in several instances. She has lied about borrowing money, and to repay the money she must lie about how she spends her household accounts and she must lie about taking odd jobs to earn extra money. She also chooses to lie about eating macaroons which her husband has forbidden her.

Torvald betrays Nora when he rejects her pleas for understanding. Torvald’s betrayal of her love is clearly shown when he doesn’t want to understand that Nora took the loan because of his own welfare. To him, she threatened his otherwise good reputation in the eyes of the society, which was an unforgivable sin to him. This was the reality that Nora requires to finally awaken from; her previous view about her husband and their marriage was just but a sham.

Mrs Linde also betrays Krogstad when she opts to marry a richer man because Krogstad was too poor to help her sustain her sick mother and needy siblings.

GROWTH AND DEVELOPMENT

In Act I, Nora is portrayed as nothing more than a “doll,” a child who has exchanged a father for a husband without changing or maturing in any way. But as the play progressed, she realized that she had no identity separate from that of her husband. Torvald owned her just as he owned their home or any other possession. She was finally forced to face the reality of the life she was living. She realized in the final act that if she had to develop an identity as an adult, she must leave her husband’s home. When Nora finally gave up her dream for a miracle and, instead, accepted the reality of her husband’s self-centredness, she finally took her first steps toward maturity. She realized the inequity of her situation; she also recognized her own self worth. Her decision to leave is a daring one that indicates the seriousness of Nora’s desire to find and create her own identity.

THE HOME

The fact that the play is called A Doll’s House means that home might be a prevalent theme. Early on in the text, the home is seen as a thing of joy, a place of comfort and shelter. The idea of home is enmeshed with the idea of the happy family, which the Helmers seem to be.

Toward the play’s conclusion, however, the imbalance of power in the family becomes an issue. Now the seemingly happy home is revealed as having been an illusion – a doll’s house – that hid the gulf between the Helmers. The Helmers’ home is really more of a prison than a shelter.

The title, A Doll’s House, implies that everything is a façade, an illusion. Just like a doll that has a plastered smile on its face, the doll’s house hides the problems in the marriage.

STYLISTIC/LITERARY DEVICES

Stylistic or literary devices are techniques (ways to do things, styles, or forms) that authors use to get the attention of the reader which include playing with words, creating imagery, comparing and contrasting, or using metaphors, just to name a few. In A Doll’s House, the author has used a variety of stylistic devices, as discussed below.

SYMBOLISM

Symbols are objects, characters, figures, or colours used to represent abstract ideas or concepts. The following are the symbols used in the play:

  1. Christmas and New Year Days

The action of the play is set at Christmas and New Year season. Christmas and New Year holidays are both associated with rebirth and renewal and several of the characters go through a kind of rebirth over the course of the play.

Both Nora and Torvald have a spiritual awakening, which could be seen as a “rebirth.” Nora’s trials and tribulations wake her up to the sorry state of her marriage. When the “wonderful thing” fails to happen, she realizes she will never be a fully realized person until she breaks away from her husband. And when she slams the door behind her, she is in a way reborn.

Nora is not alone in her spiritual awakening, however. Torvald’s last line, “The most wonderful thing of all?”(Pg 120) seems to indicate that he has also realized the complete inadequacy of his existence. By the end of the play, both Helmers have been reborn.

Krogstad and Christine are reborn as well. When these “two shipwrecked people…join forces,” (Pg 88) they each get a fresh start in life. Both of them view their renewed love affair as a chance for salvation. Krogstad hopes that it will help increase his standing with the community, and that Christine’s influence will make him a better person. Christine, on the other hand, is overjoyed that she will have someone to care for. She once again has purpose in her life.

Nora and Torvald both look forward to New Year’s as the start of a new, happier phase in their lives, a new beginning with no debts. In the New Year, Torvald will start his new job, and he anticipates with excitement the extra money and admiration the job will bring him. Nora also looks forward to Torvald’s new job, because she will finally be able to repay her secret debt to Krogstad. By the end of the play, however, the nature of the new start that New Year’s represents for Torvald and Nora has changed dramatically. They both must become new people and face radically changed ways of living. Hence, the New Year comes to mark the beginning of a truly new and different period in both their lives and their personalities.

In the end of the play, it resembles new beginnings as almost all the characters are starting new lives, Nora and Torvald separately, while Christine and Krogstad together.

  1. Christmas Tree

The Christmas tree symbolizes Nora’s role in her household. She is only a decoration to be looked at. Her function in the household is pretty much the same as the tree. She is merely decorative and ornamental. She dresses up the tree just as Torvald dresses up her for the party. It’s interesting that she tells the maid not to let the children see the tree until it’s decorated.

The Christmas tree, therefore, a festive object meant to serve a decorative purpose, symbolizes Nora’s position in her household as a plaything who is pleasing to look at and adds charm to the home.

It also symbolizes family happiness and unity, as well as the joy Nora takes in making her home pleasant and attractive.

At the beginning of Act Two, the Christmas tree has been stripped of its ornaments and is only left with burnt-down candle-ends on its disheveled branches. Nora is alone in the room, walking about uneasily. Basically, Nora is a mess and so is the tree. She’s gotten the bad news from Krogstad, and as a result her mind is just as disheveled as the poor tree.

The tree seems to mimic Nora’s psychological state. It can be interpreted as symbolic of Nora’s disintegrating web of lies. The pretty decorations that Nora used to cover up her deceit are falling away. Soon the bare, ugly truth will emerge. This represents the end of Nora’s innocence and foreshadows the Helmer family’s eventual disintegration.

  1. Macaroons

Torvald has banned Nora from eating macaroons. Although Nora claims that she never disobeys Torvald, this is proved false in the very opening of the play when Nora eats macaroons while she was alone in the living room. The macaroons are symbolic of Nora’s disobedience and deceit. She lies to Dr. Rank about having been given some by Mrs. Linde, and after giving her performance of the tarantella asks that macaroons be served at dinner, which indicates a close relationship between the macaroons and her inner passions, both of which she must hide within her marriage.

  1. The tarantella

Tarantella takes its name from a spider, a Tarantula,   which, according to the Italian legends, bites its victim to quick death. The only way to get rid of its poison is to dance so as to let the poison come out of the body with the sweat. Similarly, the wild dance of Nora is a symbolic expression of her tragic inner condition and, at the same time, a therapeutic instrument that gives her courage to face up the suicide that she plans to carry out. Nora dances the Tarantella at a time when she had accelerated anxiety, on the border of madness. So through the dance, her body was trying to express what couldn’t be said in words.

Like the macaroons, the tarantella symbolizes a side of Nora that she cannot normally show. It is a fiery, passionate dance that allows her to drop the mask of the perfect Victorian wife and express her desperate and tragic interior condition and her inner feelings.  It is a dance of recovering from the madness of her fate; Tarantella has the power to heal Nora.

After the dance, in fact, she reemerges matured and able to look death in the eyes.

It is important to note that the rehearsal of Tarantella is the first moment in which Nora doesn’t obey what Torvald commands. Her repressed feelings are not allowed to come out in her marriage, the only way she can express them is through a performance. And her performance is wild and hysteric. Through the dance Nora liberates herself from her sexual doll’s role, which is a transformation from an old existence to a new one.

  1. The Doll’s House

The title of the play A Doll’s House is also symbolic. It represents something impermanent or short-lived.

There are a few mentions of dolls early on in the play; for example, when Nora shows Torvald the dolls she bought for her daughter, and says that the fact that they are cheap doesn’t matter because she will probably break them soon anyway. This probably suggests that Nora is raising her daughter for a life similar to her own. It also foreshadows Nora breaking up her family life by leaving Torvald.

When Nora plays with her children she also refers to them as her “little darlings.” (Pg 42) However, it is not until the end of the play that the metaphor becomes explicitly clear. Nora tells Torvald that both he and her father treated her like a doll, and cites this as one of the reasons why she has become dissatisfied and disillusioned with her life with him.

  1. The dance costume

At the end of the play, Nora decides to leave Torvald. The next thing Nora does is to change out of her fancy dance dress. Torvald bought this dress for Nora to wear at a costume party because he wanted her to appear as a “Neapolitan fish girl”. As one would put clothes on a doll, Torvald dresses Nora. When she sheds this dress, she is symbolically shedding her past life with Torvald and her doll-like existence.

  1. Dr. Rank

Dr. Rank is a symbol of moral corruption within society. He has been lusting for Nora secretly. His illness is symbolic of the moral illness of the society as represented by himself, Krogstad and, by extension, Torvald.

 

 

  1. Mrs Linde

Mrs Linde is a symbol of a modern, independent woman. She arrives in town in search of a job in order to earn money and survive independently. She perhaps also symbolizes hollowness in the matriarchal role.

  1. Torvald Helmer

Torvald Helmer is a symbol of a male dominated, authoritative, and autocratic society.

  1. The slamming of the door

The slamming of the door symbolizes the finality of the relationship between Torvald and Nora Helmer.

USE OF FIGURATIVE LANGUAGE

IMAGERY

Metaphors

A metaphor is a comparison without using the terms ‘like’ or ‘as.’ Henrik Ibsen uses quite a number of metaphors in A Doll’s House. These include the following:

  1. Torvald’s pet names for Nora

-He calls her “featherhead,” “songbird,”  “squirrel,” “hunted

dove…saved from hawk’s claws,” and “skylark.” When she leaves him, he calls her a “heedless child.” All these metaphors are, on one hand, aimed at reflecting Nora’s apparently innocent, carefree nature, and on the other hand, they suggest that her husband does not think of her as a proper adult because she is a woman.

-Another metaphor is where Torvald says, “…how much it costs a man to keep such a little bird as you.” Here, Torvald is comparing Nora to a bird by saying that people would not expect her to spend as much money as she does.  The “bird” reference means that birds are typically low maintenance, but Nora is not.

  1. The doll

-In Act 3, Nora tells Torvald that both her father and Torvald have treated her like a doll-child, with no opinions of her own, and have only played with her. Both men, she says, have committed “a great sin” against her in discouraging her from growing up. Torvald’s pet names for her are prefaced by “little,” showing that he sees her as a child.

  1. Big black hat

-In Act 3. Dr. Rank has a coded conversation with Nora (designed to protect Torvald from unpleasant truths) in which he says he will attend the next fancy dress ball wearing “a big black hat” that will make him invisible. This is a way of saying that he will be dead.

Other metaphors

-Krogstad is labeled “morally diseased” because of the incriminating forged bond and the forged documents that tarnished his reputation.

-Nora and Torvald crumbling marriage and home are referred to as a “doll’s house” to mean their impermanency.

– Krogstad uses this metaphor, “I am a shipwrecked man clinging to a bit of wreckage” (Pg 87) to describe how he felt when Mrs. Linde chose to marry her late husband instead of him. Mrs. Linde replies that she had her mother and younger brothers to take care of and she needed financial stability, which Krogstad could not offer her. In this metaphor, Krogstad might be suggesting that he is still in love with Mrs. Linde.

-Torvald refers to his wife as his “frightened little songbird” and promises her that his “big broad wings” would protect her.

-The title of the play A Doll’s House is an extended metaphor. It is comparison of a small toy with that of a perfect house. It compares Nora’s relationship with every man in her life to that of a young child playing with her, merely a pretty plaything.

Similes

A simile is a comparison by use of the terms ‘like’ or ‘as.’ Similes are used in different places in the story to compare certain necessary ideas.

Examples

-“It was like being a man.”(Pg 21) This simile was used by Nora to compare the role she played in sustaining the family during their one-year stay in Italy. It made her feel like a man supporting them for all that time.

-Torvald brags that he will protect Nora “like a hunted dove that [he has] saved from the talons of a hawk.” Here, he wants to emphasize his commitment in ensuring Nora of her safety.

IRONY

Irony is a figure of speech in which words are used in such a way that their intended meaning is different from the actual meaning of the words. It may also be a situation that ends up in quite a different way than what is generally anticipated. There are three types of irony evident in A Doll’s House, namely: verbal, situational and dramatic irony.

 

 

Dramatic irony

Dramatic irony occurs when the audience is more aware of what is happening than one, some or all the characters on stage.

The full significance of a character’s words or actions is clear to the audience or reader although unknown to the character. In other words, the audience’s or reader’s knowledge of events or individuals surpasses that of the characters.

Examples

-This happens in A Doll’s House near the opening of the play when Nora eats macaroons. When Torvald then asks Nora if she has been eating sweets, she lies and says she has not. Nora and the audience know this is a lie and so know more than Torvald, making this a situation of dramatic irony.

-Torvald tells Nora, “That is like a woman! But seriously, Nora, you know what I think about that. No debt, no borrowing. There can be no freedom or beauty about a home that depends on borrowing and debt.”(Pg 3) But nevertheless, she has borrowed money from Krogstad which she has been paying for a long time without his knowledge.

-The reader is aware that Nora borrowed money from Krogstad without her husband’s permission. Nora also forged her father’s name to gain the money. She says, “You don’t know all. I forged a name.” In the following conversation between Nora and Christine it is clearly stated that Torvald does not know of Nora’s actions

Mrs. Linde: And since then have you never told your secret to your husband?

Nora. Good heavens, no! (Pg 20)

-Another example of dramatic irony in A Doll’s House is when Nora wants to practise a dance called the Tarantella. When Torvald goes to look in the letter box Nora says, “Torvald please don’t. There is nothing in there.” (Pg 80) The reader knows there is a letter in the mailbox that has been dropped by Krogstad. The reader also knows that Nora has not forgotten the dance as she claimed, she was just pretending. The reader knows this when Torvald goes to check the mail and Nora begins to play the Tarantella. Nora then says, “I can’t dance tomorrow if I don’t practise with you.” (Pg 81) The reader knows that all Nora is trying to do is keep Torvald from reading the mail which contains a letter from Krogstad.

-Dramatic irony is evident throughout the text to indicate Nora’s exit from her marriage with Torvald. Some escalating events have happened in the three acts to give clues to the audience that she has already decided to leave.

Examples

  1. i) Nora to Nurse regarding the children:

Nora: Yes, but, nurse, I shall not be able to be so much with them now as I was before.

Nurse: Oh well, young children easily get accustomed to anything.

Nora: Do you think so? Do you think they would forget their mother if she went away altogether? (Pg 50)

  1. ii) Nora to Torvald:

Nora: “Torvald, you will be sorry for not letting me stay, even

for just half an hour.”

She knows that the letter is still in the mailbox and doesn’t want Torvald to find out about the contract.

iii) Nora to Mrs Linde:

Nora: “You all think I’m incapable of doing anything serious…or of ever having to face the brutality of life.”

 

 

 

Situational irony

Situational irony occurs when something entirely different happens from what the audience may be expecting, or the final outcome is opposite to what the audience is expecting.

Examples

-Situational Irony is present when Nora is discussing Krogstad’s forgery with her husband in Act 1. Minutes before this conversation, Krogstad approached Nora about her own forgery of her father’s signature.

-There is very little hint that Nora is going to leave Torvald until the end of the play. At the beginning of the play she acts as if she loves him very much. Not until she says “Or if anything else should happen to me – anything, for instance, that might prevent me from being here” does anyone think about Nora leaving Torvald. At the end of the play she calls Torvald a “stranger” and walks out.

-It is ironic that Torvald states that he awaits the moment when Nora will be in trouble so that he can rescue her. When in fact the truth comes out and Torvald has been given his opportunity to rescue Nora, all he is concerned with is his reputation. He yells at her. He insults her by calling her feather brain. He screams at her, telling her to go to her room. He is not interested in how he can rescue her. He is interested in how he can get out of this mess without ruining his good name.

-When Krogstad returns the IOU document, Torvald exclaims that he is saved and that he has forgiven Nora. When Nora asks if she is saved, Torvald exclaims that she is, of course. Only moments earlier, he was furious with her. Ironically, he did not even consider that she had borrowed the money to in fact save him.

-Situational irony is also evident earlier on in the play during Nora’s chat with Mrs. Linde, where she talks, or rather brags about her husband getting promoted as the manager of the bank. She says, “I feel so relieved to have heaps of money and not need to have any anxiety…” (Pg 11-12) Here, Nora visualizes a happy and blissful life with Helmer. However, there is irony in what she says because later on in the play, her marital relationship will be shattered and she will leave her husband and all the “money” that she had visualized, for an uncertain future away from Torvald after realizing that the world she was living in was equivalent to the world of a puppet, or rather, a doll.

 

Verbal irony

Verbal irony occurs when a speaker’s intention is the opposite of what he or she is saying.

Examples

-Verbal irony is present when Helmer says, “Is that my little skylark twittering out there?” (Pg 2) He is not really asking if Nora is a bird. He is not even saying that she is twittering like a bird. He is just asking if it is his wife, Nora, and if she is saying something. When Torvald Helmer says, “Is it my little squirrel bustling about?” (Pg 2) he does not think that Nora is a squirrel either.

-Nora has her share of verbal irony too. When she is sitting down talking to Mrs. Linde she says, “There now, it is burning up.” The place is not literally burning up. The house is not on fire. Nora is just stating that the temperature inside the house is hot.

– When Nora is chatting with Mrs. Linde, where she says “just fancy, my husband has been made manager of the Bank!”(Pg 11) where she talks, or rather brags about her husband getting promoted as the manager of the bank. She says, “I feel so relieved to have heaps of money and not need to have any anxiety…” (Pg 11-12) The reader is tempted to think that her life and that of her family is one smooth ride. But it emerges that she is deep in debt and even has to work extra hours at night in order to keep up with the payments.

FORESHADOW

Foreshadowing refers to clues that point to events that will happen later.

Examples

-Nora’s early rebellion of eating the macaroons against Torvald foreshadows her later rebellion

-The way Torvald always called Nora “My little skylark”, “My little squirrel”, “My little singing bird,” “My pretty little pet,” “My little sweet-tooth,” and “My poor little Nora.” was a foreshadow. She ends up saying something like “I’m just your little dove” in the later Acts when she decides to leave him. She acknowledges the fact there was never actually love between them; she was just his play toy, hence the name of the play, A Doll’s House.

-In the following conversation between Nora and Anne-Marie, there is use of foreshadow.

Nora: Yes, but, nurse, I shall not be able to be so much with them now as I was before.

Nurse: Oh well, young children easily get accustomed to anything.

Nora: Do you think so? Do you think they would forget their mother if she went away altogether? (Pg 50)

Nora eventually leaves her family, which was why she asked Anne-Marie how she possibly could have done it.

-Torvald’s stubbornness about denying Krogstad the banking job has complicated Nora’s attempt to continue hiding her little secret. We know there is going to be trouble later on. Nora’s secret is bound to come out. Ibsen has foreshadowed an ironic inevitability.

-Mrs. Linde plays the role of foreshadowing the future of Nora and a mirror to Nora’s character. She delves into the mistakes Nora will make and views her for what she truly is. She is the wise woman who has hindsight of what becomes of women who spend their money and borrow. She provides exposition to the play because she is the only one Nora can discuss her history with without consequences. Talking to Mrs. Linde provides an opportunity for the audience to understand Nora’s character.

-Mrs. Linde shares with Nora that her husband had died and that, due to her habits and his unstable business, she was now poor and struggling to make ends meet. She seems to be foreshadowing Nora’s impending fate.

CONTRAST/JUXTAPOSITION

Contrast or juxtaposition involves two characters or things being placed together with a contrasting effect.

Examples

-The father-daughter relationship between Nora and her father and that of Nora and Torvald is contrasted in the final Act. Nora makes this connection that life with her father was like life with Torvald. Nora’s father would force his beliefs on her and she would comply with them lest she upset him; she would bury her personal belief under Papa’s. According to Nora, Torvald was guilty of the same things. A good example was his insistence on her wearing the fish girl costume and his frustration over her inability to grasp the tarantella.

-Krogstad and Nora are also contrasted. The more we learn of Krogstad, the more we understand that he shares a great deal with Nora Helmer. First of all, both have committed the crime of forgery. Moreover, their motives were out of a desperate desire to save their loved ones. Also like Nora, Krogstad has contemplated ending his life to eliminate his troubles but was ultimately too scared to follow through.

– Dr. Rank’s treatment of Nora is contrasted sharply with that of Torvald. Rank always treats Nora like an adult. He listens to her and affords her a dignity, which is definitely missing in Torvald’s treatment.

– Mrs Linde’s relationship with Krogstad also provides a point of comparison with that of Nora and Torvald.

-Nora and Mrs Linde are also contrasted. Whereas Mrs. Linde took responsibility for her sick parent, Nora abandoned her father when he was ill. Mrs. Linde’s account of her life of poverty underscores the privileged nature of the life that Nora leads. Her sensible worldview contrasts sharply with Nora’s somewhat childlike outlook on life.

FOIL

A foil is a literary character who contrasts another character in order to highlight certain aspects of the other character.

Examples

-Mrs Linde’s life’s journey from independence to marriage is a foil to Nora’s journey in the opposite direction.

-Dr Rank is a foil to Torvald in that he treats Nora as an intelligent human being and she in return speaks more openly to him than she does to her husband.

-Mrs. Linde is the character that really makes Nora look bad in comparison and acts as a foil for Nora. In fact, you could argue that all the characters act as foils for Nora.

MOTIFS

Motifs are recurring structures, contrasts, or literary devices that can help to develop and inform the text’s major themes.

Examples

Nora’s definition of freedom

-Nora’s understanding of the meaning of freedom recurs in the course of the play. In the first act, she believes that she will be totally “free” as soon as she has repaid her debt, because she will have the opportunity to devote herself fully to her domestic responsibilities. After Krogstad blackmails her, however, she reconsiders her outlook regarding freedom and questions whether she is happy in Torvald’s house, subjected to his orders and commands. By the end of the play, Nora seeks a new kind of freedom. She wishes to be relieved of her familial obligations in order to pursue her own ambitions, beliefs, and identity.

Use of letters

-Many of the plot’s twists and turns depend upon the writing and reading of letters. Krogstad writes two letters: the first reveals Nora’s crime of forgery to Torvald; the second retracts his blackmail threat and returns Nora’s promissory note.

-The first letter, which Krogstad places in Torvald’s letterbox near the end of Act Two, represents the truth about Nora’s past and initiates the inevitable dissolution of her marriage. The second letter releases Nora from her obligation to Krogstad and represents her release from her obligation to Torvald.

-The two letters have exposed the truth about Torvald’s selfishness, and Nora can no longer participate in the illusion of a happy marriage.

-Dr. Rank communicates his imminent death through another form of a letter: a calling card marked with a black cross in Torvald’s letterbox. By leaving his calling card as a death notice, Dr. Rank politely attempts to keep Torvald from the “ugly” truth, as he had said earlier about his best friend, Torvald.

Other letters include Mrs. Linde’s note to Krogstad, which initiates her life-changing meeting with him, and Torvald’s letter of dismissal to Krogstad.

HYPERBOLE

Hyperbole refers to extreme exaggeration of statements or claims which makes someone or something sound bigger, better or more than they are.

Examples

  1. i) Nora: Yes, that’s just it.

Helmer: Now you have destroyed all my happiness. You have ruined all my future. It is horrible to think of! I am in the power of an unscrupulous man; he can do what he likes with me, ask anything he likes of me, give me any order he pleases – I dare not refuse. And I must sink to such miserable depths because of a thoughtless woman! (Pg 104)

It is a hyperbole because although Nora may have caused a major accident of forging a signature and hiding it from Helmer, it is not obvious it will affect his future. Helmer is exaggerating that his happiness is destroyed because he feels betrayal and anger, just to show the seriousness of the shame that Nora has caused.

  1. ii) Linde: But now I am quite alone in the world – my life is so dreadfully empty and I feel so forsaken.

This is a hyperbole because Mrs. Linde is exaggerating about her situation.

iii) NORA: I should like to tear it into a hundred thousand pieces.

It is a hyperbole because Nora cannot possibly be able to tear the letter into a thousand pieces.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

REVISION QUESTIONS

CONTEXT QUESTIONS

  1. Where is A Doll’s House set?

-In a middle-class living-room; nice but not extravagant furniture; piano, fire, winter’s day; Norway

  1. Why does Nora tell the porter to hide the tree?

-So the kids don’t see it before it’s decorated

  1. What does Nora sneak at the beginning of the play?

-Macaroons

  1. What does Torvald call Nora?

-Squirrel, skylark

  1. Why does Torvald scold Nora at the beginning of the play?

-for buying things and being careless with money

  1. When is Torvald getting a new job? Why does he want Nora to wait to spend money?

-New Year; what if he gets hit in the head with a brick and therefore never gets the job?

  1. How does Nora suggest they buy things before the promotion?

-borrowing from people (Helmer thinks she’s just a typical woman who doesn’t understand debt)

  1. What does Nora want Torvald to get her for a present? Why doesn’t he want to give it to her?

-money; he calls her a spendthrift and says she’ll just waste all the money on unimportant things

  1. What does Torvald think runs in Nora’s family?

-spending too much money (like her father)

  1. Does Nora want to be like her dad? Does Torvald want this?

Yes, no

  1. What is the first thing Nora lies about?

-she denies that she would ever eat macaroons because she promised not to…but she did

  1. What did Nora do the Christmas before this one?

-locked herself away to make paper flowers to decorate the tree (the cat ripped up the flowers, though)

  1. Who is Mrs. Linde?

-Christine; Nora’s childhood friend; they haven’t seen each other in 9 or 10 years

  1. What does Nora think of Mrs. Linde’s appearance?

-she looks pale

  1. What work has Nora had to do?

-sewing, embroidery, etc.

  1. What secret does Nora tell Mrs. Linde first?

-when Torvald worked too hard in the first year of their marriage and got sick and had to go to Italy to recover, she borrowed money from her father in order to pay for it (at the time: her father was dying, she was pregnant, and she couldn’t go see him)

  1. What is Christine’s backstory?

-she never loved her husband, she just married Mr. Linde because she had a sick mother and two little brothers to support; when he died, his business fell apart and she was left with nothing; she had been working ever since; recently, her mom died and her brothers are now old enough to take care of themselves.

 

  1. What does Christine ask of Nora?

-if Torvald can give her a job at the bank he’ll be managing

  1. Why does Christine think it’s nice of Nora to care so much about her troubles?

-because she says Nora has never had any of her own

  1. What does Nora reveal to Mrs. Linde after being insulted because Christine said she didn’t really know trouble?

-she didn’t actually borrow money from her father; she borrowed it from someone else, but Torvald doesn’t know this; he didn’t even know how sick he was and that the Italy trip was to make him survive; she pretended that she just wanted to travel abroad and secretly borrowed the money to spare his pride

  1. When does Nora plan on telling Torvald the truth?

-when she’s old and he’s not attracted to her any more

  1. How has Nora been paying back the loan?

-lots of odd jobs secretly, scrimping on buying new clothes for herself (cheap clothes look good on her so Torvald doesn’t notice); last winter she copied letters late every night (perhaps she didn’t really spend all that time making tree decorations?)

  1. What does Nora dream?

-a rich old man would come and give her all the money she needs

  1. How does Mrs. Linde first claim to have known Krogstad?

-he was a lawyer in her area

  1. What is Krogstad’s backstory?

-had an unhappy marriage, now has several children, got himself into bad business troubles

  1. What does Mrs. Linde say when Dr. Rank tells her and Nora what a bad guy Krogstad is?

-people should try to help the diseased (Dr Rank says Krogstad is morally diseased)

  1. What does Torvald say when Nora asks if Christine can have a job?

-yes; she’s come at a good time

  1. What does Krogstad first ask Nora? Then what does he ask of her?

-if Mr. Helmer is giving Christine a job; if she will use her influence to get him a job (she says she doesn’t have much influence)

  1. Why is Nora no longer afraid of Krogstad when he first comes to visit?

-come New Year she’ll be out from under his thumb

  1. Why does Krogstad want to regain the community’s respect?

-for the sake of his sons

  1. Why does Nora tell Krogstad not to tell Torvald about the money?

-that would only confirm what a bad person Torvald sees Krogstad as saying it would make everything unpleasant

  1. What does Krogstad tell Nora that makes the whole thing worse?

-she forged her father’s signature; he can prove it because the signature is in Nora’s handwriting and she dated the document after her father died; basically he can prove that not only does she owe money but she broke the law (Nora says she didn’t have time to think about it because her husband was dying)

  1. When does Nora start decorating the Christmas tree?

-after Krogstad leaves after blackmailing her

  1. Why does Torvald think Krogstad was at the house?

-to get Nora to put in a good word for him

  1. What does Nora ask Torvald to do for her?

-pick out her costume for the Stenborgs’ party because she’s helpless without his taste

  1. What crime did Krogstad commit? Why does Torvald not like him?

-forgery; he never admitted his guilt, he got off through loopholes, he is corrupt and has lived a life of lies even around his family

  1. How does the Christmas tree look at the start of Act 2?

-stripped and bedraggled

  1. Who is Anne?

-nurse; she got pregnant by a bad guy and had to give up her child

  1. What costume is Nora going to wear to the Stenborgs’ party?

-Neapolitan fisher girl

  1. What dance does Torvald want Nora to do?

-tarantella

  1. What does Christine offer Nora?

-to fix up her costume (if she can come by and see Nora in it)

  1. What is Dr. Rank sick with? How did he get it?

-consumption of the spine; his promiscuous father

 

  1. What can Nora not talk about with Torvald? Why?

-her old school friends; he gets jealous

  1. Who does Christine guess lent Nora the money?

-Dr. Rank

  1. What does Christine think of Nora and Dr. Rank’s relationship?

-they should stop talking to each other so much because he’s probably interested in Nora

  1. Who does Nora think about borrowing money from to pay Krogstad?

-Dr. Rank

  1. What does Nora tell Torvald she’ll do anything he asks of her for? What does Torvald say?

-if he’ll not fire Krogstad; he’s given his job to Mrs. Linde

  1. What reasoning does Nora give as to why Torvald should not fire Krogstad?

-he writes small newspapers and he might write nasty articles about Torvald if he’s fired (like when her father got in legal trouble and was slandered by the papers); he should fire someone else and let Mrs. Linde and Krogstad both have jobs

  1. What does Torvald think of Nora’s dad’s business?

-his dealings were shady

  1. Why won’t Torvald give in to Nora’s request for a job for Krogstad?

-he already said no and told everyone at the bank he’s firing Krogstad, and he’s worried that if he changes his mind people will think his wife rules him; also, he knew Krogstad when they were kids and Krogstad insists on calling Torvald his first name which is really embarrassing given Krogstad’s corrupt past

  1. What finally makes Torvald send a letter firing Krogstad?

-Nora calls him petty for worrying about his affiliations with Krogstad

  1. Why does Torvald forgive Nora’s behavior in wanting a job for Krogstad?

-shows how much she loves him; but if trouble comes from the firing, he can handle it

  1. What does Dr. Rank ask of Nora?

-to keep Torvald away from Dr. Rank’s sick room, because he is sensitive to such unpleasantness and Rank doesn’t want to upset him

  1. How will Dr. Rank inform Nora of his death?

-business card with black cross in their mailbox when he starts to die (Nora says it’s morbid and depressing)

  1. Who does Dr. Rank think will replace him as Nora’s friend after he dies?

-Mrs. Linde

  1. How does Nora react when Dr. Rank tells her he loves her?

-she says it’s inappropriate; it makes her uncomfortable, he shouldn’t have said that

  1. How does Nora respond when Rank tells her that she seems like she loves him more than Torvald?

-she says the people you have fun with aren’t the same as the ones you love; when she was little she loved her father the most, but had fun gossiping with the maids

  1. What relationship does Nora liken her relationship with Torvald to?

-her relationship with her father

  1. Why is Krogstad surprised Torvald would fire him at first?

-knowing what power he has over them

  1. What does Krogstad offer?

-that the three of them can settle the matter, and no one else has to be involved

  1. How does Krogstad plan to use the blackmail?

-to blackmail Torvald into giving him a promotion; in a year it’ll be him running the bank instead of Torvald

  1. What does Nora threaten to do? What does Krogstad say?

-commit suicide; doesn’t believe she’ll do it, and it wouldn’t do any good because he would still be able to ruin her reputation which Torvald would never let happen

  1. What does Nora expect Torvald to do when he finds out?

-take all the blame on himself; a wonderful, yet terrible thing

  1. How does Mrs. Linde reassure Nora?

-says she will go and convince Krogstad to ask Torvald for the unopened letter back because they used to be close

  1. What does Nora ask Torvald to help her with?

-practising the tarantella (she does it, and he says she has a lot of practising to do, which she agrees with and says he will have to help her every moment between then and the party)

  1. What does Torvald guess when Nora says he will have no time to open letters since he’s helping her practise? What does he agree to do?

-that there’s a letter from Krogstad; wait till after the party to open it

  1. Why is Nora kind of glad that Torvald’s going to find out?

-the wonderful thing, the miracle will happen – he’ll jump to her rescue and take all the blame

  1. Where does Mrs. Linde wait for Krogstad? Why?

-at the Helmers’ house while they’re at the party; there’s no private entrance at the place where she’s staying

  1. What is Krogstad and Mrs. Linde’s past?

-she broke up with him abruptly for the man she married because she needed his money to support her family

  1. What does Krogstad say he has been like since Christine left him? What does she say?

-a man lost at sea on a wreck; she feels the same, and the shipwrecks should get together

  1. Why does Mrs. Linde tell Krogstad she came to town? What does he say?

-for him: she needs someone to work for, to help, or she feels like she has no purpose in life; Krogstad calls her hysterical and says she’s just looking for a chance at self-sacrifice

  1. What does Mrs. Linde say when Krogstad asks if she only got back together with him for Nora?

-she sold herself in the past and would never do it again

  1. Why does Christine tell Krogstad not to get his letter back?

-all the lies in the Helmers house need to come to light

  1. Did Nora want to leave the party so early?

-no

  1. Why does Mrs. Linde say she is at their house?

-to see Nora’s costume

  1. What did Torvald think of Nora’s performance at the party?

-a bit too realistic, she was a bit too much like a Neapolitan fisher girl for his taste (but the other guests loved it)

  1. What does Torvald think Mrs. Linde should do instead of knitting?

-embroidery (more tasteful since knitting needles looks Chinese)

  1. What does Torvald pretend when the Helmers go to parties?

-Nora’s not his wife, but his secret lover

  1. What does Dr. Rank say he will be at the party next year?

-invisible

  1. Why does Dr. Rank tell Torvald he had a right to drink a lot at the party?

-he did medical research all day and has found something for sure and was very productive (but really, he found out he’s definitely going to die)

  1. What does Torvald notice about the mailbox? What is Nora’s response?

-someone tried to pick the lock; one of Nora’s hairpins is jammed into it; she blames it on the kids

  1. What does Torvald find in the mailbox from Rank?

-two cards with black crosses on them

  1. What does Torvald tell Nora after he says he’s glad to have her?

-he sometimes wishes she were in trouble so he could save her

  1. What does Torvald do after reading Krogstad’s letter?

-screams at Nora, says she is just as disgusting as her father was; says his happiness is now destroyed because he will have to obey Krogstad

  1. What does Torvald say in response to Nora’s threatening to commit suicide?

-it won’t do any good because Krogstad will still have power and suspect him as an accomplice (he doesn’t really care that she’d be dead?!)

  1. What is Torvald’s solution?

-Nora can still live in the house to keep up appearances, but the relationship is over and she’s not allowed near the children because she’ll corrupt them

  1. How does Krogstad fix the problem?

-sends another letter with the forgery and says he’s ashamed he tried to blackmail them

  1. Why does Torvald say his love for Nora is even deeper now?

-after having forgiven her from the bottom of his heart; his possession of her has grown even greater; she shouldn’t worry because hewill continue to guide her through life as if she were a child

  1. What does Nora complain about after Torvald forgives her?

-she says they have never had a serious conversation before now

  1. Who does Nora accuse of treating her like a doll?

-Torvald and her father; they dressed her up and made her into what they wanted her to be

  1. Why does Nora say she’s leaving Torvald and the kids?

-she has a duty to herself that she’s never fulfilled; she realizes she’s never been happy with Torvald; she will spend the night at Mrs. Linde’s

  1. Why does Torvald admonish Nora for wanting to leave?

-he says she’s forsaking her sacred duties to her husband and children; then he says what Jesus would do

  1. What does Nora need to learn by leaving?

-whether she’s just too ignorant to understand society, as Torvald says, or if society’s wrong

  1. Why does Nora realize she never loved Torvald?

-she realizes he isn’t the man she thought he was when the “miracle” of him taking the blame from her didn’t happen

  1. When does Nora say she would come back?

-if they had a true marriage instead of just living together

  1. What does Helmer end with?

“The most wonderful thing of all”

ESSAY QUESTIONS

  1. What is important about the title? Who is the “doll” Ibsen refers to?
  2. Who is the more significant female character in terms of plot development, Nora or Christine? Explain your answer.
  3. Do you think Christine’s decision not to prevent Krogstad from revealing the truth to Torvald is a betrayal of Nora? Does this act ultimately hurt or benefit Nora?
  4. How does Henrik Ibsen reveal character in A Doll’s House? Is Nora a sympathetic character? Did your opinion of Nora change from the beginning of the play to its conclusion?
  5. Does the play end the way you expected? Do you think this was a happy ending?
  6. A Doll’s House is generally considered a feminist work. Do you agree with this characterization? Why or why not?
  7. What does the Tarantella dance symbolize in A Doll’s house?

 

 

 

 

 

Gathuthuma Secondary School’s CBE Subjects, Pathways, Contacts, Location, Fees, Admission {Full Details}

Gathuthuma Secondary School is a public Mixed, (Boys’ and Girls’) Sub-County Level Day School. The Senior School (Secondary School) is physically located at Kirinyaga Central Subcounty in Kirinyaga County of the Central Region, Kenya. Placement in the school is done by the Ministry of Education of the Republic of Kenya. To be placed to join this school, a grade 9 learner has to select the school online and placement is done based on the available grade 10 vacancies.  We have also provided the School’s Official Phone Number Contact. Reach the Principal directly on:  0720802354. or (+254) 0722894259.

Continue reading: New list of all National Schools in Central Region {CBE Senior Schools}

This well researched article provides the latest and accurate on the school’s School Physical Location, Postal Address, Mobile Number, Telephone Number, Email Address and School Website. Also available is the school’s Category, type, level, accomodation type, Knec Code and Performance at KNEC EXAMS.

Key Details about the school.

  • Country where found: Kenya.
  • Region: Central.
  • County: Kirinyaga County.
  • Subcounty: Kirinyaga Central Subcounty.
  • School Type/ Ownership: A Public School.
  • Nature os School/ CBE Level: Senior School (SS).
  • Category: Regular School
  • School’s Official Name: Gathuthuma Secondary School 
  • Sex: Mixed, (Boys’ and Girls’)  School.
  • School Cluster/ Level: Sub-County School whose Classification is C4.
  • Accomodation Type: Day  School.
  • Knec Code:  9203439
  • School’s Official Phone Number:  0720802354. or (+254) 0722894259.
  • Official Email Address for the School: bmkibiru@gmail.com.
  • Postal Address: P.O Box 753-10300 Kerugoya, Kenya
  • Total Number of Subjects Combinations Offered at the School: 4 Subjects’ Combinations in various Pathways.

Important Links with Schools’ details:

Fees paid at Gathuthuma Secondary School 

Fees paid at the school is determined by the Ministry of Education and is uniform for all Senior Schools; countrywide. National Senior schools have the highest fees set; followed by Extra-County, County and Sub-County schools in that order.

Gathuthuma Secondary School Profile & Information

Complete overview of academic programs and school details

GATHUTHUMA SEC

LocationKIRINYAGA
SexMIXED
CategoryREGULAR
ClusterC4

2

STEM

2

Social Sciences

4

Total Combinations

Subject Combinations Offered at Gathuthuma Secondary School

View all available subject combinations at this school

SOCIAL SCIENCES

2
HUMANITIES & BUSINESS STUDIESCode: SS2019
Christian Religious Education,Geography,History & Citizenship
3 SubjectsSOCIAL SCIENCES
LANGUAGES & LITERATURECode: SS1080
Business Studies,Fasihi ya Kiswahili,Literature in English
3 SubjectsSOCIAL SCIENCES

STEM

2
APPLIED SCIENCESCode: ST2049
Agriculture,Business Studies,General Science
3 SubjectsSTEM
PURE SCIENCESCode: ST1026
Core Mathematics,Business Studies,General Science
3 SubjectsSTEM

How to get the School’s Knec Results.

To check the school’s Knec results, visit Knec Portal and search for the school by its name.

📍 How to get more Information about the School

For more information about admission requirements, facilities, and application procedures, contact the school directly. Use the official phone number indicated above to get information about the school’s fees, uniform, meals and performance.

Continue reading:

How to Contact the Ministry of Education, MoE- Kenya..

Contact the Ministry by using the following details;

1. State Department of Early Learning and Basic Education:

  • Postal Address: P.O Box 36260-00200 Nairobi, Kenya
  • Physical Address: Jogoo House B, Taifa Road
  • Telephone Number: +254-020-3318581
  • Email: psbasic@education.go.ke

2. MINISTRY OF EDUCATION’s HEAD OFFICE.

  • Physical Location: Jogoo House “B” Taifa Road
  • Postal Address: P.O. Box 300400-00100 Nairobi.
  • Email: info@education.go.ke
  • Phone : +254(0) 3318581

How to Select Grade 10 Subjects and schools

To select Grade 10 schools and subjects under the Competency-Based Curriculum (CBC) in Kenya, Grade 9 learners should first choose a career pathway (STEM, Social Sciences, or Arts & Sports Science). Then, they’ll select three subject combinations within that pathway and finally, choose four schools for each combination, totaling 12 schools. To select preferred Grade 10 Schools and Subject Combinations, use the Ministry of Education portal selection.education.go.ke.

1. How you can Choose a Career Pathway:

  • Identify your interests and potential career aspirations.
  • Select one of the three pathways: STEM, Social Sciences, or Arts & Sports Science.
  • Confirm your choice to proceed with the pathway.

2. Select Subject Combinations:

  • The portal will provide you with a list of subject combinations available within your chosen pathway.
  • Choose three subject combinations that align with your interests and strengths.

3. Select Preferred Senior Schools:

  • For each subject combination, select four schools from the available clusters.
  • This ensures a diverse range of options and equal representation from different categories of schools.
  • A total of 12 schools will be selected: 4 for the first subject combination, 4 for the second, and 4 for the third.

TSC List of Approved TSC Teaching Subject Combinations for Secondary Schools

Science Combinations

  • Chemistry & Biology
  • Chemistry & Physics
  • Agriculture & Geography
  • Biology & Geography
  • Home Science & Biology
  • Physics & Computer
  • Agriculture & Biology
  • Agriculture & Chemistry

Language Combinations

  • English & Literature
  • Kiswahili & CRE (Christian Religious Education)
  • Kiswahili & History
  • Kiswahili & Geography
  • German & Geography
  • German & CRE
  • Kiswahili & IRE (Islamic Religious Education)
  • Kiswahili & Home Science
  • Kiswahili & P.E. (Physical Education)
  • French & Geography
  • French & Business Studies
  • French & CRE
  • French & History
  • Arabic & IRE
  • German & History

Mathematics Combinations

  • Mathematics & Business Studies
  • Mathematics & Geography
  • Mathematics & Computer
  • Mathematics & Chemistry
  • Mathematics & Biology
  • Mathematics & Physics

Humanities Combinations

  • Geography & CRE
  • History & CRE
  • Geography & History
  • Geography & IRE
  • History & IRE
  • History & P.E.
  • Geography & P.E.
  • Geography & Business Studies
  • Home Science & History
  • Home Science & CRE
  • SNE & P.E.
  • Music & P.E.

Technical Combinations

  • Metal Work & Mathematics
  • Metal Work & Physics
  • Woodwork & Mathematics
  • Woodwork & Physics
  • Electricity & Mathematics
  • Electricity & Physics
  • Fine Arts & History (Fine)
  • Art & Geography
  • Art & Design & History
  • Art & Design & Geography
  • Kiswahili & Home Science
  • Kiswahili & P.E.

SELECTION OF PATHWAYS AND SENIOR SCHOOLS

• Determination of pathways per senior school
• Determination of vacancies for boarding and day schooling in senior schools
• Selection of pathways, subjects’ combination and schools by grade 9 learners Selection based on pathway

The learner will select 12 schools for their chosen pathway as follows.
– Four 4 schools in first choice track and subject combination
– Four (4) schools in second choice subject combination
– Four (4) schools in third choice subject combination (Total 12 schools) Selection based on accommodation

Out of the 12 schools selected based on pathway;

  • 9 will be boarding schools; 3 from the learners’ home county, 6 from outside their home county/county of residence.
  • Three (3) day schools in their home sub county/sub county of residence. (Total 12 schools)
    Pre selection – A school that does not allow open placement can apply to be pre-select if it meets the criteria defined by the Ministry of Education.

How Placement of learners to Senior Schools (SS) at Grade 10 will be done

It will be based on:

  • Top 6 learners per gender in each STEM track per sub-county will be placed for Boarding in  schools of choice
  • Top 3 learners per gender in each Social Science track per sub-county will be placed for Boarding in schools of choice
  • Top 2 learners per gender in each Arts and Sports Science track per sub-county be placed to Boarding schools of their choice
  • Placement of Candidates with Achievement Level of averaging 7 and 8 per track to boarding schools of their choice

Admission and replacement process

  • Placement results will show where learners have been placed
  • Admission letters/joining instructions shall be accessed online using the learner’s assessment Number.
  • All schools, both public and private shall admit Grade 10 through KEMIS
  • Leaners will be entered to the KEMIS System only after reporting physically to the school.
  •  Daily online reporting shall be monitored through KEMIS

Replacements

  •  Schools with vacancies will declare through their respective county directors of education
  •  Learners who wish to change schools will make requests through the heads of junior schools at least 2 weeks before the official grade 10 reporting date
  • Priority shall be given to those who had earlier selected the schools they are requesting for  Approvals by MoE shall be based on senior schools’ documented capacity
  • Upon a request’s approval, the joining instructions shall be accessed online; at no time shall a school issue printed letters for replacement cases
  • Replacement will be done once and shall be irreversible

LIST OF ALL SENIOR SCHOOLS PER COUNTY.

Senior School Subjects and Pathways selection Form.
Senior School Subjects and Pathways selection Form.

Senior School Selection Form educationnewshub.co.ke

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Del Monte Mixed Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

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Ngatho Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Mugumo Mixed Day Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Kyaume Mixed Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Iembeni Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Muti Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Thungururu Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Gititu Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Ngelelya Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Swani Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

St Benedict Ithanga Township Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Ithanga High School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Mwanawikio Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Matunda Mixed Day Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Kanderendu Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Karega Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Wamahiga Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Mutunguru Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Kigumo Mixed Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Marumi Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Mugumoini Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Kigumo Bendera High School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Mununga Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Mathareini Mixed Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Kinyona Mixed Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

St Francis Mukuyuini Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Mairi Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Makomboki Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Muthithi Mixed Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Matu Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Mumbu Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Mwarano Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Mariira Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Ikumbi Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Gikigie Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Thamara Mixed Day Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Rarakwa Girls High School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

St John The Baptist Kirie Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Gatumbi Baptist Mixed Day Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Ndugamano Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Njora Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Kigumo Girls High School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Nguku Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Ack Kahumbu Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Kamukabi Mixed Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Turuturu Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Njiiri Boys Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Kiugu Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Karinga Mixed Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Githima Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Bishop Gatimu Kinyona Girls Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Kiaguthu Boys School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Kianderi Girls High School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

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Mukangu Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

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Kamaguta Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Gitare Mixed Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

Dr. Gitau Matharite Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

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Kiumba Mixed Day Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

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Dr. Kiano Girls Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

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Kahuhia Mixed High School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

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Kahatia Secondary School’s CBE Subjects, Pathways, Contacts, Location {Full Details}

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DP Kithure Kindiki issues stern warning to School Principals.

DP Kithure Kindiki’s stern warning to School Principals.

Deputy President Kithure Kindiki has issued a stern warning to principals who are still holding on ad refusing to release the Kenya National Examinations Council (Knec) examination certificates to students because of school fee arrears. The DP indicated that such Principals risk losing their jobs.

Speaking in Nakuru on Thursday, May 15, the DP urged parents to report any “stubborn” head of an institution who refuses to release the certificate to the assistant chief for action to be taken.

“There are so many schools that have refused to release certificates to students who finished their national examinations. I want to warn that principals who continue defying government directives are risking their jobs,” he said.

The DP said that after a principal releases a certificate, the government will follow up and develop a mechanism to ensure that the fee balance is sorted.

“All heads of institutions must release certificates for all the children of Kenya, and then after that, we will sort out other fee-related issues. It is not a joke, I know the minister has already said, but I’m saying this as the deputy president because we are not requesting you, we are directing,” he said.

“I’m speaking on behalf of the president and the government, please, principals, let us not play monkey games. I would like to tell parents to report any principal who refuses to issue these certificates to the assistant chief,” he said.

According to the government, withholding certificates from students is unlawful, and a principal is at risk of facing legal repercussions.

Government spokesperson Isaac Mwaura, in a statement on Monday, April 14, said denying students the document undermines the country’s education system.

“Denying any student access to their academic credentials undermines the very foundation of our education system and compromises the rights of the child. The government will not tolerate this continued impunity, and stern actions will be taken against schools that continue to defy this directive,” he stated.

Speaking on Monday, March 17, the Education CS, Julius Ogamba, revealed the government’s plan to ensure that students collect their certificates from sub-county education offices rather than their former schools.

Ogamba further announced a crackdown on heads of institutions withholding national examination certificates from students.

“The other day, I issued a directive that teachers holding certificates should release them to all the students, and I gave a deadline, and the deadline has already passed, so from next week, we will be taking action against teachers who are still withholding the certificates,” he said.

Despite the stern warning from the Government, schools are still holding on to the Kenya Certificate of Secondary Education, KCSE, Exams’ certificates.

A parent who sought anonymity narrated to this reporter how a school Principal is holding her girls’ KCSE Certificate; dimming her hopes of securing a job.

“I have visited the school three times in a row to collect my daughter’s certificate. Unfortunately, the Principal wants me to clear a fee balance of Ksh6,500 and which I don’t have at the moment”. Said the parent.

According to the parent, her daughter sat the KCSE 2024 exams and scored a C- (minus). The single parent has approached a good Samaritan who is willing to assist the girl with College sponsorship.

The parent went on to say that her pleas, to be given even photocopies of the Certificates, fell on deaf ears.

KMTC Kitale Campus: Courses, Fees, Location, Contacts and Programs plus Requirements

KMTC Kitale Campus: Courses, Fees, Location, Contacts and Programs plus Requirements

KMTC Kitale Campus

Background information

KMTC Kitale Campus was started in 2014 as an initiative of the KMTC Board of Directors, Trans Nzoia County Government and Kitale County Hospital.

Programmes and Courses Offered

  • Diploma in Clinical Medicine and Surgery
  • Diploma in Nursing

Facilities

  • Tuition block
  • Well-stocked library,
  • Well-equipped skills lab
  • Hostel block
  • Dinning hall

Student population

The Campus has 540 students with this number expected to increase in the next intake.

Clinical experience sites

  • County hospitals
  • Rural facilities
  • Psychiatric units
  • Community diagnosis

Clubs, societies and sports

  • Christian Union
  • Seventh Day Adventist
  • Young Catholic Society
  • Choir
  • Drama
  • Sports activities

 Campus contact information

The Principal

KMTC Kitale Campus

P.O. BOX 3187 – 30200

Kitale

Tel: 0722307988

Email: kitale@kmtc.ac.ke

AON latest list of Outpatient Service Providers Per County

OutPatient Service Providers
BARINGO
OutPatient Service Providers Baringo Baringo Central Karbarnet Direct Bliss Gvs Healthcare Ltd Kabarnet
OutPatient Service Providers Baringo Baringo Central Karbarnet Direct Reale Hospital Ltd Karbarnet
OutPatient Service Providers Baringo Baringo Central Karbarnet Direct Tionybei Chemist Diagnostic Lab And Medical Clinic K
OutPatient Service Providers Baringo Baringo Central Karbarnet Direct Tionybei Chemist Diagnostic Lab And Medical Clinic K
OutPatient Service Providers Baringo Baringo North kabartonjo Direct Tionybei Chemist Diagnostic Lab And Medical Clinic K
OutPatient Service Providers Baringo Baringo South Marigat Direct Tionybei Chemist Diagnostic Lab And Medical Clinic M
OutPatient Service Providers Baringo East Pokot Chemolingo Direct Parkerra Medical Clinic
OutPatient Service Providers Baringo Kabarnet Kambi ya moto Direct Familia Bora Medical Centre
Type County Sub County Town Access Facility Name
OutPatient Service Providers Baringo Koibatek Eldama ravine Direct Ravine Glory Healthcare Services
OutPatient Service Providers Baringo Mogotio Mogotio Direct Berur Medical Clinic
BOMET
OutPatient Service Providers Bomet Bomet Central Bomet Direct Siloam Hospital Limited Bomet
OutPatient Service Providers Bomet Bomet Central Tenwek Referral Tenwek Hospital Tenwek
OutPatient Service Providers Bomet Bomet East Longisa Direct Legacy Family Health Option
OutPatient Service Providers Bomet Konoin Mogogosiek Direct Chelymo Medical Center
OutPatient Service Providers Bomet Sotik Kaplong Direct St Clares Kaplong Mission Hospital
BUNGOMA
OutPatient Service Providers Bungoma Bungoma Central Chwele Direct Nzoia Medical Centre Chwele
OutPatient Service Providers Bungoma Bungoma East Webuye Direct Bliss Gvs Healthcare Ltd Webuye
OutPatient Service Providers Bungoma Bungoma South Bungoma town Direct Life Care Hospital Limited Bungoma
OutPatient Service Providers Bungoma Bungoma South Nzoia Direct Nzoia Medical Centre Nzoia
OutPatient Service Providers Bungoma Bungoma West Bungoma town Direct Bungoma West Hospital
OutPatient Service Providers Bungoma Kimilili Kimilili Direct Khamulati Medical Clinic
OutPatient Service Providers Bungoma Kimilili Kimilili Direct Bliss Medical Centre Kimilili
OutPatient Service Providers Bungoma Malakisi Mayanja Direct Nabuala Medical Centre
OutPatient Service Providers Bungoma Mt Elgon Mt elgon Direct Transelgon Healthcare Ltd
OutPatient Service Providers Bungoma Mt Elgon Mt elgon Direct Scovic Healthcare Services
OutPatient Service Providers Bungoma Select Sub County Bungoma town Direct Bungoma County Referral Hospital
OutPatient Service Providers Bungoma Select Sub County Soysambu Direct Waraba Medical Centre
OutPatient Service Providers Bungoma Select Sub County Mayanja Direct Nabuala Medical Centre
OutPatient Service Providers Bungoma Webuye Webuye Direct Calvery Hope Medical Centre
BUSIA
OutPatient Service Providers Busia Butula Butula Direct Butula Nursing Home
OutPatient Service Providers Busia Funyula Funyula Direct Nassi Hospital
OutPatient Service Providers Busia Matayos Busia Direct Stirling Healthcare Consultants Busia
OutPatient Service Providers Busia Matayos Busia Direct Bliss Medical –  Busia
OutPatient Service Providers Busia Nambale Nambale Direct Nambale Medical Center
OutPatient Service Providers Busia Nambale Nambale Direct Stirling Healthcare Consultants Nambale
OutPatient Service Providers Busia Teso South Malaba Direct Raimu Hospital Limited
OutPatient Service Providers Busia Teso South Malaba Direct Pynet Healthcare Adungosi
ELGEYO MARAKWET
OutPatient Service Providers Elgeyo marakwet Keiyo North Iten Direct Bliss Gvs Healthcare Ltd Iten
Type County Sub County Town Access Facility Name
Elgeyo/Marakwet
OutPatient Service Providers Elgeyo/marakwet Marakwet West Kapsowar Direct Aic Kapsowar Mission Hospital
EMBU
OutPatient Service Providers Embu Embu East Ena Direct Acef Ena Health Centre
OutPatient Service Providers Embu Embu West Embu Direct Bliss Gvs Healthcare Ltd Embu
OutPatient Service Providers Embu Embu West Embu Direct Embu Childrens Hospital
OutPatient Service Providers Embu Embu West Embu Direct Greenspan Medical Centre
OutPatient Service Providers Embu Embu West Direct Direct Imaramediplus Medical Centre
Garissa
OutPatient Service Providers Garissa Dadaad Hagadera Direct Asad Medical Centre
OutPatient Service Providers Garissa Garissa Township Garissa township Direct Antaliya Hospital Ltd
OutPatient Service Providers Garissa Garissa Township Garissa township Direct Bliss Gvs Healthcare Ltd Garissa
OutPatient Service Providers Garissa Ijara Ijara Direct Shamsu Healthcare Limited
Homa Bay
OutPatient Service Providers Homa bay Homabay Town Homabay town Referral Oasis Specialist Hospital Homabay
OutPatient Service Providers Homa bay Homabay Town Homabay town Direct Homabay Teaching And Referral Hospital
OutPatient Service Providers Homa bay Mbita Mbita town Direct Prince Richard Healthcare
OutPatient Service Providers Homa bay Mbita Mbita town Direct Sori Lakeside Nursing And Maternity Mbita
OutPatient Service Providers Homa bay Ndhiwa Ndiwa Direct Sori Lakeside Nursing And Maternity Ndhiwa
OutPatient Service Providers Homa bay Select Sub County Kadel centre Direct Termary Healthcare
OutPatient Service Providers Homa bay Select Sub County Sindo Direct Sanura Medical Centre
OutPatient Service Providers Homa bay Select Sub County Oyugis Direct Hawi Family Hospital
OutPatient Service Providers Homa bay Select Sub County Homabay town Direct Kendu Adventist Hospital
OutPatient Service Providers Homa bay Select Sub County Oyugis town Direct Matata Nursing And Maternity Home
HOMABAY
OutPatient Service Providers Homabay Homabay Homabay town Direct Baypharm Optics
OutPatient Service Providers Homabay Homabay Homabay town Direct Bliss Gvs Healthcare Limited Homabay
OutPatient Service Providers Homabay Homabay Homabay town Direct St Pauls Mission Hospital
OutPatient Service Providers Homabay Mbita Mbita town Direct Osani Community Health And Development Centre
OutPatient Service Providers Homabay Ndhiwa Ndhiwa town Direct Tudor Healthcare Ltd Ndhiwa
OutPatient Service Providers Homabay Rachuonyo East Rachuonyo Direct Osman Medical Services
OutPatient Service Providers Homabay Rachuonyo North Nyangweso Direct Homabay Community Medical Clinic
OutPatient Service Providers Homabay Rachuonyo North Gendia centre Direct Janeiro Nursing Home
OutPatient Service Providers Homabay Rachuonyo South Oyugis town Direct Evans Healthcare Clinic
OutPatient Service Providers Homabay Suba Sindo Direct Tudor Healthrecare Ltd Sindo
Type County Sub County Town Access Facility Name
OutPatient Service Providers Homabay Suba Sindo Direct Sori Lakeside Nursing And Maternity Sindo
ISIOLO
OutPatient Service Providers Isiolo Isiolo Isiolo town Direct Galaxy Hospital
OutPatient Service Providers Isiolo Isiolo Isiolo town Direct Hexagon Nursing Home
OutPatient Service Providers Isiolo Select Sub County Merti Referral Hussein Tarole Medical Centre
Kajiado
OutPatient Service Providers Kajiado Kajiado Central Ngong Direct Matasia Nursing Home
OutPatient Service Providers Kajiado Kajiado Central Isinya Direct Favour Medical Services
OutPatient Service Providers Kajiado Kajiado Central Namanga Direct Tulah Medical Service Namanga
OutPatient Service Providers Kajiado Kajiado Central Ngong Direct King David’s Hospital
OutPatient Service Providers Kajiado Kajiado East Kajiado Direct Kitengela Medical Services Kajiado
OutPatient Service Providers Kajiado Kajiado North Kiserian Direct Meridian Hospital Kiserian
OutPatient Service Providers Kajiado Kajiado North Kajiado Direct City Point Group Of Hospitals- Kajiado
OutPatient Service Providers Kajiado Kajiado North Rongai Direct Wananchi Jamii Maternity & Nursing Home
KAKAMEGA
OutPatient Service Providers Kakamega Butere Butere Direct St James Memorial Hospital/ammenity Hospital
OutPatient Service Providers Kakamega Ikolomani Ikolomani Direct Khayega Medical Clinic
OutPatient Service Providers Kakamega Ikolomani Musoli Direct Inaya Afia Nursing And Maternity Home
OutPatient Service Providers Kakamega Kakamega Ikolomani Direct Sheywe Community Hospital
OutPatient Service Providers Kakamega Kakamega East Khayega Direct Khayega Medical Clinic
OutPatient Service Providers Kakamega Likuyani Matunda Direct Matunda Maternity Home
OutPatient Service Providers Kakamega Lugali Lugali Direct Lugari Medical Centre
OutPatient Service Providers Kakamega Lurambi Kakamega Direct Bliss Gvs Healthcare Ltd Kakamega
OutPatient Service Providers Kakamega Lurambi Kakamega Direct Kakamega Central Maternity And Nursing Home
OutPatient Service Providers Kakamega Lurambi Shirere Direct Jumuia Hospital Kakamega Clinic
OutPatient Service Providers Kakamega Lurambi Shirere Direct Nala Maternity And Nursing Home
OutPatient Service Providers Kakamega Lurambi Kakamega town Referral Oasis Doctors Plaza Kakamega
OutPatient Service Providers Kakamega Lurambi Kakamega town Direct Tropical Clinic
OutPatient Service Providers Kakamega Malava Malava Direct Liason Manyatta Medical Clinic
OutPatient Service Providers Kakamega Malava Malava Direct Leben Medical Services
OutPatient Service Providers Kakamega Matungu Koyonzo Direct Capital Care Hospital
OutPatient Service Providers Kakamega Mumias East Mumias Direct Alphond Medical Centre
OutPatient Service Providers Kakamega Mumias/matungu Mumias Direct Medicross Medical- Mumius
Type County Sub County Town Access Facility Name
OutPatient Service Providers Kakamega Navakholo Navakholo Direct Navakholo Health Centre
OutPatient Service Providers Kakamega Shinyalu Shinyalu town Direct Virhembe Community Nursing Home
OutPatient Service Providers Kakamega Shinyalu Shinyalu town Referral St Gallen Ongoloy Network
OutPatient Service Providers Kakamega Shinyalu Shinyalu town Direct Shinyalu Healthcare Ltd
Kericho
OutPatient Service Providers Kericho Ainamoi Kericho Direct St. Leonard’s Hospital
OutPatient Service Providers Kericho Ainamoi Kericho Direct Bliss Gvs Healthcare Ltd Kericho
OutPatient Service Providers Kericho Ainamoi Kericho Direct Kericho Medical Clinic Ltd
OutPatient Service Providers Kericho Ainamoi Kericho Direct Greenview Nursing Home
OutPatient Service Providers Kericho Ainamoi Kericho Direct Siloam Hospital Limited Kericho
OutPatient Service Providers Kericho Kipkelion East Chepseon Direct Chesinende Health Services
OutPatient Service Providers Kericho Kipkelion West Kipkelion Direct Siloam Hospital Limited -kipkelion
KIAMBU
OutPatient Service Providers Kiambu Gatundu South Gatundu town Direct St Marys Mission International Dispensary
OutPatient Service Providers Kiambu Githunguri Githunguri town Direct Ivory Health Solutions Githunguri
OutPatient Service Providers Kiambu Juja Juja town Direct Medicross Kenya Juja Mall
OutPatient Service Providers Kiambu Kabete Kabete Direct Kabete Gardens Hospital
OutPatient Service Providers Kiambu Kiambaa Rwaka Direct Garden Park Medical Centre
OutPatient Service Providers Kiambu Kiambu Kiambu town Direct St. Mary’s Mother And Child
OutPatient Service Providers Kiambu Kiambu Kiambu township-telkom bl Direct Bliss Gvs Healthcare Ltd Kiambu
OutPatient Service Providers Kiambu Kikuyu Kikuyu town Referral Life Care Hospital Kikuyu
OutPatient Service Providers Kiambu Kimende Town Direct Direct Limuru Nursing
OutPatient Service Providers Kiambu Lari Kijabe Direct Aic Cure Hospital
OutPatient Service Providers Kiambu Limuru Limuru town Direct Limuru Cottage Hospital
OutPatient Service Providers Kiambu Thika Thika town Direct Bliss Gvs Healthcare Ltd Thika
OutPatient Service Providers Kiambu Thika Makongeni Direct Medicross Kenya Ananas Mall
OutPatient Service Providers Kiambu Thika Thika town Direct Winellis Medicare Co.ltd(dr Maina)
OutPatient Service Providers Kiambu Thika Town Thika town Referral Apex Specialist And Diagnostic Center
OutPatient Service Providers Kiambu Thika Town Thika town Referral Mediva Wellness Centre
KILIFI
OutPatient Service Providers Kilifi Bamba Bamba Direct Bamba Maternity And Nursing Home
OutPatient Service Providers Kilifi Ganze Ganze Direct Chamalo Medical Clinic
OutPatient Service Providers Kilifi Kaloleni Kaloleni Direct Bombolulu Diagnostic Medical Centre
Type County Sub County Town Access Facility Name
OutPatient Service Providers Kilifi Kilifi Mariakani Direct Mephi Health Center
OutPatient Service Providers Kilifi Kilifi North Malindi town Direct Malindi Sub County Hospital
OutPatient Service Providers Kilifi Kilifi North Tezo Direct Jambo Jipya Medical Centre And Maternity
OutPatient Service Providers Kilifi Kilifi South Mtwapa Direct Bliss Gvs Healthcare Ltd Mtwapa
OutPatient Service Providers Kilifi Malindi Malindi town Direct Maimoon Hospital
OutPatient Service Providers Kilifi Malindi Malindi Direct Bliss Gvs Healthcare Ltd Malindi
OutPatient Service Providers Kilifi Malindi Magarini Direct Kijanaheri Medical Centre Magarini
OutPatient Service Providers Kilifi Malindi Marafa Direct Kijanaheri Medical Centre Marafa
OutPatient Service Providers Kilifi Malindi Malindi Direct Meridian Hospital Galana
OutPatient Service Providers Kilifi Select Sub County Kilifi town Direct Kilifi County Referral Hospital
KIRINYAGA
OutPatient Service Providers Kirinyaga Kirinyaga  East Mwea Direct Silvercrest Medcare Hospital
OutPatient Service Providers Kirinyaga Kirinyaga Central Kerugoya Direct Bliss Gvs Healthcare Ltd Kerugoya
OutPatient Service Providers Kirinyaga Kirinyaga Central Kerugoya Direct Kerugoya Medical Centre
OutPatient Service Providers Kirinyaga Select Sub County Kagio Direct Ndia Medical Centre
OutPatient Service Providers Kirinyaga Select Sub County Sagana Direct Fortress Medical Centre
KISII
OutPatient Service Providers Kisii Bomachoge Ogembo Direct Neocare Specialist Hospital
OutPatient Service Providers Kisii Gucha South Tabaka Direct Tabaka Mission Hospital
OutPatient Service Providers Kisii Kenyenya Kenyenya Direct Kenyenya Medical And Diagnostic Centre
OutPatient Service Providers Kisii Keroka Keroka Direct St Catherines Ichuni Dispensary
OutPatient Service Providers Kisii Kisii Kisii town old uchumi build Direct Bliss Gvs Healthcare Ltd _stage
OutPatient Service Providers Kisii Kisii Kisii town opp co-op bank Direct Bliss Gvs Healthcare Ltd Kisii
OutPatient Service Providers Kisii Kisii Central Sansora building Direct Kisii Family Medical Centre
OutPatient Service Providers Kisii Kisii Central Kisii town Direct Ram Hospital Ltd
OutPatient Service Providers Kisii Kitutu Chache Gesonso Direct Nyangena Hospital
OutPatient Service Providers Kisii Nyaribari Chache Kisii town Direct Christmarianne Mission Hospital
OutPatient Service Providers Kisii Sameta Sameta Direct Lenmek Hospital Ltd
OutPatient Service Providers Kisii Select Sub County Migori Direct Ropment Medical Centre
OutPatient Service Providers Kisii Select Sub County Kenyanya Direct Kenyenya Medical And Diagnostic Centre
OutPatient Service Providers Kisii Select Sub County Kisii town Referral Kisii Teaching And Referral Hospital
KISUMU
OutPatient Service Providers Kisumu Kisumu Central Kisumu town Direct Bliss Gvs Healthcare Ltd Almiran
Type County Sub County Town Access Facility Name
OutPatient Service Providers Kisumu Kisumu Central Kisumu town Direct Bliss Gvs Healthcare Ltd Nakumatt Mega Mall
OutPatient Service Providers Kisumu Kisumu Central Kisumu town Direct Medicross Kenya Mega City
OutPatient Service Providers Kisumu Kisumu Central Kisumu town Direct Medicross Kenya Tuff Foam
OutPatient Service Providers Kisumu Kisumu Central Kisumu town Referral Nightingale Medical Centre
OutPatient Service Providers Kisumu Kisumu Central Kisumu Direct Kannika International Ltd. Specialist Medical And Diagn
OutPatient Service Providers Kisumu Kisumu Central Kisumu Referral Bloom Surgical Center
OutPatient Service Providers Kisumu Kisumu Central Kisumu town Direct Jaramogi Oginga Odinga Teaching And Referral Hospita
OutPatient Service Providers Kisumu Kisumu East Manyatta Direct Manyatta Community Medical Clinic
OutPatient Service Providers Kisumu Kisumu West Maseno Direct Maseno Mission Hospital
OutPatient Service Providers Kisumu Nyakach Katito Direct Katito Medical Centre
OutPatient Service Providers Kisumu Nyakach ?katito Direct House Of Hope Medical Centre
OutPatient Service Providers Kisumu Nyando Ahero Direct Medicross Kenya Abala
OutPatient Service Providers Kisumu Nyando Awasi Direct The Hope Medical Centre Awasi
KITUI
OutPatient Service Providers Kitui Ikutha Ikutha Direct Ikutha Medicare Clinic
OutPatient Service Providers Kitui Kitui Central Kitui Direct Bliss Gvs Healthcare Ltd Kitui
OutPatient Service Providers Kitui Kitui Central Kitui Direct Kitui Level 4 Hospital
OutPatient Service Providers Kitui Kyuso Kyuso Direct Mephibosheth Medical Clinic
OutPatient Service Providers Kitui Nzambani Nzambani Direct Zombe Medical Clinic
OutPatient Service Providers Kitui Select Sub County Mulango Direct Neema Hospital
OutPatient Service Providers Kitui Select Sub County Mulango Direct Mwingi Family Nursing Home
OutPatient Service Providers Kitui Select Sub County Tseikuru Direct Muumoni Nursing Home
KWALE
OutPatient Service Providers Kwale Likoni Likoni Direct Diani Beach Hospital Likoni
OutPatient Service Providers Kwale Likoni Likoni Direct St Thomas Maternity And Nursing Home
OutPatient Service Providers Kwale Matuga Kwale Direct Bliss Gvs Healthcare Ltd Kwale
OutPatient Service Providers Kwale Msambweni Diani Direct Bliss Gvs Healthcare Ltd Diani
OutPatient Service Providers Kwale Msambweni Diani Direct Diani Beach Hospital Diani Main
OutPatient Service Providers Kwale Msambweni Lunga lumga Direct Precious Light Medical
LAIKIPIA
OutPatient Service Providers Laikipia Laikipia North Nanyuki town Direct Bliss Gvs Healthcare Ltd Nanyuki
OutPatient Service Providers Laikipia Laikipia West Rumuruti Direct Unison Medical Centre
OutPatient Service Providers Laikipia Laikipia West Nanyuki town Direct Nyanyuki Teaching And Referral Hospital
Type County Sub County Town Access Facility Name
OutPatient Service Providers Laikipia Nyahururu Nyahururu town Direct Charity Medical Hospital
OutPatient Service Providers Laikipia Nyahururu Ngarua Direct Ngarua Catholic Dispensary
OutPatient Service Providers Laikipia Nyahururu Sipili Direct Sipili Maternity & Nursing Home
OutPatient Service Providers Laikipia Select Sub County Nyahururu town Direct Oljabet Hospital
OutPatient Service Providers Laikipia Select Sub County Nyahururu town Direct Nyahururu County Referral Hospital
OutPatient Service Providers Laikipia Select Sub County Nyahururu town Direct Pope Benedict Xvi Hospital
LAMU
OutPatient Service Providers Lamu Lamu West Mpeketoni town Direct Siha Medical Home
MACHAKOS
OutPatient Service Providers Machakos Kangundo Kangundo town Direct Kangundo Redeemed Medical Centre
OutPatient Service Providers Machakos Machakos Machakos town Direct Bliss Gvs Healthcare Ltd Machakos
OutPatient Service Providers Machakos Machakos Machakos town Direct Bishop Kioko Catholic Hospital
OutPatient Service Providers Machakos Masinga Masinga Direct Reliable Healthcare
OutPatient Service Providers Machakos Yatta Matuu Direct Matuu Cottage Medical Centre
MAKUENI
OutPatient Service Providers Makueni Kibwezi Kambu Direct Kambu Catholic Dispensery Mission
OutPatient Service Providers Makueni Kibwezi Kibwezi town Direct Kibwezi Catholic Dispensary
OutPatient Service Providers Makueni Kilome Emali town Direct Bliss Gvs Healthcare Ltd Emali
OutPatient Service Providers Makueni Makindu Makindu town Direct Mulatya Nursing Home
OutPatient Service Providers Makueni Makueni Wote town Direct Bliss Gvs Healthcare Ltd Wote
OutPatient Service Providers Makueni Makueni Makueni Direct Makueni General Hospital
OutPatient Service Providers Makueni Mbooni Mbooni Direct Mbooni Subcounty Hospital
MANDERA
OutPatient Service Providers Mandera El Wak El wak Direct Elwak Cottage Nursing Home
OutPatient Service Providers Mandera El Wak El wak Direct Midrar Medical Centre
OutPatient Service Providers Mandera Lafey Lafey town Direct Lafey Care Nursing Home
OutPatient Service Providers Mandera Mandera East Mandera Direct Shamaal Hospital
OutPatient Service Providers Mandera Mandera North Rhamu Direct The Rhamu Hospital
OutPatient Service Providers Mandera Mandera South Wargadud Direct Wargadud Medical Centre
OutPatient Service Providers Mandera Takaba Takaba Direct Mandera West Nursing Home
MARSABIT
OutPatient Service Providers Marsabit Marsabit Central Marsabit Direct I.s.m.c Clinic And Hospital
OutPatient Service Providers Marsabit Moyale Moyale Direct Afya Nursing Home Moyale
OutPatient Service Providers Marsabit Moyale Moyale town Direct Moyale Nursing Home
MERU
OutPatient Service Providers Meru Igembe North Laare market Direct Tuuru Cottolengo Health Centre
Type County Sub County Town Access Facility Name
OutPatient Service Providers Meru Igembe South Maua town Direct Bliss Gvs Healthcare Ltd Maua
OutPatient Service Providers Meru Igembe South Maua town Direct Maua Methodist Hospital
OutPatient Service Providers Meru Igembe South Maua town Direct Robins Healthcare Limited
OutPatient Service Providers Meru Nkubu Nkubu Direct Jekim Hospital Nkubu
OutPatient Service Providers Meru North Imenti Meru town Direct United Health Group
OutPatient Service Providers Meru North Imenti Meru town Direct Bliss Gvs Healthcare Ltd Meru
OutPatient Service Providers Meru North Imenti Kiirua Direct St Theresas Mission Hospital Kiirua
OutPatient Service Providers Meru North Imenti Meru town Direct Medicross Kenya Meru
OutPatient Service Providers Meru North Imenti Makutano town Direct Mediwell Healthcare K Ltd
OutPatient Service Providers Meru North Imenti Kambakia Direct Meru Jordan Hospital Ltd
OutPatient Service Providers Meru North Imenti Meru town Referral Fortis Diagnostics Centre & Hospital
OutPatient Service Providers Meru North Imenti Ruiri town Direct Sanitas Family Hospital
OutPatient Service Providers Meru South Imenti Nkubu Direct Bliss Gvs Healthcare Ltd Nkubu
OutPatient Service Providers Meru South Imenti Nkubu town Direct Jekim Maternity And Nursing Home
OutPatient Service Providers Meru South Imenti Mitunguu market Direct Mitunguu Hospital
OutPatient Service Providers Meru Tigania East Tigania Direct St John Of God Tigania
OutPatient Service Providers Meru Tigania East Mikinduri Direct The Avenue Medical & Diagnostic Centre Mikinduri
MIGORI
OutPatient Service Providers Migori Awendo Awendo town Direct Bliss Gvs Healthcare Ltd Awendo
OutPatient Service Providers Migori Awendo Awendo town Direct Sare Family Medical Centre
OutPatient Service Providers Migori Kuria Isibania Direct St. Akidiva Mindira Hospital Isibania
OutPatient Service Providers Migori Kuria  West Kehancha town Direct Kehancha Mother And Child Hospital
OutPatient Service Providers Migori Kuria  West Kehancha centre Direct Ladopharma Nursing Home Ltd Kehancha
OutPatient Service Providers Migori Kuria  West Mabera Direct St. Akidiva Mindira Hospital Mabera
OutPatient Service Providers Migori Kuria  West Isibania Direct Nyabohanse Medical Centre
OutPatient Service Providers Migori Kuria East Komotobo Direct Mother Sollbritt Health Centre
OutPatient Service Providers Migori Kuria East Kegonga Direct Getende Nursing Home
OutPatient Service Providers Migori Nyatike Nyakweri Direct Sori Lakeside Maternity And Nursing Home Nyatike
OutPatient Service Providers Migori Nyatike Mihuru bay Direct Frontal Medical Clinic
OutPatient Service Providers Migori Rongo Rongo town Direct Royal Medical Clinic & Marternity Home
OutPatient Service Providers Migori Rongo Rongo Direct Ladopharma Nursing Home Rongo
Type County Sub County Town Access Facility Name
OutPatient Service Providers Migori Rongo Rongo Direct Ladopharma Nursing Home Annex
OutPatient Service Providers Migori Select Sub County Ogwedhi town Direct Enockhan Hospital
OutPatient Service Providers Migori Suna East Migori town centre Direct Bliss Gvs Healthcare Ltd Migori
OutPatient Service Providers Migori Suna East Isibania Direct Getontira Medical Centre
OutPatient Service Providers Migori Suna East Migori Direct Ladopharma Medical Centre Ltd Migori
OutPatient Service Providers Migori Suna West Masara Direct Ladopharma Medical Centre Ltd Masara
Mombasa
OutPatient Service Providers Mombasa Changamwe Mombasa town Direct Jocham Hospital
OutPatient Service Providers Mombasa Changamwe Mombasa town Direct Nairobi Womens Hospital Mombasa
OutPatient Service Providers Mombasa Changamwe Mikindani Direct Mikindani Medical Centre
OutPatient Service Providers Mombasa Changamwe Mombasa town Direct Tudor Healthcare Ltd Magongo
OutPatient Service Providers Mombasa Jomvu Mombasa town Direct Tudor Healthcare Ltd Miritini
OutPatient Service Providers Mombasa Kisauni Kisauni town Direct Bliss Gvs Healthcare Ltd Mombasa
OutPatient Service Providers Mombasa Kisauni Bombolulu Direct Meditrust Healthcare Services
OutPatient Service Providers Mombasa Kisauni Mombasa Referral Aga Khan Hospial Mombasa
OutPatient Service Providers Mombasa Kisauni Mombasa town Direct Tudor Healthcare Ltd Kingeleni
OutPatient Service Providers Mombasa Kisauni Mombasa town Direct Tudor Healthcare Ltd Bamburi
OutPatient Service Providers Mombasa Mvita Mvita Direct Mewa Hospital
MURANGA
OutPatient Service Providers Muranga Kagundu Gaichanjiru Direct Gaichanjiru Mission Hospital
OutPatient Service Providers Muranga Kangema Kangema town Direct Kangema Medical Services
OutPatient Service Providers Muranga Muranga East Muranga town Direct Bliss Gvs Healthcare Ltd Muranga
OutPatient Service Providers Muranga Muranga South Kiria-ini Direct Kiria-ini Mission Hospital
OutPatient Service Providers Muranga Muranga South Maragwa Direct Virgin Hospital
OutPatient Service Providers Muranga Select Sub County Kangari Direct Aic Githumu Mission Hospital
OutPatient Service Providers Muranga Select Sub County Kenol Direct Kenol Hospital
NAIROBI
OutPatient Service Providers Nairobi Dagoreti North Ngong road Direct The Nairobi Womens Hospital Adams Branch
OutPatient Service Providers Nairobi Dagoreti North Hurlingham Direct The Nairobi Womens Hospital Hurlingham Branch
OutPatient Service Providers Nairobi Dagoreti North Upperhill Direct Mediheal Hospital Upperhill
OutPatient Service Providers Nairobi Dagoretti Dagoretti Direct Bliss Gvs Healthcare Ltd Kibera
OutPatient Service Providers Nairobi Dagoretti Upperhill Referral Nairobi Radiotherapy & Cancer Centre
OutPatient Service Providers Nairobi Dagoretti Upperhill Referral Fertility Point Ltd
Type County Sub County Town Access Facility Name
OutPatient Service Providers Nairobi Dagoretti Kawangware Direct Medicross Kenya Kawangware
OutPatient Service Providers Nairobi Dagoretti North Hurlingham Direct Bliss Gvs Healthcare Ltd Day Star Valley Road
OutPatient Service Providers Nairobi Embakasi Umoja_wanandege plaza Direct Bliss Gvs Healthcare Ltd Embakasi
OutPatient Service Providers Nairobi Embakasi Mombasa road Direct Bliss Gvs Healthcare Ltd Panari
OutPatient Service Providers Nairobi Embakasi Pipeline Direct Bliss Gvs Healthcare Ltd Pipeline
OutPatient Service Providers Nairobi Embakasi Bee centre spine road Direct Bliss Gvs Healthcare Ltd Umoja
OutPatient Service Providers Nairobi Embakasi Komarock Direct Komarock Modern Healthcare Komarock
OutPatient Service Providers Nairobi Embakasi Utawala Direct Komarock Modern Healthcare Utawala
OutPatient Service Providers Nairobi Embakasi Central Kayole Direct St Patrick Health Care Centre
OutPatient Service Providers Nairobi Kamukunii Eastleigh Direct Anka Hospital
OutPatient Service Providers Nairobi Kamukunji Industrial area Direct Bliss Gvs Healthcare Ltd Enterprise
OutPatient Service Providers Nairobi Kamukunji Hailesalasie Direct Bliss Gvs Healthcare Ltd Haille Selasie
OutPatient Service Providers Nairobi Kamukunji Eastleigh Direct Mediheal Hospital Eastleigh
OutPatient Service Providers Nairobi Langata Langata Direct Bliss Gvs Healthcare Ltd Mfi
OutPatient Service Providers Nairobi Langata Nairobi west Direct Bliss Gvs Healthcare Ltd Nairobi West
OutPatient Service Providers Nairobi Langata Langata Direct Medicross Kenya Langata Clean Shelf
OutPatient Service Providers Nairobi Makadara Jogoo road Direct Bliss Gvs Healthcare Ltd Jogoo Road
OutPatient Service Providers Nairobi Makadara Buruburu Direct Bliss Gvs Healthcare Ltd Buruburu
OutPatient Service Providers Nairobi Njiru Ruai Direct Ruai Family Hospital
OutPatient Service Providers Nairobi Roysambu Dykaan college kahawa wen Direct Bliss Medical –  Githurai
OutPatient Service Providers Nairobi Roysambu Dykaan college kahawa wen Direct Bliss Gvs Healthcare Ltd Kahawa Wendani
OutPatient Service Providers Nairobi Starehe Starehe Direct Bliss Gvs Healthcare Ltd Teleposta
OutPatient Service Providers Nairobi Starehe huruma Direct Jumuia Hospital Huruma
OutPatient Service Providers Nairobi Starehe Koinange street cbd Direct Bliss Gvs Healthcare Ltd College House
OutPatient Service Providers Nairobi Westlands Westlands Referral Spine And Specialist Hospital
OutPatient Service Providers Nairobi Westlands Westlands Direct Bliss Gvs Healthcare Ltd Westlands
OutPatient Service Providers Nairobi Westlands Westlands Direct Chiromo Lane Mental Hospital
OutPatient Service Providers Nairobi Westlands Parklands Direct Mediheal Hospital Parklands
NAKURU
OutPatient Service Providers Nakuru Molo Molo town Direct St Joseph Hospital
OutPatient Service Providers Nakuru Naivasha Naivasha Direct Aic Kijabe Hospital – Naivasha Medical Centre
Type County Sub County Town Access Facility Name
OutPatient Service Providers Nakuru Naivasha Naivasha Direct Bliss Gvs Healthcare Ltd Naivasha
OutPatient Service Providers Nakuru Naivasha Naivasha town Direct Naivasha Quality Healthcare Services Ltd
OutPatient Service Providers Nakuru Naivasha Naivasha town Direct The Nairobi Womens Hospital Naivasha
OutPatient Service Providers Nakuru Naivasha Gilgil Direct Ndonyo Healthcare-gilgil
OutPatient Service Providers Nakuru Naivasha Kwa-muhia Direct Ndonyo Healthcare-kwa Muhia
OutPatient Service Providers Nakuru Nakuru North Pipeline Direct Mediheal Hospital & Fertility Centre-nakuru
OutPatient Service Providers Nakuru Nakuru Town East Nakuru town Direct Nakuru Specialist Hospital
OutPatient Service Providers Nakuru Nakuru Town East Nakuru town Direct Alexandria Cancer Centre Palliative Care Hospital Naku
OutPatient Service Providers Nakuru Nakuru Town East Nakuru town Direct Mediheal Hospital & Fertility Centre-cbd
OutPatient Service Providers Nakuru Nakuru West Kenlands/shabaab Direct Medicross Kenya Shabaab Nakuru
OutPatient Service Providers Nakuru Njoro Egerton Direct Benmac Health Clinic
OutPatient Service Providers Nakuru Njoro Njoro Direct New Point Medical Centre
OutPatient Service Providers Nakuru Select Sub County Nakuru town Direct Nairobi Womens Hospital – Hyrax
OutPatient Service Providers Nakuru Subukia Subukia town Direct Jancam Medical Centre
Nandi
OutPatient Service Providers Nandi Aldai Kobujoi Direct Kobujoi Medical Centre
OutPatient Service Providers Nandi Chesumei Kapsabet town_safari hotel Direct Bliss Gvs Healthcare Ltd Kapsabet
OutPatient Service Providers Nandi Chesumei Kapsabet town Direct Reale Hospital Ltd Kapsabet
OutPatient Service Providers Nandi Chesumei Baraton Direct Jeremic Baraton Community Medical Centre
OutPatient Service Providers Nandi Chesumei Mosoriot Direct Mosoriot Medical Services
OutPatient Service Providers Nandi Mosop Kimngoror Direct Ack Kimngoror Health
OutPatient Service Providers Nandi Mosot Kabiyet Direct Eagle Medical Centre Kabiyet
OutPatient Service Providers Nandi Nandi Nandi Direct Alpha Hill Medical Centre
OutPatient Service Providers Nandi Nandi Hills Nandi hills Direct Bethesda Health Care Services
NAROK
OutPatient Service Providers Narok Narok North Narok town_suswa commer Direct Bliss Gvs Healthcare Ltd Narok
OutPatient Service Providers Narok Narok North Narok town Direct Medicross Kenya Narok
OutPatient Service Providers Narok Narok North Narok Direct Narok Cottage Hospital
OutPatient Service Providers Narok Transmara East Emuurua dikkir Direct Njipiship Hillview Medical Centre
OutPatient Service Providers Narok Transmara East Emuurua dikkir Direct Olchobosei Medical Centre
NYAMIRA
OutPatient Service Providers Nyamira Manga Kimera Direct Holsoms Medical Centre
OutPatient Service Providers Nyamira Masaba North Keroka town Direct Medstops Healthcare Kenya
Type County Sub County Town Access Facility Name
OutPatient Service Providers Nyamira Nyamira South Township Direct Bliss Gvs Healthcare Ltd Nyamira
OutPatient Service Providers Nyamira Nyamira South Township Direct Oasis Doctors Plaza Nyamira
OutPatient Service Providers Nyamira Select Sub County Nyasiongo Direct St. Joseph Mission Hospital
NYANDARUA
OutPatient Service Providers Nyandarua Central Ol kalou Direct Bliss Gvs Healthcare Ltd Ol Kalou
OutPatient Service Providers Nyandarua Kinangop Kinangop Direct Globe Medical Centre
OutPatient Service Providers Nyandarua Ol-jorok Oljororok Direct Brooke Of Cherith Medical And Nutrition Centre
NYERI
OutPatient Service Providers Nyeri Karatina Giakaibei Direct St Patrick’s Nursing Home
OutPatient Service Providers Nyeri Kieni Kieni Direct St. John Divine Nursing Home
OutPatient Service Providers Nyeri Kieni East Chaka Direct Chaka Medical Centre
OutPatient Service Providers Nyeri Kieni East Narumoro Direct Naromoru Medical Services
OutPatient Service Providers Nyeri Kieni West Mweiga Referral Mary Immaculate Hospital Mweiga
OutPatient Service Providers Nyeri Kieni West Mweiga Direct Muruguru Medical Clinic Mweiga
OutPatient Service Providers Nyeri Mathira Kirimukuyu Direct Tumutumu Mission Hospital
OutPatient Service Providers Nyeri Mukurwe-ini Mukurwe-ini Direct The Avenue Medical & Diagnostic Centre Mukurwe-ini
OutPatient Service Providers Nyeri Nyeri Central Nyeri Direct Bliss Gvs Healthcare Ltd Nyeri Jenkim Plaza
OutPatient Service Providers Nyeri Nyeri Central Nyeri Direct Bliss Gvs Healthcare Ltd Nyeri Sunguest Hotel
OutPatient Service Providers Nyeri Nyeri Central Nyeri Direct Thuti Medical Clinic
OutPatient Service Providers Nyeri Othaya Othaya Direct Mathingira Medical Centre
SAMBURU
OutPatient Service Providers Samburu Samburu Central Maralal Direct Bliss Gvs Healthcare Ltd Maralal
OutPatient Service Providers Samburu Samburu East Wamba Direct Catholic Hospital Wamba
OutPatient Service Providers Samburu Select Sub County Maralal Direct Samburu Doctors Plaza
SIAYA
OutPatient Service Providers Siaya Bondo Bondo town Direct Havens Hospital
OutPatient Service Providers Siaya Gem Yala Direct Olympus Medical Centre
OutPatient Service Providers Siaya Select Sub County Mbita town Direct Home Pearl Medical  Clinic
OutPatient Service Providers Siaya Select Sub County Sega Direct Gedmed Medical Centre And Nursing Home – Sega
OutPatient Service Providers Siaya Siaya Siaya Direct Bliss Gvs Healthcare Ltd Siaya
OutPatient Service Providers Siaya Siaya Siaya Direct Gedmed Medical Centre And Nursing Home
OutPatient Service Providers Siaya Ugenya Ukwala centre Direct Matibabu Foundation Hospital Ukwala
OutPatient Service Providers Siaya Ugenya Ungunja Direct Matibabu Foundation Hospital Ungunja
TAITA TAVETA
OutPatient Service Providers Taita taveta Mwatate Mwatate Direct Horesha Medical Clinic
Type County Sub County Town Access Facility Name
OutPatient Service Providers Taita taveta Taveta Taveta Direct Dawida Maternity And Nursing Home Taveta
OutPatient Service Providers Taita taveta Taveta Taveta Direct Taveta Meditec Centre
OutPatient Service Providers Taita taveta Voi Kaloleni Direct Tsavo Comprehensive Clinic
OutPatient Service Providers Taita taveta Voi Voi Direct Moi District Hospital
OutPatient Service Providers Taita taveta Wundanyi Wundanyi Direct Dawida Maternity And Nursing Home Wundanyi
Tana River
OutPatient Service Providers Tana river Bura Bura Direct Bura Tana Nursing Home
OutPatient Service Providers Tana river Galole Holla south Direct Siha Nursing Home
OutPatient Service Providers Tana river Galole Holla south Direct Hola County Hospital
OutPatient Service Providers Tana river Tana Delta Garsen Direct Suheyla Medical Centre
THARAKA NITHI
OutPatient Service Providers Tharaka nithi Maara Chogoria Direct Holiday  Medical Centre
OutPatient Service Providers Tharaka nithi Tharaka North Gatunga Direct Gatunga Catholic Health Centre
OutPatient Service Providers Tharaka nithi Tharaka North Mukothima Direct Mukothima Health Centre
OutPatient Service Providers Tharaka nithi Tharaka South Kathwana Direct Komarock Modern Hospital
TRANS NZOIA
OutPatient Service Providers Trans nzoia Kwanza Kitalale town Direct St. Jude Thaddeus Medical Centre Kamungei
OutPatient Service Providers Trans nzoia Saboti Kitale town Direct Bliss Gvs Healthcare Ltd Kitale
OutPatient Service Providers Trans nzoia Saboti Kitale town Direct Crystal Cottage Hospital
Turkana
OutPatient Service Providers Turkana Select Sub County Lodwar Direct Lodwar Tumaini Medical Services
OutPatient Service Providers Turkana Turkana South Lokichar Direct Canaan Medical Clinic
OutPatient Service Providers Turkana Turkana West Kakuma Direct Bliss Gvs Healthcare Ltd Kakuma
OutPatient Service Providers Turkana Turkana West Lokichoggio Direct Eureka Loki Medical Clinic
OutPatient Service Providers Turkana Turkana West Kakuma Direct Kakuma Mission Hospital
UASIN GISHU
OutPatient Service Providers Uasin gishu Ainabkoi Eldoret Direct Alexandria Cancer Centre Palliative Care Hospital Eldor
OutPatient Service Providers Uasin gishu Ainabkoi Zion mall Direct Bliss Gvs Healthcare Ltd Zion Mall
OutPatient Service Providers Uasin gishu Ainabkoi Burnt forest Direct Burnt Forest Medical Clinic
OutPatient Service Providers Uasin gishu Kapseret Pioneer Direct Bliss Gvs Healthcare Ltd Pioneer
OutPatient Service Providers Uasin gishu Kapseret Racecourse Direct Racecourse Hospital
OutPatient Service Providers Uasin gishu Kapseret Elgon view Direct Reale Hospital Ltd
OutPatient Service Providers Uasin gishu Kesses Kesses Direct Kaiyanet Medical Centre
OutPatient Service Providers Uasin gishu Kesses Naiberi Direct Mediheal Hospital & Fertility Centre-eldoret
OutPatient Service Providers Uasin gishu Soy Eldoret Direct Top Hill Hospital, Brain &spine Centre
Type County Sub County Town Access Facility Name
OutPatient Service Providers Uasin gishu Turbo Kahoya Referral Gynocare Womens & Fistula Hospital
OutPatient Service Providers Uasin gishu Turbo Turbo Direct St.raphael Arch Angel Medical Clinic
OutPatient Service Providers Uasin gishu Turbo Jua kali Direct Neema Medical Services
OutPatient Service Providers Uasin gishu Wareng Eldoret Referral Eldo Eye Centre
Vihiga
OutPatient Service Providers Vihiga Hamisi Kaimosi Direct Kapsanganyi Medical Centre
OutPatient Service Providers Vihiga Hamisi Kaimosi Direct Jumuia Friends Hospital Kaimosi
OutPatient Service Providers Vihiga Sabatia Shamakhokho Direct Shamakhokho Neema Medical Centre
OutPatient Service Providers Vihiga Vihiga Mbale Direct Bliss Gvs Healthcare Ltd Mbale
OutPatient Service Providers Vihiga Vihiga Mbale Direct Mbale Specialist Health Center
Wajir
OutPatient Service Providers Wajir Tarbaj Tarbaj Direct Al-amanah Hospital
OutPatient Service Providers Wajir Tarbaj Wajir Direct Wajir Fokas Nursing Home
OutPatient Service Providers Wajir Wajir East Wajir Direct Wajir Maternity And Nursing Home
OutPatient Service Providers Wajir Wajir East Wajir Direct Camel Hospital
OutPatient Service Providers Wajir Wajir North Bute Direct Bute Nursing Home
OutPatient Service Providers Wajir Wajir West Griftu Direct Wajir West Roadside Medical Centre
WEST POKOT
OutPatient Service Providers West pokot Pokot West Kapenguria Direct Bliss Gvs Healthcare Ltd Kapenguria
OutPatient Service Providers West pokot West Pokot Makutano Direct Shalom Medical Clinic

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