` Mathematics CONTENTS `

- EMBU COUNTY COMMON EVALUATION
- KISII CENTRAL DISTRICT JOINT EVALUATION
- ALLIANCE GIRLS HIGH SCHOOL
- STAREHE BOYS CENTER
- ALLIANCE BOYS HIGH SCHOOL TRIAL EXAMINATION
- NAMBALE DISTRICT JOINT EXAMINATION
- MERU CENTRAL COMMON EVALUATION
- NYERI NORTH REGION JOINT EXAMINATION
- THIKA(FORM FOUR MID YEAR CONTINUOUS ASSESMENT TEST)
- KITUI WESTFORM FOUR EVALUATION EXAMINATION
- NYERI SOUTH FORM FOUR JOINT EVALUATION

121/1

MATHEMATICS

PAPER 1

2 ½ HOURS

July/August 2011

THIKA DISTRICT

FORM FOUR MID YEAR CONTINUOUS ASSESSMENT TEST.

Kenya Certificate of Secondary Education

MATHEMATICS

PAPER 1

2 ½ HOURS

INSTRUCTIONS TO CANDIDATES

(a) Write your name and admission number in the spaces provided above.

(b) Sign and write the date of examination in the spaces provided above.

(c ) This paper consists of TWO sections; Section I and section II.

(d) Answer all the questions in section I and ONLY FIVE questions from section II.

(e) All answers and working must be written on the question paper in the spaces provided below each question.

(f) Show all the steps in your calculations, giving your answers at each stage in the spaces provided below each question.

(g) Marks may be given for correct working even if the answer is wrong.

(h) Non-programmable silent electronic calculators and KNEC mathematical tables may be used except where stated otherwise

(i) Candidates should check the question paper to ascertain that all pages are printed as indicated and that no questions are missing.

FOR EXAMINER’S USE ONLY

SECTION I

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Total

SECTION II

17 18 19 20 21 22 23 24 Total

SECTION 1 ( 50 MARKS )

Answer all questions in this section

- Without using a calculator, evaluate
-5 (-23 + 41) – (-10)

-3 + (-8) ÷ 2 x 4

Leaving your answer as a mixed fraction. ( 3 marks ) - A year ago, Musango was four times as old as Wambua. In three years time,

Musango will be three times as old as Wambua. Find their present ages ( 4 marks ) - Find the integral values of X satisfying the inequalities

½ (X + 3 ) > 2X + 4 ≥ X + 1 ( 3 marks ) - A translation maps a point P(1, 2) onto P1 (-2, 2). Point R is mapped onto R1 (-3, 3)

under the same translation. Find the length of PR. ( 3 marks ) - Nairobi and Mtito Andei are 240km apart. A bus left Nairobi at 9.00 a.m and travelled

towards Mtito–Andei at 80km/h. 45 minutes later, a car left Mtito-Andei for Nairobi

at a speed of 100km/h. Calculate the distance covered by the bus before meeting

the car. ( 3 marks ) - Solve for x in 9x + 32x – 1 = 53. ( 3 marks )
- The line 4ax = 5y + 12 and 7bx – 3y = 2 are parallel. Find the value of a

b ( 3 marks ) - If the expression given is a perfect square, find the value of R.

X2 – RX + 2R – 3. ( 3 marks ) - A Kenyan bank buys and sells foreign currencies as shown below.

Buying ( in Ksh ) Selling (in Ksh)

I S.A Rand 7.88 7.91

I Saudi Riyal 19.75 20.00 Mr. Brown arrived in Kenya with 102000 S.A rand and changed the whole amount to Kenya

shillings. While in Kenya, he spent Ksh. 203,760 and changed the balance to Saudi Riyal

before leaving for Saudi Arabia. Calculate, without using Mathematical tables, the

amount in Saudi Riyal that he received. ( 3 marks ) - Solve for P in the equation 3 log 8P + 5 = 2 logP8. ( 4 marks )
- In the figure below, triangle PQR is similar to triangle PST and QR//TS.

Given that PQ : PT = 2 : 5, find the ratio of the area of triangle PQR to that of

the trapezium QRST. ( 3 marks )`P T S Q R</code></pre></li>Using tables of square roots and reciprocals, evaluate 0.01960.246 Giving your answer correct to three significant figures. ( 3 marks )</code></pre></li>The sums of interior angles of two regular polygons of sides n-1 and n are in`

the ratio 2 : 3. Calculate; (i) The value of n. ( 2 marks ) (ii) The size of interior angle of each polygon. ( 2 marks ) Find the least number of steps in staircase if, when I go up 2 steps at a time, 3 steps

at a time or 4 steps at a time, there is always 1 step remaining at the top. ( 3 marks )

```
.
Solve for Ө in the range 00≤ Ө ≤3600 given that sin Ө = - ½ ( 2 marks )
Triangle ABC in the figure below is transformed to a triangle A1B1C1 by a rotation
```

of -600 about the point O.
O.
A
B
C</code></pre>Construct ∆A1B1C1 and measure the length of AA1. ( 4 marks )
SECTION II ( 50 MARKS )

Answer ONLTY five questions from this section
The marks scored by a certain number of students in a maths contest are as shown below.
Marks 45-49 50-54 55-59 60-64 65-69 70-74 75-79

Number of students 10 11 14 41 27 18 19
(a) Using an assumed mean of 62, calculate to 2 d.p. the mean of the marks. ( 4 marks )
(b) Calculate the median. ( 3 marks )
(c ) Calculate the standard deviation. ( 3 marks )
.
A ship leaves port K for port M through port L. L is 200km on a bearing of 2300 from K.

M is 400km on a bearing of 1500 from L.
(a) Using a scale of 1 : 4000000, draw a diagram to show the relative positions

of the three ports K, L and M. ( 3 marks )
(b) If the ship had sailed directly from K to M at an average speed of 40 knots, find how

long it would have taken to arrive at M. (Inm = 1.853km) ( 2 marks )
(c ) By further drawing on the same diagram, determine:
(i) How far M is to the South of K. ( 2 marks )
(ii) The shortest distance from L to the direct path KM ( 2 marks )
(d) What is the bearing of K from M? ( 1 mark )
The figure below represents Musau’s piece of land PQR divided into two triangular plots.
Q
320
125m 84m
70m
P 65m S R </code></pre>(a) Calculate the area of the plot PQR in square metres correct to two decimal

places. ( 4 marks)

(b) In the year 2008, Musau used 65% of the plot QSR for grazing and 40%

of the plot QPS for horticultural farming. Calculate, to the nearest square

metre, the piece of land left unutilized. ( 2 marks )
(c ) In the year 2009, the land grazing changed in the ratio 5 :3 while that for

horticultural farming changed in the ratio 2 : 3. Calculate, to two decimal

places the total area of land used for both grazing and horticultural farming. ( 4 marks )
Vector OA = 3 and OB = 7

6 -6
Point C is on OB such that CB = 2 OC and D is on AB such that AD = 3DB.
(a) Express CD as a column vector. ( 4 marks )
(b) Length CD ( 3 marks )
(c ) A point Q divides AB in the ratio 5 : -3. Find the position vector q of point Q. ( 3 marks)
In the figure below, the circle with centre A, radius 18cm represents the number of

people in a certain town who are infected with HIV/AIDs. The circle with centre B,

radius 12cm represents the number of people in the same town infected with TB.
Calculate to two decimal places

(a) The area of the sector subtending angle 650 at the centre ( π= 3.142) ( 2 marks )
(b) The area of the sector subtending angle 1000 at the centre. ( 2 marks )
(c ) The area representing people with both HIV/AIDs and TB. ( 4 marks )
(d) The minor arc length subtended by 1000. ( 2 marks )
The diagram below shows the speed –time graph of a bread delivery van traveling between

Emali and Sultan – Hamud. The van starts from rest and accelerates uniformly for 200 seconds.

It then travels at a constant speed for 350 seconds and then decelerates uniformly for 300 seconds.
Speed (m/s)
Time (sec)</code></pre>Given that the distance between the two towns is 15km, calculate the

(a) Maximum speed in km/h the van attained. ( 3 marks )

(b) Acceleration. ( 2 marks )
(c ) Distance the van travelled during the last 150 seconds. ( 2 marks )
(d) Time the van takes to travel the first half of the journey. ( 3 marks )
(a) Make a table of values for the function

y = 2x2 – 3x – 5 for integral values of x in the range -2 ≤ x ≤ 3. ( 2 marks )
(b) Plot the graph of y = 2x2 – 3x – 5 for -2 ≤ x ≤ 3

Use the scale 1cm rep. 2 units vertically

2cm rep. 1 unit horizontally ( 3 marks)
(c ) Use your graph to solve the equation

2x2 – 3x – 5 = 0 ( 1 mark )
(d) By drawing a suitable straight line on the same axes, solve the equation

4x2 – 2x – 6 = 0 . ( 4 marks )
A rectangular tank whose internal dimensions are 2.4m by 2.5m by 3.7m is two thirds full of juice.

(a) Calculate the volume of the juice in litres. ( 3 marks )
(b) The juice is packed in small packets in a shape of a right pyramid with equilateral

triangle sides of 20cm. The height of each packet is 15cm. Full packets obtained are sold at Ksh. 50 per packet.
Calculate;
(i) The volume in cm3 of each packet to the nearest whole number. ( 3 marks )
(ii) The number of full packets of juice. ( 2 marks )
(iii) The amount of money realized from the sale of the juice. ( 2 marks )
121/2

MATHEMATICS

PAPER 2

2 ½ HOURS

July/August 2011

THIKA DISTRICT

FORM FOUR MID- YEAR CONTINUOUS ASSESSMENT TEST

Kenya Certificate of Secondary Education

MATHEMATICS

PAPER 2

2 ½ HOURS
INSTRUCTIONS TO CANDIDATES

(a) Write your name and admission number in the spaces provided above.

(b) Sign and write the date of examination in the spaces provided above.

(c ) This paper consists of TWO sections; Section I and section II.

(d) Answer all the questions in section I and ONLY FIVE questions from section II.

(e) All answers and working must be written on the question paper in the spaces provided below each question.

(f) Show all the steps in your calculations, giving your answers at each stage in the spaces provided below each question.

(g) Marks may be given for correct working even if the answer is wrong.

(h) Non-programmable silent electronic calculators and KNEC mathematical tables may be used except where stated otherwise

(i) Candidates should check the question paper to ascertain that all pages are printed as indicated and that no questions are missing.
FOR EXAMINER’S USE ONLY
SECTION I

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Total
SECTION II

17 18 19 20 21 22 23 24 Total
This paper consists of 21 printed pages

Please Turn Over
SECTION I ( 50 MARKS )

Answer all questions in this section
Use logarithms correct to 4 decimal places to evaluate.
0.0249 x 5.825 3 0.0006748 ( 4 marks )</code></pre></li>Make X the subject of the formula ( 2 marks )

K = xt + 1

x
The nth term of G.P is given by 5 x 2n-2

(i) Write down the first 4 terms of the G.P ( 1 mark )
2.
(ii) Calculate the sum of the first 6 terms. ( 2 marks )
A student converts the fraction 5/9 into decimal and truncates the result to

2 decimal places. Calculate the relative error in this approximation

(Give your answer in fraction form ) ( 3 marks )
The figure below shows a circle centre O, of radius 7cm. TP and TQ are tangents to

the circle at points P and Q respectively. OT = 16cm
P
7cm
O 16cm

T
Q
Calculate the length of chord PQ. ( 3 marks)
3.
Triangle ABC is such that ABC = 900, AB = ( √3 + 1 )cm and AC = (√3 + 2 )cm.

Find cos A in simplified surd form. ( 4 marks )
If variables M, D, R are related by the equation M = KDR3 where K is constant,

find the percentage change in M if D is reduced by 10% and R is increased by 10%. ( 3 marks )
The table below is part of a taxation table for monthly income for the year 2010.
Monthly taxable income in Ksh. Tax rate (%) per Ksh.

1 – 9680 10

9681 – 18800 15

18801 – 27920 20
Kyele paid sh. 2476 as tax during the month of February in this year. Calculate his
monthly income. ( 3 marks )
Determine the equation of the tangent to the circle with equation.
(x-5)2 +(y-5)2 = 5 at the point (6, 7) ( 4 marks )
The cost of three books and five pens is sh. 135. The cost of two similar books and

seven similar pens is sh. 145. Use the matrix method to find the cost of each item. ( 4 marks )
A point R (2,2) is mapped onto R1 (8,2) by a shear, x –axis invariant. Find the shear

factor hence or otherwise write down the shear matrix. ( 3 marks )
The graph in the figure below shows the variation of depth of water d(metres) in a

river with time t(hours) after a heavy down pour.
.

Use the graph to find the rate of decrease in depth at t = 1.25 hours

( Give your answer correct to 2 significant figures) ( 3 marks )
The number of overtime hours worked per week by 130 men of the maintenance

department of a power and lighting company are as shown.
(a) Find X ( 1 mark )

(b) Find the mean number of overtime hours giving your answer correct to 2dp. ( 3 marks )
No. of overtime hours 0 – 2.5 2.6 – 5.0 5.1 – 7.5 7.6 – 10.0 10.1 – 12.5 12.6 – 15.0

Frequency (f) 31 48 26 14 8 x
P
B Ө 90m 60m
C
A </code></pre>In the diagram A, B, C are all in a horizontal plane. Flag pole BP is 90m from A and 60m

from C. Angle ABC = 900. The angle of elevation of P from A is Ө where tan Ө = 1/3.

Calculate the height BP and the angle of elevation of P from C. ( 4 marks )
A square whose vertices are the points A(0, 4) B (1, 1), C (4, 2), D (3, 5) is mapped

onto the plane figure A1 B1 C1 D1 by matrix M = -1 0

4 3

Find the area of A1 B1 C1 D1. ( 3 marks )
Find the 4th term of the expansion of 1 + 1 x 6 ( 2 marks )

5
SECTION II ( 50 MARKS )

Answer only five questions in section II
A bag contains 10 similar pens of which 6 are red and the rest are blue in colour.

Three pens are picked at random, one at a time from bag without replacement.
(a) Draw a tree diagram to show the various outcomes. ( 2 marks )

(b) Find the probability that

(i) None of the pens picked is red. ( 2 marks)
(ii) At least one of the pens picked is red. ( 2 marks )
(iii) Only one blue pen is picked. ( 2 marks )
(iv) The first two pens picked are of the same colour. ( 2 marks )
(a) On the grid provided, draw a graph of the function y = 3x2 + 8 for 0 ≤ x ≤ 5 ( 3 marks)
(b) Calculate the mid-ordinates for 4 strips between x = 1 and x = 5 and hence use the mid-ordinate rule to approximate the area under the curve between x = 1, x = 5 and the x – axis . ( 3 marks )
(c ) Assuming the same area determined by integration to be the actual area, calculate the percentage error in using the mid-ordinate rule. ( 4 marks )
.
(a) Complete the table below for the functions y = 2 cos x and y = sin 2x. ( 2 marks )
x0 -1800 -1500 -1200 -900 -600 -300 0 300

2x0 -300 -120 0

2 cos x 1.73 2

Sin 2x 0.87 0
x 600 900 1200 1500 1800

2x0 180 360

2 cos x 1 2

Sin 2x -0.87 0
(b) On the same axes, draw the graphs of y = 2 cos x and y = sin 2x for -1800 ≤ x ≤ 1800. ( 4 marks)
( c ) Use the graphs in (b) above to find;
( i) The values of x such that 2cos x – sin 2x = 0 . ( 1 mark )
(ii) State the amplitude and period of each graph. ( 2 marks )
(iii) Find the difference in the values of y when x = -45. ( 1 mark )
Use ruler and a pair of compasses only in this question.

(a) Construct triangle ABC with AB = 5cm, BC = 6cm and AC = 7cm. ( 2 marks )
(b) Construct the locus of a point O such that OA = OB = OC. ( 2 marks )

(c ) With O as centre and radius OB draw a circle. ( 1 mark )
(d) At B construct the tangent YBX such that X is on the same side of BO produced as C. ( 2 marks)
(e) Construct the locus of points L through O such that the distance from L to

XY is constant. ( 2 marks )
(f) Locate point P such that P is on L and ABY = BPA. ( 1 mark )
The derived function of a curve is 2x + 2

(a) Find the equation of the curve if it passes through ( -2, -3 ) ( 2 marks )
(b) Find the x intercepts of the curve. ( 3 marks )
(c ) The curve passes through point (O, y). Find the value of y. ( 1 mark )
(d) Find the value of y at the point where the gradient of the curve is O (zero). ( 2 marks )
E) In the space provided below, sketch the curve. ( 2 marks )
A drug company specializing in food supplements makes up a product by mixing two ingredients A and B each containing both zinc and iron. The mineral composition per gram of each ingredient is given in the table below.
Ingredient Zinc (Units per gram) Iron (Units per gram)

A 20 50

B 40 75
The product is to be packaged in small bottles each containing x grams of A and y grams of B.
The following conditions must be satisfied
I: The mass of the product in each bottle must not exceed 10 grams
II: Each bottle is to contain at least 200 units of zinc
III: Each bottle is to contain at least 450 units of iron
(a) Write down three inequalities in terms of x and y to express the above conditions. ( 3 marks )
(b) On the grid provided, draw the inequalities by shading the unwanted region. ( 3 marks )
(c ) Given that the costs per gram of ingredients A and B are sh. 60 and

sh. 100 respectively, draw a search line and use it to determine the minimum cost of one bottle of the product. ( 4 marks )
A plane flying at 200 knots leaves airport P(200N, 340W) and flies due South to an

airport Q(600S, 340W).
(a) Calculate the distance covered by the plane in nautical miles. ( 3 marks )
(b) The plane stops at Q for 30 minutes before flying due East to an airport R(600S, 60E)

at the same speed. Calculate the total time taken to complete the journey from

airport P to airport R through airport Q. ( 5 marks )
(c ) If at the time of arrival, the local time at R is 6.00pm, what is the local time at P then ? (2 marks)
The figure below shows a right pyramid VABCD with a square base of side 6cm.

VA = VB = VC = VD = 12cm.
(a) Calculate

(i) The height of the pyramid correct to 2 d.p . ( 3 marks )
(ii) The angle between the planes VAD and VBC correct to 2 significant figures. ( 3 marks )

(b) B1 and C1 are points on VB and VC respectively such that

VB1 : VB = VC1 : VC = 1 : 3. Calculate the angle between planes ABCD

and AB1C1D. ( 4 marks )
121/1

MATHEMATICS

PAPER 1

July/August 2011

THIKA DISTRICT

FORM FOUR MID YEAR CONTINUOUS ASSESSMENT TEST

MARKING SCHEME
-5 (18) + +10

-3 + -4 x 4 M1 √ Removal of brackets
-90 + 10

-3 + -16

-80

-19 M1 √ Order of operations
80
19 44/19 A1 3</code></pre></li>Musango now - x yrs

Wambua now - y yrs

x – 1 = 4 (y-1) M1

x + 3 = 3 ( y + 3 ) M1

x – 4y = -3

x – 3y = 6 M1 √ attempt to eliminate

-y = 9

x = 33
Musango = 33 yrs
Wambua = 9 yrs A1
4
½ ( x + 3 ) > 2x + 4 ………….. (i)

2x + 4 ≥ x + 1 ………….. (ii)
x + 3 > 4x + 8

-3x > 5

x < 5/-3

x < -12/3 B1
2x + 4 ≥ x + 1

x ≥ -3 B1
-3 ≤ x < -12/3
Integral values = -2, -3 B1 3</code></pre></li>
4 Let T = x

y
1 + x = -2 M1
2 y 2
x = -3
y 0
a + -3 = -3
b 0 3 M1
a = 0
b 3
PR = (1 – 0)2 + (2-3)2
= 12 + (-1)2
= 2 = 1.414 A1
3
Dist covered by bus before car starts

= ¾ x 80 M1

= 60km

Rem. Dist. = ( 240 – 60 )km

= 180km
Time taken to meet = 180 = 1hr

180
Dist. Covered by bus = 60 + 80 M1

= 140km. A1
32x + 32x – 1 = 53 M1 √ Expressing in

terms of 3

2 x 32x = 54

32x = 27

32x = 33

2x = 3 M1 √ Equating powers

x = 3/2 A1
3
5y = 4ax – 12

y = 4ax – 12

5 5
3y = 7bx – 2

y = 7bx - 2 B1 equations in the form

3 3 y = mx + c
Parallel lines have equal gradients

4a = 7b

5 3 M1 equating gradients
12a = 35b

12a = 35

b
a = 35
b 12 A1 03</code></pre></li>Last term = (Middle term )2

4 x 1st term
2R – 3 = R2 . X2

4 . X2 M1
4(2R – 3) = R2
R2 – 8R + 12 = 0
R2 – 6R – 2R + 12 = 0
R ( R – 6) -2 (R -6) = 0

(R – 6) (R -2) = 0 M1

R = 2 or R = 6 A1
102000 x 7.88 M1 Conversion

= Ksh. 803760
Rem. = 803760 – 203760

= Ksh. 600,000
In Riyal = 600000 M1

20

= 30000 Riyal A1
03
3log8P + 5 – 2logP8 = 0

3log8P + 5 – 2 1 = 0

log8P M1
Let log8P = t

3t + 5 – 2/t = 0
3t2 + 5t – 2 = 0

3t2 + 6t – t -2 = 0

3t (t+2) – 1(t+2) = 0

(3t – 1) (t + 2) = 0 M1

t = 1/3 or t = -2
But log8P = t

Log8P = 1/3 or log8P = -2 M1
1/3

8 = P or 8-2 = P

P = 2 or P = 1/64 A1
L.S.F of ∆s = 2 :5

A.S.F = 4 : 25 B1

Area of trap = 25 – 4 M1

= 21

Ratio 4 : 21 A1

3
1 = 0.4065 x 10

0.246 = 4.065 B1
0.14 x 4.065 M1

= 0.569 A1 C.A.O

03
[ 2(n-1) -4] x 900 2

(2n-4) x 900 = 3 M1 √ Expression for sum

2n – 6 = 2

2n – 4 3
6n – 18 = 4n – 8

2n = 10

n = 5 A1
n = 5 (2 x 5 – 4) x 900

5
= 5400
5
= 1080 B1
n -1 = 4 (2 x 4 – 4) x 900

= 3600

4

= 900 B1
4
L.C.M = 22 x 3

= 12 B1

No. of steps = 12 + 1

= 13 B1
02
Acute angle = 300 B1

Ө = 2100, 3300
16 B1 02
(a) B1

Marks (x) f t = x – 62 ft

47 10 -15 -150

52 11 -10 -110

57 14 -5 -70

62 41 0 0

67 27 5 135

72 18 10 180

77 19 15 285

f = 140 ft = 270
X = 62 + 270

140 M1
= 62 + 1.93
= 63.93 A1
C.F: 10, 21, 35, 76, 103, 121, 140 B1
(b) Median = 59.5 + 140 35 x 5

2 M1

41
= 59.5 + 35/41 x 5
= 59.5 + 4.27
= 63.77 A1
(c )

d = x – m d2 f Fd2

-16.93 286.62 10 2866.2

-11.93 142.32 11 1565.52

-6.93 48.02 14 672.28

-1.93 3.72 41 152.52

3.07 9.42 27 254.34

8.07 65.12 18 1172.16

13.07 170.82 19 3245.58

f = 140 fd = 9928.6
S = 9928.6

140 M1
= 70.9186
= 8.421 A1
10
1cm rep. 40km B1 √ Scale interpretation
B1 K to L (√ bearing
and √ distance ) B1 L to M (√ bearing
and √ distance )</code></pre>(b) km = 12 x 40 = 259.04nm B1 (12.0 0.1)

1.853
Time taken = 259.04 = 6.476 hrs B1
40
.
(c ) (i) 11.9 x 40 = 476km 4km B1 point X √ly located
B1 distance 476km
(ii) 4.2 x 40 = 168km 4km B1 point ly located
B1 distance 168km.
(d) 3540 10 B1 bearing
10
(a) ∆PQS , S = 130 B1
Total A = 130 (130 – 125) (130 – 65) ( 130 – 170) M1
+ ½ x 70 x 84 sin 320 M1 = 2535000 + 1557.96
= 3150.13m2 A1 C.A.O </code></pre>(b) Unutilised = 35 x 1557.96 + 60 x 1592.17

100 100
= 1500.588
= 1501m2 A1
(c ) 5 x 65 x 1557.96 + 2 x 40 x 1592.17 M1M1

3 100 3 100
= 2112.368667 A1
= 2112.37m2 B1 10</code></pre></li>A (3, 6 )
a 3
B1
D
O
1 b 1
C
2 B (7, -6) Q</code></pre>.

(a) CD = CO + OA + ¾ AB
= - 1/3b + a + ¾ ( b – a ) M1
= - 1 7 + 3 + 3 4 M1
3 -6 6 4 -12
= -7/3 + 3 + 3
2 6 -9 = 32/3
-1 A1</code></pre>(b) CD = (32/3)2 + (-1)2 M1
= 14.44 M1 = 3.8 A1</code></pre>(c ) q = n a + m b

m + n m + n
= -3 3 + 5 7
2 6 -2 -6 M1 = -9/2 + 35/2 M1
-9 -15
= 13
-24 A1
10</code></pre></li>(a) A = 65 x 3.142 x 18 x 18 M1

360

= 183.807

= 183.81cm2 A1 C.A.O
(b) A = 100 x 3.142 x 12 x 12 M1

360

= 125.68cm2 A1 C.A.O
(c ) For A, area of ∆ = ½ x 18 x 18 sin 650

= 146.82
Area of segment = 183.81 – 146.82 M1
= 36.99cm2
For B, area of ∆ = ½ x 12 x 12 sin 100
= 70.91cm2
Area of segment = 125.68 – 70.91 M1
= 54.77cm2
Total area = 36.99 + 54.77 M1
= 91.76cm2 A1 C.A.O
(d) Length = 100 x 3.142 x 24 M1

360

= 20.95cm A1
10
(a) ½ h ( 850 + 350) = 15000 M1 √eqn.

1200h = 30000

h= 25 m/s A1
In km/h = 3.6 x 25
= 90 km/h B1
(b) a = 25 m/s M1

200s
= 1/8 m/s2 A1 or 0.12m/s2
(c ) S = ½ x 150 x 12.5 M1

= 937.5m A1
(d) ½ x 25 ( 200 + t + t ) = ½ x 15000 M1 eqn.
2t + 200 = 7500
12.52t = 400
t = 200 A1 Total time = 200 + 200
= 400 sec. B1
x -2 -1 0 1 2 3

y 9 0 -5 -6 -3 4 B2 all values √

allow B1 for 4 value √
S1 √ Scale P1 All pts √ly plotted
√ if B1 earned
C1 Smooth curve</code></pre></li>
.
(c ) x = -1 or 2.5 B1 (both )
(d) O = 4x2 – 2x – 6
O = 2x2 – x – 3 M1 Subtraction
y = 2x2 – 3x – 5
y = -2x – 2 A1 eqn of line
L1 √ line drawn
x = -1 or x = 1.5 B1 (both)
10
(a) V = 2/3 x 2.4 x 2.5 x 3.7m3 M1
= 14.8m3 M1
= 14,800 litres A1
(b) (i)
20 20 h
600
10 10
h = 20 sin 60
= 17.32 M1Volume = 1/3 x ½ x 20 x 17.32 x 15 M1
= 866cm3 A1
(ii) No. of pockets = 14.8 x 1000000
866 M1 = 17,090 A1(iii) Money = sh. 50 x 17,090 M1
= Sh. 854,500 A1 10</code></pre></li>
121/2

July/August 2011

THIKA DISTRICT

FORM FOUR MID YEAR CONTINUOUS ASSESSMENT TEST

MATHEMATICS

PAPER 2

Kenya Certificate of Secondary Education

MARKING SCHEME
No. Log

0.0249 2.3962

5.825 0.7653 + M1 All logs √

1.1615

3√ 0.0006748 1/3 x 4.8292 M1 √attempt to add/sub

= -6 + 2.8292

3 2.9431 M1 √ attempt to divide by 3
1.6535 0.2184 A1 C.A.O
04
kx = xt + 1

kx – xt = 1 M1 Xmult and collect

like terms

x ( k – t) = 1
x = 1
k – t A1 02</code></pre></li>(i) nth term = 5 x 22-n

n = 1 = 5 x 2 = 10

n = 2 = 5 x 20 = 5

n = 3 = 5 x ½ = 2.5

n = 4 = 5 x ¼ = 1.25
10, 5, 2.5, 1.25 …. B1 √ all terms
(ii) Sn = a (1 – rn )

1 – r
S6 = 10 (1 – 24)
1 – 2 M1 √ substitution = 10 ( 1 - 16 )
-1
= 10 x -15
-1
= 150 A1 √
3</code></pre></li>5 becomes 0.55 = 55 B1 for 0.55

9 100
Rel. error = 5 - 55

9 100 M1

5/9
= 1 ÷ 5
180 9 = 1
100 A1
03</code></pre></li>PT = (16)2 – (7)2
= 256 - 49
= 207 = 14.39cm
Cos Ө = 14.39 = 0.5756 M1
25 Ө = cos -1 ( 0.5756)
= 54.860Sin 35.14 = h
7
PQ = 2h = 2 ( 7 sin 35.14 ) M1
= 2 ( 4.029)
= 8.058cm A1
03
A
√3 + 1 √3 + 2
B C</code></pre>Cos A = √3 + 1
√3 + 2 M1 √ Fraction
= √3 + 1 x (√3 - 2√3 + 2 (√3 – 2) M1= 3 – 2√3 + √3 – 2 M1 removing surd from
3 – 4 denominator
= √3 - 1 A1 04</code></pre></li>M = KDR3

M1 = 90/100 x D x ( 110/100 x R )3 x K M1 √ new M
M1 = K x 0.9D x ( 1.1r)3

= 1.1979 KDR3
Change in M = 1.1979KDR3 – KDR3 M1 √change in M
= 0.1979KDR3
% change in M = 0.1979KDR3 x 100
KDR3
= 19.8% A1
1st bracket : 10/100 x 9680 = 968
Remaining tax = 2476 – 968 = 1508 M1 1st and 2nd brackets
2nd bracket: 15/100 x 9120 = 1368
3.

Remaining tax = 1508 – 1368 = 140
20/100x X = 140 M1
x = sh. 700
Income = 700 + 9120 + 9680 = sh. 19500 A1
03
Centre O is (5, 5), D is (6, 7)
Gradient of OD = 7 – 5

6 - 5 M1 √ gradient of OD

= 2
The gradient of the tangent at D = - ½ M1 √ Gradient of tangent

at D
Choosing a point on the tangent P(x, y)

D (6, 7) and P (x, y)

Gradient = y – 7

x – 6
y – 7 = - ½
x – 6 M1 √ expression
-1(x – 6) = 2 ( y – 7 )
-x + 6 = 2y – 14
2y + x = 20 A1 √ tangent equation
3x + 5y = 135

2x + 7y = 145 M1 √ equations
3 5 x = 135

2 7 y 145
Inverse = 1 7 -5

11 -2 3 B1 inverse
x = 1 7 -5 135
y 11 -2 3 145 M1
x = 20
y 15 x = 20, y = 15 A1
04 </code></pre></li>Alternatively

Shear factor = 8 – 2 M1 1 K 2 = 8 M1

2 0 1 2 2

= 3 A1 2 +2k = 8

k = 3 A1
Matrix = 1 3 1 3

0 1 B1 0 1 B1
03
Rate of decrease

Tangent drawn at t = 1.25 B1 see graph
Rate = 2.2 – 0.8 M1 check for √ reading

0.5 – 1.55
= 1.4
-1.05
= -1.333
Rate of decrease = 1.3 m/h A1 Accept 1.2, 1.4

Accept answer with

negative sign

03
(a) x = 130 – 127

= 3 B1
Hours Mid pt f fx

0 – 2.5 1.25 31 38.75 B1 √ fx column

2.6 – 5.0 3.75 48 180.0

5.1 – 7.5 6.25 26 162.5

7.6 – 10.0 8.75 14 122.5

10.1 – 12.5 11.25 8 90.0

12.6 – 15.0 13.75 3 41.25

f =130 fx = 635
Mean (x) = fx = 635 M1 √ substitution
f 130
= 4.88 (2dp) A1 √ A.O 14. Tan Ө = 1/3 </code></pre> 1/3 = BP

AB M1 √ relation
1 = BP

3 90
3BP = 90
BP = 30m
Tan C = BP

60
Tan C = 30
60 M1 √expression for PCB
Tan C = 0.5
C = 26.60 A1 03</code></pre></li>Length of side of the square = √ 10 M1

The area of ABCD = 10 sq units

Area scale factor = det. = -3 – 0

= 3

Area of A1B1C1D1

= 10 x 3 M1

= 30 sq units A1
20.13 . ( 1/5x)3 M1

20 x 1/125 x3

20/125x3
4/25x3 A1 02</code></pre></li>R …..RRR
4/8 R 4/8 B ……RRB B1 first two sets
5/9
R …. RBR
5/8
R 4/9 B</code></pre>6/10 3/8 B ….RBB B1 Last set

R R ….BRR

4/10 B 6/9 5/8

3/8

B …..BRB
3/9 R …..BBR
B 6/8
2/8
B …..BBB
(b) (i) P (no red) = P (BBB)

= 4/10 x 3/9 x 2/8 M1
= 1/30 A1
(ii) P (at least red) = 1 – P ( no red)
= 1 = -1/30 M1 = 29/30 A1</code></pre>(iii) P(one blue) = P (RRB or RBR or BRR)

= ( 6/10 x 5/9 x 4/8 ) + ( 6/10 x 4/9 x 5/8) + (4/10 x 6/9 x 5/8) M1

= ½ A1
(iv) P (first two same) = P(RRR or RRB or BBR or BBB)

= (6/10 x 5/9 x 4/8) + (6/10 x 5/9 x 4/8) + (4/10 x 3/9 x 6/8) + (4/10x3/9x2/8) M1

= 7/15 A1

10
(a)

x 0 1 2 3 4 5

y 8 11 20 35 56 83
(b)

x 1.5 2.5 3.5 4.5

Y 14.75 26.75 44.75 68.75
A = 1 ( 14.75 + 26.75 + 44.75 + 68.75) M1

= 155 sq. units A1
(c ) ∫ (3x2 + 8) dx

1

= 5

= x3 + 8x M1

1
= {125 + 40 } - { 9 }
= 156 A1
% error = 156 – 155 x 100 M1
156 = 1/156 x 100
= 0.006410 x 100
= 0.6410% A1
10</code></pre></li>
.
(a)

x0 -1800 -1500 -1200 -900 -600 -300 00 300 600 900 1200 1500 1800

2x -3600 -3000 -2400 -1800 -1200 -600 0 600 1200 1800 2400 3000 3600

2 cos x -2 -1.73 -1 0 1 1.73 2 1.73 1 0 -1 -1.73 -2

Sin 2x 0 0.87 0.87 0 -0.87 -0.87 0 0.87 0.87 0 -0.87 -0.87 0
B2 all values B1 10 values
(i) x = -900 and x = 900 B1
(ii) y = sin 2x; amplitude = 1

period = 1800 B1
y = 2 cos x ; amplitude = 2
period = 3600 B1
(iii) Difference 2.4 0.1 B1
10
(b)
(a) dy = 2x + 2

dx
y = x2 + 2x + C M1 √ Integration
at ( -2, -3 )
-3 = 4 + ( -4) + c
c = -3
y = x2 + 2x – 3 A1 √ equation
(b) x2 + 2x – 3 = 0 M1

( x + 3) ( x – 1) = 0 M1 Factorisation

or equivalent

x = -3 and x = 1 A1 (both)

(-3, 0) and (1, 0)
(c ) y = 0 + 0 – 3

y = -3 B1
(d) 2x + 2 = 0

x = -1 B1 for x = -1

y = (-1)2 – 2 – 3

y = -4 B1 for y = -4
(e) y x -3 -1 0 1 -3 -4 (-1, -4)
B2 Curve thro’ all
Intercepts and
stationary point
(a) x + y ≤ 10 …………. (i) B1

20x + 40y ≥ 200 ………. ..(ii) B1

Or x + 2y ≥ 10
50x + 75y ≥ 450 ………. (iii) B1
Or 2x + 3y ≥ 18
(b)
B1 x + y ≤ 10 B1 x + 2y ≥ 10
B1 2x + 3y ≥ 18</code></pre>(c ) 60x + 100y = c B1 objective function

x = 6, y = 2 or ( 6, 2) L1 Search line drawn

C = 60 x 6 + 100 x 2

= sh. 560 B1
(a) Ө = 60 + 20 = 800 B1

PQ = 60 x 80 M1

= 4800nm A1
(b) Ө = 6 + 34 = 400 B1
QR = 60 x 40 cos 600 M1
= 1200nm A1
Total time = 1200 + 4800 + ½ M1
200
= 30 ½ A1
(c ) Time difference = 40 x 4 = 160 min B1
Time at P = 6.00
2.40 -
= 3.20pm B1
10
(a) (i)

C

AC = √ 62 + 62 M1

O

= √ 72
A B AO = ½ √ 72

VO = 122 – ( ½ √ 72)2 M1
= 144 – 72/4 VO = 11.22cm A1 C.A.O (ii)
V Ө
11.22cmM 3cm O Tan Ө = 3
11.22 M1 follow through if Ө = 14.970 A1 Cosine rule is used.
Reqd angle = 2 x 14.97
≃ 300 B1 C.A.O </code></pre>(b) l.s.f = 1/3
M1 2cm N1 7.48cm M1N1 = 1/3 x 6 = 2cm B1 angle Ө identified
Ө</code></pre>M 4cm x 2cm N XN1 = 2/3 x 11.22 = 7.48 B1 MX and XN1 obtained

MX = 4cm ( any two sides of

∆MXN1)
Tan Ө = 7.48
4 M1 Ө = 61.860 A1
10</code></pre></li>
NAME……………………………………………………. INDEX NO……………….
SCHOOL……………………………………………. CANDIDATE SIGN………………
121/1

MATHEMATICS

PAPER I

TIME: 2 ½ hours

JULY/AUGUST 2011

KITUI WEST

FORM FOUR EVALUATION EXAMINATION 2011

Kenya Certificate Of Secondary Education
INSTRUCTIONS TO CANDIDATES
a) Write your name and index number in the spaces provided.

b) This paper consists of two sections I and II.

c) Answer all the questions in section I and only five from section II.

d) All answers and working must be written on the question paper in the space provided below each question.

e) Show all the steps in your calculations giving your answer at each stage in the space blow each question.

f) Marks may be given for correct working even if the answer is wrong.

g) Use calculators and KNEC mathematical tables except where stated otherwise.
FOR EXAMINERS USE ONLY

SECTION 1

TOTAL 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
SECTION 11

17 18 19 20 21 22 23 24 TOTAL
SECTION A

Answer all questions in this section in space provided.
Use logarithm table to evaluate. (4 marks)
3 (0.0246)2 x 142
0.002 x 1.14
A two digit number is such that the sum of the ones and the tens digit is ten. If the

digits are reversed, the new number formed exceeds the original number by 54.

Find the number.

(3 marks)
A straight line L1 passes through P(2,1) and is perpendicular to straight line L2,

whose equation is 2y – x + 4 = 0. Find the equation of L1. (3 marks)
Alice chepchumba on her cycling practice cycled on a bearing of 120o for 5.5km, then

on a bearing of 200o for 8km finally he turned northwards for 13.5km, by scale

drawing determine her final position from starting point. (4 marks)
Make t the subject of the formula. (3 marks)
x = 1 + kt

kt – 1
The volume of a rectangular tank is 256cm3. The dimensions are as in the figure.
¼ x
x-8

16cm
Find the value of x (3 marks)
Solve for x
125-x x 52(x-2) = 25 (x+2) (3 marks)
Find the equation of the normal to the curve
X2 = 4y at the point (6,9) (3 marks)
The acceleration of a particle in M5-2 is given by the expressions 3t –4

Find:-

(i) an expression for velocity Vms1 (1 mark)
(ii) an expression for distance 5 metres from a fixed point O. Given that S=0
when V=3 and t=0 (2 marks)
10 On Monday this currency exchange rate was

1 Euro (E) = Kshs.95.65

1 US dollar($) = Ksh.76.50

A gentle man Tourist decided to exchange half of his 2400E into Dollars.

Calculate to 2 decimal places the number of dollars he received. (3 marks)
Two coils which are made by winding copper wire of different gauges and length have the same mass. The first coil is made by winding 270 metres of wire with cross sectional diameter 2.8mm while the second coil is made by winding a certain length of wire with cross-sectional diameter 2.1mm. Find the length of wire in the second coil . (4 marks)
a) Find the greatest common divisor of the term. (1 mark)
144x3y2 and 81xy4
b) Hence factorise completely this expression 144x3y2-81xy4 (2 marks)
a) Find the range of values x which satisfied the following inequalities simultaneously. (2 marks)
4x – 9 < 6 + x

8 – 3x < x + 4
b) Represent the range of values of x on a number line. (1 mark)
Find the value of x in the equation.
Cos(3x - 180o) = √3 in the range Oo < x < 180o (3 marks)
2
The length and width of a rectangle are stated as 18.5cm and 12.4cm respectively. Both measurements are

given to the nearest 0.1cm.
a) Determine the lower and upper limit of each measurement. (1 mark)
b) Calculate the percentage error in the area of the rectangle. (3 marks)
A surveyor recorded the measurement of field in a field book using lines AB = 260m as shown below.
B

130 R40

70 Q10

50 P20

S50 10

A
a) Use a suitable scale to draw the map of the field. (2 marks)
b) Find the area of the field. (2 marks)
SECTION B (50 MARKS)

Answer only five questions in this section in the spaces provided.
A trader sold an article at sh.4800 after allowing his customer a 12% discount on the marked price of the

article. In so doing he made a profit of 45% .
a) Calculate

(i) the marked price of the article. (3 marks)
(ii) the price at which the trader had bought the article (2 marks)
b) If the trader had sold the same article without giving a discount. Calculate the percentage profit he would have made. (3 marks)
c) To clear his stock, the trader decided to sell the remaining articles at a loss of 12.5% (Calculate the price at which he sold each article. (2 marks)
Three quantities P, Q and R are such that P varies directly as the cube of Q and inversely as the square of R.
a) Given that P = 16 when Q = 2 and R = 3. Determine the value of R when P= 288 and Q = 4

(5 marks)
b) Q decreases by 30% while R increases by 40%. Find the percentage decrease or increase in P.

(5 marks)
All employees of silver springs enterprises pay income tax at the rate shown in the table below.
Taxable income (p.a) Rate sh. Per K₤

1 – 3780

3781-7560

7561-11,340

11,341-OVER 2

3

4

5
Mr. Mooka earns a basic salary of sh.12,150 and a house allowance of sh.2800 per month. He is

entitled to a family relief of sh.450 per month. A part from income tax the following deductions are

also made from his monthly pay.
a) Servicing loan payment sh.450

b) Hospital fund sh.260

c) Sacco contribution sh.120
Determine Mr. Mooka’s net monthly income. (10 marks)
A passenger train travelling at 25km/h is moving in the same direction as a truck travelling at 30km/h. The

railway line runs parallel to the road and the truck takes 1 ½ minutes to over take the train completely.

a) Given that the truck is 5 metres long determine the length of the train in metres. (6 marks)
b) The truck and the train continue moving parallel to each other at their original speeds. Calculate the distance between them 4 minutes and 48 seconds after the truck overtakes this train. (4 marks)
22.5cm
The diagram represent a solid frustum with base radius 21cm and top radius 14cm. The frustum is 22.5cm high and is made of a metal whose density is 3g/cm3 π = 22/7.
a) Calculate

(i) The volume of the metal in the frustrum. (5 marks)
(ii) The mass of the frustrum in kg. (2 marks)
b) The frustrum is melted down and recast into a solid cube. In the process 20% of the metal is lost. Calculate to 2 decimal places the length of each side of the cube. (3 marks)
The table below shows the masses to the nearest kg of all the students of marigu-ini secondary. School.
Masses (kg) No. of students

30-34 5

35-39 7

40-44 10

45-49 10

50-54 19

55-59 20

60-64 20

65-69 6

70-74 2

75-79 1
a) Taking the assumed mean A=52kg

Calculate:
(i) The actual mean mass of the students. (3 marks)
(ii) The standard deviation of the distribution. (3 marks)
b) Draw a cumulative frequency curve and use it to estimate the number of students whose masses lie

between 44.5kg and 59.5kg. (4 marks)
1
a) Given the transformation matrices
T1 = 2 1 and T2 = 3 1
-1 -2 1 3
and that transformation T1 followed by T2 can be replaced by a single transformation T, write
down the matrix for T. (3 marks)
b)Find the inverse of matrix T (2 marks)</code></pre></li>
c) The points A11(7,-11), B11(-7,-13), C11(-8,16) and D11(8,8) are the images of points A, B, C and

D respectively under transformation T1 followed by T2

Write down the co-ordinates of A, B, C, and D. (5 marks)
24. Draw the graph of
y = 2x2 + x – 2 and use it to solve the equations
(10 marks)
a) 2x2 + x – 2 = 5
b)2x2 + x – 5 = 0
c)2x2 +2x – 3 = 0
Name ………………………………………..……………………………. Index no. ……………………………..
121/2

MATHEMATICS

PAPER 2

JULY / AUGUST2011

2 ½ HOURS
KITUI WEST
FORM FOUR EVALUATION EXAMINATION 2011

Kenya Certificate of Secondary Education

MATHEMATICS

PAPER 2

2 ½ HOURS
INSTRUCTIONS TO CANDIDATES

(a) Write your name and Index number in the spaces provided above

(b) This paper consists of two sections. Section I and section II.

(c ) Answer ALL the questions in section I and only FIVE questions in section II.

(d) All answers and working must be written on the question paper in the spaces provided below each question.

(e) Non-programmable silent calculators and KNEC maths tables may be used except

here stated otherwise.
FOR EXAMINER’S USE ONLY
SECTION I
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Total
SECTION II
17 18 19 20 21 22 23 24 Total
Use mathematical tables to evaluate.
3 4.68 x 0.13242
5 log 7 ( 3 marks )
Make y the subject of the formular

m = d + ky

d – ky ( 2 marks )
Expand ( 1 – 2x)6 upto the term in x3. Use your expansion to estimate 0.986. ( 3 marks )
In a geometric sequence, the third term is 20 and the fourth term is 5. Find

(a) The two possible values of the common ratio. ( 2 marks )
(b) The first term. ( 1 mark )
Let x = 8

3 - √5
(a) Write x in the form a + b √5 where a and b are integers. ( 2 marks )
(b) Use √5 = 2.236068 to write x correct to 6 decimal places. ( 2 marks )
Solve the equation ( log x)2 – log x – 2 = 0 ( 3 marks )
Three representatives are to be selected randomly from a group of 7 girls and 8 boys.

Calculate the probability of selecting two girls and one boy. ( 3 marks )
Two variables x and y are such that x varies partly as y and partly as the square root of y.

Given that x = 30, when y = 9 and x = 14 when y = 16, find x when y = 36. ( 3 marks )
.
Given that cos 2x0 = 0.8070. Find x where O0 x 3600. ( 3 marks )
Given that A = 3i + 2j + 6k and B = 2i – 2j – 5k. Find

(a) AB ( 2 marks )
(b) AB ( Leave your answer in root form ) ( 2 marks )
.
A minor sector of a circle of radius 28cm includes an angle of 1350 at the centre.

Convert 1350 into radians. Hence or otherwise find the area of the sector. ( 3 marks )
The rate of depreciation of a car is 12% per annum. Find its value 3 years ago if its

present value is 320,000 shillings. Give your answer to the nearest shillings. ( 3 marks )
In the figure below, find the length of UT. (3 marks )
X 3cm
Y
4cm
T
5cm
V U
The equation of a certain circle drawn in a Cartesian plane is given by x2 + 2x + y2 – 16y = 42.

Find the radius and the coordinates of the centre of the circle. ( 4 marks )
A rectangular card measures 3.5cm by 1.4cm. Find

Absolute error in the area of the card. ( 3 marks )
Solve for the values of x and y given that

x – y = 5

x2 – y2 = 5 ( 3 marks )
SECTION II
An engineer wants to design a roof for a CDF sponsored classroom in Kathithyamaa

secondary school. The roof takes the shape shown below.

F

E 10cm
D 8cm C
A 12cm

B

Calculate

(a) The angle between faces CDEF and ABCD. ( 3 marks )

(b) The space occupied by the roof. ( 3 marks )

(c ) The angle between plane BEC and ABCD. ( 4 marks )
is projected at time t = 0 along the x-axis so that at time t seconds its velocity is

Vm/s and its displacement from the origin O is x metres, where x = 24t + 9t2 – 2t3 + 77, t 0
Find
(a) An expression for V in terms of t. ( 1 mark )
(b) The value of t when the particle is at rest and the displacement of P from O at this time. ( 4 mks )
(c ) An expression for acceleration a m/s2 of P in terms of t. ( 1 mark )
(d) The distance traveled by P during the fifth second. ( 2 marks )
(e) Show that the particle P passes through O when t = 7 and find the acceleration
of P at this time. ( 2 marks )
In the figure below TA is a tangent to a circle. AB and CD are parallel. AC and BD meet

at X BAS = 770 and DCB = 1230.

C
1230
B
x
D
770T
A S
(a) Find ABD. ( 2 marks )

(b) Find DAT. ( 2 marks )

(c ) Show that ∆XCD is isosceles. ( 3 marks )

(d) Let TB = xcm and TD = ycm. Find TA in terms of x and y. ( 3 marks )
(a) Complete the table for y = ½ sin ( x – 200) and y = cos 2x

x0 -90 -60 -30 0 30 60 90 120 150

½ sin (x-200) -0.49 -0.17 0.49

Cos 2x -1 1 -0.50
(b) On the same set of axis on the grid provided draw the graphs of y = ½ sin ( x – 200)

and y = cos 2x. Use a scale of 2cm to represent 1 unit on y –axis and 1cm to

represent 300 on the x-axis. ( 4 marks )
(c ) Use your graph to solve the equation ½ sin (x – 20)0 – cos 2x = 0 ( 2 marks )
(d) For what range of values of x is the function cos 2x ½ sin ( x – 200) ( 1 mark )
R is a transformation defined by +900 turn about (0, 0), M is a transformation represented

by the matrix 1 0 . Triangle PQR where P (3, -1) Q (5, -1) and R ( 5, -4) is mapped onto

0 -1

P1 Q1 R1 under R.
(a) Plot PQR and P1Q1R1 on the same axes. ( 3 marks )

(b) Triangle PQR is mapped onto P2Q2R2 under the transformation M.

Plot P2Q2R2 under the same grid. ( 2 marks )

(c ) P3Q3R3 is the triangle whose vertices are P3(-3, -1) Q3(-7, -1) and R3 (-7, 5).

Draw P3Q3R3 on the same grid. ( 1 mark )

(d) Obtain a 2 x 2 matrix N, that maps PQR onto P3Q3R3, hence describe fully the

transformation represented by matrix N. ( 4 marks )
The figure below shows a rectangle and a smaller rectangle inside it.
6cm 3cm
4cm
8cm</code></pre>(a) If a point is selected at random inside the bigger rectangle. What is the probability

that it lies in the shaded region? ( 2 marks )

(b) A bag contains 4 yellow balls and 3 black balls. Two balls are picked at random

one after the other from the bag without replacement. By using a tree diagram,

find the probability that the second ball picked from the bag is black ball. ( 4 marks )

(c ) A certain form 3 in a mixed class has 18 girls. If the probability that a student

selected at random from this class is a girl is 2/5.

(i) How many students are there in the class? ( 2 marks )

(ii) What is the probability of a student selected at random for the class being a boy. ( 2 mks )
.
The table show the income tax rates

Total income k£p.a Rate in sh. per k£

1 – 2100 2

2101 – 4201 3

4201 – 6300 5

6301 – 8400 7

8401 and over 9
Norah is salaried employee. She earns a basic salary of Kshs. 11,550 per month and a house

allowance of Kshs. 2,045 monthly. She claims Kshs. 275 relief per month.
Calculate
(a) Norah’s taxable income per annum in k£. ( 3 marks )

(b) Her P.A.Y.E per month to the nearest 10ct. ( 5 marks )

(c ) Her net monthly income ( 2 marks )
The height to the nearest cm, of 40 boys is as shown in the distribution table
Height (cm) 130 – 139 140 – 149 150 – 159 160 – 169 170 – 179 180 - 189

Frequency 1 3 7 13 10 6
Using assumed mean of 164.5 or otherwise, find
(a) The mean ( 5 marks )
(b) The standard deviation of the distribution. ( 5 marks )
JULY/AUGUST 2011
FORM FOUR MID-TERM EXAMINATION 2011

121/1

MATHEMATICS

PAPER 1

MARKING SCHEME
Q WORKING MRKS COMMENTS

1 No.

(0.00246)2

142
0.002

1.14
3.3527 Std. form

(2.46 x 10-2)2

1.42 x 102
2.0 x 10-3

1.14 x 10o
3.3527 x 10o

Log

2.3909

x 2

4.7818

+2.1523

2.9341
3.3010

+0.0569

3.3579
2.9341

-3.3579

1.5762

1.5762

3
0.5254
1M
1M
1M
A1
Correct logs addition
Correct logs addition
Correct logs substractions
Correct answer

4

2 Let the digits be x and y

The number becomes xy

= 10x +y

and x + y = 10

Reserved yx = 10y + x

(10y + x) – (10x + y) = 54

10y + x – 10x – y = 54

9y – 9x = 54

y – x = 6

y – x = 6

y + x = 10

2y = 16

y = 8
x = 8-6 = 2

.

. . The number is 28
M1
1M
A1
Splitting of ones & tens and the reverse
Solving of the simultaneous eqn.
Answer

3

3 Eqn of L2 2y = x – 4

Y = ½ x – 2

Hence M1 = ½

And ½ x M2 = -1

M2 = -1 x 2 = -2

Y = mx + c

Y = -2x + c

But L1 passes through P(2,1)

Hence 1 = -2 x 2 + c

C = 5
Y = -2x + 5
1M
1M
Gradient of L2
Use of gradient of L2 to get gradient of L1

3

4 1cm represent 1km
N D

208o

4KM
120o
5.5km
B 200o
13.5km
8KM
C
Bearing 030o 4km from starting point
1M
1M
1M
A1
Bearing of starting point A
Use of scale correctly and plotting of points
Use of bearing correctly

5 X2 = 1+kt

Kt –1
x2(kt-1) = 1 + kt

ktx2 – x2 = 1 + kt2

ktx2 – kt = 1 + x2

t(kx2 – k) = 1 + x2
t= 1 + x2

kx2 - k M1
M1
A1 Squaring both sides
factorisation

3
6 Volume ¼ x (x-8) 16=256

¼ x (x-8) = 16

x2 – 8x – 64 = 0

use of formula
= -(-8) + (-8)2 – 4x1x-64

2 x 1
= 8 + 64 + 256

2

= 8 + √320 = 12.9

2

M1
M1
A1

Volume in terms of x(quadratic eqn.)
Use of inte
7 125-x x 52(x-1) = 25x+2
5-3x x 52x –2 = 52x + 4

-3x + (2x-2) = 2x + 4

-x –2 = 2x + 2
-4 = 3x
-4/3 = x
M1
M1
A1 Simplification
Addition of power

8 Curve equation x2 = 4y

Y = x2

4

dy = 2x = ½ x

dx 4

But at the point x = 6

dy = ½ x 6 = 3

dx

m = 3, p (6,9)

Assume Q (xy)

y-9 = 3

x – 6
= y = 3x -9
M1
M1
A1
Defferention
Substitution of x
Equation

9 dv = 3t – 4

dt
V= ∫3t – 4dt

= 3t2 - 4t + c

2

Since V= 3 when t = 0
Then 3(0)2 - 4(0) + C = 3

2

C = 3

S = ∫vdt = 3t2 – 4t + 3) dt

2

= t3 – 2t2 +3t + k

2

= S = 0, when t = 0

k = 0

s = t3 – 2t2 + 3t

2 M1
1M
1M
A1 Use of gradient in intergration
Substitution of t=0
Integration

4

10 ½ of 2400E = 1200E

In ksh. = 1200E x 95.65

= Ksh.114,780

Number of dollar = Kshs.114,780

76.50

= sh1500.39

M1
M1
A1

3

11 Mass = Density x volume

But Density is constant.

x y
Vol (270000 x 2.8): x2.1
= 270000 x 2.8 = 360m

2.1 M1
M1
M1
A1

4
a) G.C.D of 81xy4 and

144x3y2 is 9xy2

144x3y2 – 81xy4

9xy2(16x2 – 9y2)

9xy2(4x – 3y) (4x + 3y)

M1
M1
A1

3
a) 4x – 9 < 6 + x

x < 5

8 – 3x < x + 4

1 < x

b) 1 < x < 5

M1
M1

A1
a) The least unit of measurement is 0.1cm

.

. . error 0.1 = + 0.05

2

Limits of length = 18.5 + 0.05

Upper limit = 18.5 + 0.05 = 18.55cm

Lower limit = 18.5 – 0.05 = 18.45cm

Limits of width = 12.4 + 0.05

Upper limits = 12.4 + 0.05 = 12.4cm

Lower limits = 12.4 + 0.05 = 12.35CM
a) Working area = 18.5 x 12.4 = 229.4cm2

Max.Area = 18.55 x 12.45

= 230.9475cm2
Min. Area = 18.45 x 12.35

= 227.8575cm2

Error area = 330 .9475 – 229.4

= 1.5475cm2
Average error = 1.5415 + 1.5425

2

= 1.545cm2
. .

. % error = 1.545 x 100

22.94
= 0.67% (2 d.pls)
B1
B1
M1
A1
ALL Sides
Correct average error
Expression of error in %
Let 3x – 180 =

0 < x < 180

cos = √3

2

= 30o

0 – 180< < 180 x 3 – 18

-180 < < 360

= 30o, 330,

30 = 3x – 180

210 = 3x

70, = x

330 = 3x – 180

510 = x

3

170 = x

x = 70, 170
M1
M1
A1
3

16 a) 1cm rep. 10m R
P d
4cm
B Q C
A 1CM
A 1cm 4cm 2cm 6cm 3cm B

F e

5cm
b) Area

A = ½ x 5 x 2 = 5cm2

B = ½ X 2(2+ 1) = 3cm2

C = ½ x 6 (1+4) = 15cm2

D = ½ x 3 x 4 = 6cm

E= ½ x 5 x 15 = 75/2

= 37.5
f = ½ x 1 x 5 = 2.5

Total = 69cm2
Area = 69 x 1000000

10000
= 690m2
B1
B1
M1
A1 Correct scales
Correct drawing

4

17 SECTION B (50 MARKS)
Selling price = 88/100 of marked price

(a)

(i) 4800 = 88/100 of m.p

4800/88 x 100 = m.p

= sh.5454.54
(ii) 145/100 of buying price = 4800

buying price = 48000 x 100

145

= 3310.34
(b) 5454.54 – 3310.34 x 100
3310.34
= 0.6477 x 100
= 64.77%
C) 87.5 of 3310.34

100
= 2,896.55
18 (a) P Q3

R2
P = KQ3

R2
K = PR2 = 16 x 32 = 18

Q 3 23
R = KQ3 = 18x(4)3 = 2

P 288
B) P = KQ3

R2
Q decreases to 0.7

R increases to 1.4

P = (0.7)3 x 2 = 0.35

(1.4)2
P decreases by 65% M1
M1
M1
M1

A1
M1
M1

M1

M1
M1

10
Taxable income

Salary sh12,150

House allowance sh 2800

Total sh 14950
Sh. 14950 x 12 = 179400

8970 p.a
1-3780 = 3780 x 2 = 7580 x 2 = 7560

3781 – 7560 = 3780 x 3 = 11340

7561 – 11,340 = 1410 x 4 = 5640 +

Total sh. 24540
Tax per month

Sh. 24540 :- 12

= SH2045

450 –

sh

Deductions

Loan sh 450

Hospital sh 260+

Sacco sh 120

830

Tax 1595

2,425
net salary

sh. 14,950

2,425

sh 12,525
a) Distance moved by front of truck

= length of train + length of truck

=relative of truck

= (30 – 25) = 5km/hr.

Distance = speed x time

5km/hr x 1.5

60
= 0.125km

= 125m

length of the train

= 125 – 5 = 120m
b) Relative speed remain constant

c) Time = 4min & 48 sec.

d) 5km/hr = 50 1.39m/sec

36
distance = speed x time

= 1.39 x 288 = 400.32m

M1
M1
M1

A1
M1
A1
M1

A1
M1
A1
x x
14
22.25cm 2.25cm
a) I) Volume

Ratio x/14 = 22.5 + x

21
21x = 14(22.5 + x)

21x = 3.15 + 14x

21x – 14x = 315

7x = 315

x = 45
volume of whole cone

1/3 x 22/7 x 21 x 21 x 67.5
= 31,185cm

volume of small cone

1/3 x 22/7 x 14 x 14 x 45
= 9,240cm3
volume 31,185 – 9240

= 21945cm3
ii) mass of frustrum

mass = 21945cm3 x 3g

= 65835g cm3

1000
a(ii) mass in kg
65,835g
1000
= 65.835kg
b) 20% of 65.835kg

= 14.167kg
65,835 – 13.167kg

= 52.668kg.
52.668 c 1000g = V x 3g

52,668 X cm3 = V cm3
V= 17,556cm3

b3 = 3 √ 17,556 = 25.99cm
Length 25.99cm 10
Mass

(kg) y x x-52 d=x-52

10 fd fd2 cf

30-34

35-39

40-44

45-49

50-54

55-59

60-64

65-69

70-74

75-79 5

7

10

10

19

20

20

6

2

1

Eƒ100 32

37

42

47

52

57

62

67

72

77 20

15

10

5

0

5

10

15

20

25 -4

-3

-2

-1

0

-1

-2

-3

-4

-5

-20

-21

-20

-10

0

20

40

18

8

5

Eƒd=20 80

63

40

10

0

20

80

54

32

25

Eƒd2=

404

5

12

22

32

51

71

91

97

99

100

6 marks
a(i) mean = A + cEfd

Ef
= 52 + 5 x 20

10 = 52kg
(ii) S.D = C Efd2 - Efd

Ef Ef
= 5 x 404 – 20
100 100
= 5 4.04 – 0.2
= 5 3.84 = 5 x 1.959
= 5 x 1.96
= 9.8
M1
M1
M1
A1
23
T=T2T1
= 3 1 2 1

1 3 -1 -2
= 6 + -1 3 + -2

2 + -3 1 + -6
= 5 1 = 1 -5 -1

1 -5 5 x –5 –1 -1 5
= 1 -5 -1

24 -1 5
= 5/24 1/24
1/24 -5/24
c) 5/24 1/24 7 -7 -88
1/24 -5/24 -11 -13 168
= 24/24 -48/24 -24/24 48/24
62/24 58/24 58/24 -32/24
= 1 -2 -1 2
31/24 29/24 -11/3 -4/3
A(1, 31/24) B(-2, 29/24) C(1, -11/3) D(2, -4/3)
M1
A1
M1
M1
A1
M1
M1
A1
B2
Bo if Bracket not included.

24 a) y = (2 x 2 + x-2) 2x2 + x – 2 = 5

y = 5
At y=5 x=-2.1 or x = 1.6
c) 2x2 + x – 5 = 0

Add three both sides

Y =3.
At these points

X = 1.9 or x = 1.3 (1d.p)
d) Add (t-2) to both sides

Y = -x + 1

At the points
X = -1.8 or x = 0.8 (1d.p)

M1
A1
M1
A1
M1
A1
121/2
FORM FOUR MID TERM EXAMINATION 2011

Kenya Certificate of Secondary Education

MATHEMATICS

PAPER 2
MARKING SCHEME
No Log

4.68 0.6702

0.13224 1.1209 x 2 2.2418 + M1 All logs correct

2.9120
5 0.6990
0.8451 +
1.5441
3.3679 x 1/3 M1 1.1226 </code></pre>.1326

0.1326 A1
3
m ( d – ky) = d + ky

md – mky = d + ky M1
-mky – ky = d – md

Mky + ky = md – d

y ( mk + k) = d(m-1)
y = d (m-1) A1

mk + k
02
( 1 – 2x) 6
(1)6 (2x)0 – (1)5 (2x) + 14(2x)2 – (1)3 (2x)3 ….
1 – 2x + 4x2 – 8x3 + ………….. B1
(0.98)6 = ( 1 – 0.02)6

1 – 2 (0.02) + 4(0.02)2 – 8(0.02)3 M1
1 – 0.04 + 0.0016 – 0.000064
0.961536 A1 3</code></pre></li>(A) 3rd term = ar2

5th term = ar4
ar2= 20 ……. (i)
ar4 = 5 ……. (ii)a = 20
r2 Subsistituting
20 (r4) = 5 M1
r2
20r2 = 5
r2 = 5/20 r2 = ¼
r = ½ or – ½ A1 </code></pre>(b) a = 20

r2
a = 20 = 20
( ½ )2 ¼ 20 x 4/1 = 80 B1
3</code></pre></li>(a) 8 3 + √5
3 - √5 3 + √5 M1
24 + 8 √5
9 – 5
24 + 8 √5
4
6 + 2√5 A1
(b) 6 + 2 ( 2.236068 M1

6 + 4.472136

10.472136 A1
4
(Log x)2 – log x -2 = 0

Let log x = t
t2 – t – 2 = 0 M1
t2 + t – 2t – 2 = 0

t ( t + 1) -2 (t + 1) = 0
(t – 2 ) ( t + 1 ) = 0 M1

t = 2 or t = -1 A1
3
Sample space = 8 + 7 = 15

P(2 girls ) = 2/15

P ( 1 boy) = 1/15 B1
P ( 2 girls and 1 boy )

2/15 x 1/15 = 2/225 M1A1
3
x X y x = ky

x X √y x = m√y
x = ky + m√y

30 = 9k + 3m ……. (i) x 4

14 = 16k + 4m ……(ii) x 3 B1
120 = 36k + 12m

42 = 48k + 12m M1 for any method used

78 = -12k
K = -78
12
K = -13
2
30 = 9(-13/2) + 3m
60 = -117 + 6m

177 = 6m

m = 177

6
x = -13/2y + 177/6 √y

x = -13 (36) + 177 6

2 6
-13 (18) + 177
- 234 + 177
- 57 A1 3</code></pre></li>Cos 2x0 = 0.8070

2x0 = cos-1 (0.8070)

= 36.1960 B1

2x = 36.1960

≂ 36.200, 323.800

396.2, 683.8 B1
x = 18.10, 161.90. 198.10
341.90 A1 3</code></pre></li>(a) AB = B – A

(I – 2j - 5k) – ( 3i + 2j + 6k) B1

I – 2j – 5k – 3i – 2j – 6k

-2i - 4j – 11k B1
(b) AB = √i2 + j2 + k2
= √ (-2)2 + (-4)2 + (-11)2 M1√ 4 + 16 + 121
√ 20 + 121
√141 A1
4</code></pre></li>1C = 57.290

? = 1350
135
57.29 = 2.35C B1
Area = Ө r2

360

4 1

135 x 22 x 28 x 28 M1

360 7

90 1
27 3

135 x 22 x 28

90 18 2

11

3 x 22 x 28

2 1 33 x 28 = 924cm2 A1
3
A = P ( 1 + r/100)n
A = 320,000 [ 1 + 12/100)3 M1M1
= 320,000 [1.12)3= 449,577 A1
3</code></pre></li>Let UT = a
TY x TX = TU x TV

4 x 7 = a x ( 5 + a) M1

28 = 5a + a2

a2 + 5a – 28 = 0
a = b √b2 – 4ac

2a
a = -5 √52 -4 (1) (-28)

2(1)
a = -5t √25 + 112
a = -5 √137
a = -5 11.7
a = 6.7cm M1
2
a = 3.35cm A1 3</code></pre></li>x2 + 2x + (2/2)2 + y2 – 16y + (-16/2)2 = 16 + 1 + 64 M1
(x + 1)2 + (y-8)2 = 16 + 1 + 64 = 81 A1 for the equation
Radius √81 units
=9 units B1
Cordinates of centre
= ( -1, 8 ) B1 4</code></pre></li>3.45 3.5 3.55

1.35 1.4 1.45
Max area = 3.55 x 1.45

= 5.1475

M1

Min area = 3.45 x 1.35

= 4.6575
A.E = 5.1475 – 4.6575 M1
2 = 0.245 A1
3</code></pre></li>x = 5 + y M1

( 5 + y)2 – y2 = 5 M1

25 + 10y + y2 – y2 = 5

25 + 10y = 5

+ 10y = -20

y = -2

x = 5 – 2

x = 3, y = -2 A1 for both x and y
SECTION II ( 50 MARKS )
(a)

B F

10m
D 8m Ө
C
12mA B</code></pre>The required angle is BCF = Ө

C2= f2 + b2 – 2 fb cos C
Cos C = f2 + b2 – C2

2fb
= 102 + 122 – 82
2 x 10 x 12 M1
180 = 0.75
240 M1
Ө = cos -1 ( 0.75) = 41.410 A1
(b) Cross sectional area of the roof

½ x 10 x 12 sin 41.410

½ x 10 x 12 x 0.6614 M1
Space occupied by the roof ( volume)
½ x 10 x 12 x 0.6614 x 24 M1
= 952.4m3 A1
(c )

E F
D C
x
H G A B</code></pre></li>
The required angle is HGE

EH = FG = 10 sin Ө

Tan x = EH

AB
Tan x = 10 sin 41.41
24
= 0.2756
x = tan -1 ( 0.2756) M1
15.410 A1
10
x = 24t + 9t2 – 2t3 + 77
(a) √ = dx = 24 + 18t – 6t2 B1

dt
(b) When √ = 0

24 + 18t – 6t2 = 0 B1

6t2 – 18t – 24 = 0

( t + 1) ( t – 4) = 0 M1
t = 4 secs A1
When t = 4

x = ( 24 x 4) + ( 9 x 16) – (2 x 43) + 77

= 96 + 144 – 128 + 77

= 189m B1
(c ) a = d√ = 18 – 12t B1

dt
(d) When t = 5

x = ( 24 x 5) + ( 9 x 25) – (2 x 125) + 77

= 120 + 225 – 250 + 77

= 172m B1
Distance traveled = ( 189 – 172) m
= 17m B1
(e) When t = 7, x = 0

x = ( 24 x 7) + ( 9 x 49) – (2 x 73) + 77

= 168 + 441 – 686 + 77

= 0m B1
a = 18 – 12t = 18 – ( 12 x 7) = -66m/s2 B1 10</code></pre></li>(a) BAD = ( 180 – 123) = 570 ( opposite B1B1

angles of a cyclic quadrateral
DAT = 180 – ( SAB + BAD)
= 1800 – ( 77 + 57)0 B1
1800 – 1340 = 460 B1
Angles on a straight line
(c ) ∆ X CD

XDC = DCX which are base angles of the ∆ B1

therefore line DX = XC hence B1

triangle XCD is isosceles B1
(d) If TB = xcm and TD = ycm

Then TA2 = TD x TB M1
TA2 = y X x M1
TA = √ yx A1 10</code></pre></li>x -90 -60 -30 0 30 60 90 120 150
½ sin (x – 20) -0.47 -0.49 -0.38 -0.17 0.09 032 0.47 0.49 0.38

Cos 2x -1 -0.5 0 1.0 .05 -0.50 -1 -0.5 0.5
.
(a) R 1 0

0 -1
P Q R P1 Q1 R1 M1
1 0 3 5 5 = 3 5 5
0 -1 -1 -1 -4 1 1 4
(d) a b 3 5 5 = -3 -7 -7

C d -1 -1 -4 -1 -1 5 M1
3a – b = -3
- 5a – b = -7
-2a = 4 M1
a = 4
-2
a = -2
3a – b = -3
( 3x – 2) –b = -3 M1

-6 = b = -3
-6 + 3 = b
b = = -3
3c – d = -1
5c – d = -1
-2c = 0
c = 0
3c – d = -1
-d = -1
d = 1 A1
(a) 48 – 12

48 M1

36 or ¾

48 A1

(b)

Y

3/6

B

Y 3/6 B1
4/7
4/6 Y3/7 B
2/6 B B14/7 x 3/6 + 3/7 x 2/6
12/42 + 6/42
18/42 or 3/7 A1
(c) (i) Let the no. of students be x M1

18 = 2

x 5 = 45 A1

x = 18 x 5 = 45

2
(ii) 45 – 18
45 M1 27/45 = 3/5 A1
10</code></pre></li>(a) Taxable income per month M1

= employee’s salary + allowance

11,550 + 2045

= Ksh. 13,595
Annual taxable income in kt B1
13,595 x 12
20
= kt 8157
A1
(b) 1st 2100 = 2100 x 2 = Ksh. 4200 M1

Next 2100 = 2100 x 3 = Kshs 6,300 M1

Next 2100 = 2100 x 5 = Kshs 10,500

Rem 1857 = 1857 x 7 = Kshs. 12999

Income tax = Kshs. 33,999

Per month = Kshs. 33999

12 M1
= Kshs. 2833.25 B1
Less relief = Kshs. 275.00
P.A.Y.E = Kshs. 2,558.25 A1
(c ) Net monthly income = Taxable
Income – tax
= ( Salary + house allowance – tax
Kshs. 13595 – 2833.25
Kshs. 10,761.75 A1
10
Class Mid-points t –x – 164.5 f ft ft2
B2
130 – 139 135.5 -3 1 -3 9

140 – 149 144.5 -2 3 -6 12

150 – 159 154.5 -1 7 -7 7

160 – 169 164.5 0 13 0 0

170 – 179 174.5 1 10 10 10

180 - 189 184.5 2 6 12 24

40 f 6 ft2 = 62
t = ft = 6 = 0.15

f 40 M1
x = t + C + A
= 0.15 x 10 + 164.5 M1
= 166.0cm A1
(b) Variance of t = ft2 – (f)2

f
= 62 - ( 0.15)2
40 = 1.55 – 0.0225 M1
= 1.5275Varies of x = variance of t x C A4
1.5275 x 100

= 15.28

Standard deviation = √variance

= √ 15.28 M1

= 3.909 A1

10