Instructions to candidates
 Write your name and index number in the spaces provided above.
 Sign and write the date of examination in the spaces provided above.
 This paper consists of TWO sections: Section I and Section II.
 Answer ALL the questions in Section I and only five questions from Section II.
 Show all the steps in your calculations, giving your answers at each stage in the spaces provided below each question.
 Marks may be given for correct working even if the answer is wrong.
 Nonprogrammable silent electronic calculators and KNEC Mathematical tables may be used, except where stated otherwise.
 Candidates should check the question paper to ascertain that all the pages are printed as indicated and that no questions are missing.
 Candidates should answer the questions in English.
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 









QUESTIONS
 A rectangular block has a square base whose side is exactly 8cm. Its height measured to the nearest millimetre is 3.1cm. Find in cubic centimetres the greatest possible error in calculating its volume (3mks)
 Simplify (3mks)
 Find the area bounded by the curve y= x3 + 5, the x axis and lines x=1 and x = 3 (3mks)
 A bag contains 2 green balls, 3 red balls and one blue ball. Another bag contains 4 green balls, 5 red balls and 3 blue balls. A ball is chosen at random from a bag. Find the probability that the chosen ball is blue. (3mks)
 P varies partly as the square of V and partly as the cube of V. When V= 2, P = 20 and when V = 3, P = 135. Find the relationship between P and V. (3mks)
 The second term of a G.P is 6, and the fifth term is 48, find the common ratio and the 3rd term of the G.P. (3mks)
 Nancy pays for a car on hire purchase in 15 monthly instalments. The cash price of the car is Ksh. 300,000 and the interest rate is 15% p.a. A deposit of Ksh 75,000 is made. Calculate her monthly repayments. (4mks)
 Solve for θ in the equation 2 sin (2θ + 10) = 1 for 0º ≤ θ ≤ 360º. (3mks)
 The diagram below shows a garden XYZ
 Draw the locus of points equidistant from sides XY and XZ. (1mk)
 Draw the locus of points equidistant from points X and Z. (1mk)
 A bead is lost in the garden within a region which is nearer to point X than to point Z and closer to side XZ than to side XY. Show by shading the region where the bead can be located. (1mk)
 Expand the given binomial up to the term with x4: (1+3x)^{6}. Use your expansion to evaluate (1.3)^{6} correct to 4 decimal places. (4mks)
(1 + 3x)^{6}
= 1(1)^{6} + 6(1)^{5}(3x)^{1} + 15(1)^{4}(3x)^{2} + 20(1)^{3}(3x)^{3} + 15(1)^{2}(3x)^{4} + ........................
= 1 + 18x + 135x2 + 540x2 + 1215x4
(1.3)^{6} = (1 + 0.3)^{6} = (1 + 3x)^{6}
3x = 0.3 : x = 0.1
=> (1 + 0.3)^{6} = 1 + 18(0.1) + 135(0.1)^{2} + 540(0.1)^{3} + 1215(0.1)^{4}
= 4.8115  Line PQ is the diameter of a circle. Find the equation of the circle in the form ax^{2}+bx+cy^{2}+dye =0, given the coordinates P(0,2) and Q(6,2). (3mks)
 Determine the amplitude, the period and phase angle for the graph y=3sin(1/4x90º). (3mks)
 Make x the subject of the formular. (3mks)
 Under a transformation represented by a matrix , a triangle of area 10cm^{2} is mapped onto atriangle of area 110cm^{2}. Find x. (3mks)
 Evaluate without using Mathematical tables or a calculator. (3mks)
 A body starts from rest and after t seconds its velocity in ms1 was recorded as shown below;
T in (sec)
0
1
2
3
4
5
6
Velocity
0
0.29
5.4
7.7
9.7
11.4
12.7

 Complete the table below to 2 decimal places. (2mks)
X
0^{0}
30^{0}
60^{0}
90^{0}
120^{0}
150^{0}
180^{0}
210^{0}
240^{0}
270^{0}
300^{0}
330^{0}
360^{0}
– Cos x
– 1
– 0.5
0.5
0.87
0.87
0.5
0.87
Sin ( x – 30^{0})
0.0
0.5
0.87
0.5
– 0.5
– 0.87
– 0.5
 Draw the graphs of y = sin (x – 30º) and y = – Cos x on the same axes, for 0º ≤ x ≤ 360º. (5mks)
 Use your graph to solve the equation sin (x  300) + Cos x = 0. (3mks)
 Complete the table below to 2 decimal places. (2mks)
 Figure below is a pyramid on a rectangular base. PQ=16cm, QR = 12cm and VP = 13cm.
Find The length of QS. (2mks)
 The height of the pyramid to 1 decimal place. (2mks)
 The angle between VQ and the base. (2mks)
 The angle between plane VQR and the base. (2mks)
 The angle between planes VQR and VPS (2mks)
 The following table shows the distribution of marks obtained by 50 students in a test.
Marks
45 – 49
50 – 54
55 – 59
60 – 64
65 – 69
70 – 74
75 – 79
No. of students
3
9
13
15
5
4
1
 The mean (5mks)
 The variance (3mks)
 The standard deviation (2mks)

 Hellen’s earnings are as follows:
Basic salary sh. 38000 per month
House allowance sh. 14000 per months
Travelling allowance sh.8500 per month and
Medical allowance Ksh.3300 per month.
She is given a personal relief of Ksh. 12672 per annum
The table for payable tax is shown below
Income in K£ p.a Payable tax rate in Kshs per K£
06000 2
600112000 3
1200118000 4
18001 24000 5
2400130000 6
3000136000 7
3600142000 8
4200148000 9
Over 48000 10
Calculate Hellen’s taxable income in K£ p.a (2mks)
 Her P.A.Y.E (5mks)
 Hellen is deducted the following items per month
NHIF Ksh.320
Cooperative shares Ksh.2000
Loan repayment Ksh5000
Determine her net salary per month (3mks)
 Hellen’s earnings are as follows:
 In the figure below, O is the centre of the circle. A, B, C and D are points on the circumference of the circle. A, O, X and C are points on a straight line. DE is a tangent to the circle at D. Angle BOC= 48º and angle CAD = 36º.
 Giving reasons or otherwise, find the value of the following angles:
 Angle CBA (1mk)
 Angle BDE (2mks)
 Angle CED (3mks)
 It is also given that AX = 12 cm, XC = 4 cm, DB = 14 cm and DE = 20 cm. Calculate:
 DX (2mks)
 AE (2mks)
 Giving reasons or otherwise, find the value of the following angles:
 The position of two towns A and B on the earths surface are (36ºN, 49ºE) and (36ºN, 131ºW) respectively.
 Find the difference in longitude between town A and town B. (2mks)
 Given that the radius of the earth is 6370, calculate the distance between town A and B.
 In nm. (2mks)
 In kilometers. (2mks)
 Another town C is 840km East of town B on the same latitude as town A and B. Find the longitude of town C. (4mks)

 Using a ruler and compasses only, construct triangle ABC such that AB=4cm, BC=5cm and ∠ABC = 120º. Measure AC. (4mks)
 On the diagram, construct a circle which passes through the vertices of the triangle ABC. Measure the radius of the table. (4mks)
 Construct a perpendicular from the centre of the circle to the line BC. Measure the length of the perpendicular. (2mks)
 A metal sphere has a radius 5cm and a density of 2.4g/cm³.
 Calculate the mass of the ball in kg (4mks)
 The ball is dropped into a cylindrical container which is partially filled with water. The ball is fully submerged. If the cylinder has a base radius of 8cm, calculate the change in the water level. (3mks)
 The sphere is melted down to form a metal cylinder of same radius. Calculate the height of the cylinder formed. (3mks)
MARKING SCHEME
 A rectangular block has a square base whose side is exactly 8cm. Its height measured to the nearest millimetre is 3.1cm. Find in cubic centimetres the greatest possible error in calculating its volume (3mks)
Maximum volume = 64 x 3.15 = 201.6cm^{3}
Actual volume = 64 x 3.1 = 198.4cm^{3}
Error = 201.6 – 198.4
= 3.2cm^{3}  Simplify (3mks)
 Find the area bounded by the curve y= x3 + 5, the x axis and lines x=1 and x = 3 (3mks)
35.25 – 5.25
= 30 sq. units  A bag contains 2 green balls, 3 red balls and one blue ball. Another bag contains 4 green balls, 5 red balls and 3 blue balls. A ball is chosen at random from a bag. Find the probability that the chosen ball is blue. (3mks)
 P varies partly as the square of V and partly as the cube of V. When V= 2, P = 20 and when V = 3, P = 135. Find the relationship between P and V. (3mks)
P = a V^{2} + b V^{3} : 4a + 8b = ‑20
9a  27b = 135
36a + 72b = 180
 36a – 108b = 540
180b = 720
b =  4
20 = 4a  32
4a = 12
a = 3
P = 3V^{2} + 4V^{3}  The second term of a G.P is 6, and the fifth term is 48, find the common ratio and the 3rd term of the G.P. (3mks)
ar = 6 a = 6/r
ar^{4 } = 48
6/r x r^{4} = 48
6 r^{3} = 48
r^{3} = 8
r = 2.
2a = 6
a = 3
3^{rd} term = ar^{2}
= 3 x 2^{2} = 12  Nancy pays for a car on hire purchase in 15 monthly instalments. The cash price of the car is Ksh. 300,000 and the interest rate is 15% p.a. A deposit of Ksh 75,000 is made. Calculate her monthly repayments. (4mks)
P = 300,000 – 75,000
= 225,000
A=225,000 x 1.15^{1.25}Monthly Instalment = 225,000 x 1.15 ^{1.25}
15
= 225,000 x 1.191 = 267975
15 15
= Kshs17,865  Solve for θ in the equation 2 sin (2θ + 10) = 1 for 0º ≤ θ ≤ 360º. (3mks)
2sin(2θ + 10) = 1
sin(2θ + 10) = 0.5
2θ+10 = 210,330,570, 690
2θ=200, 320,560,680
θ= 100^{o},160^{o},280^{o},340^{o}  The diagram below shows a garden XYZ
 Draw the locus of points equidistant from sides XY and XZ. (1mk)
 Draw the locus of points equidistant from points X and Z. (1mk)
 A bead is lost in the garden within a region which is nearer to point X than to point Z and closer to side XZ than to side XY. Show by shading the region where the bead can be located. (1mk)
 Draw the locus of points equidistant from sides XY and XZ. (1mk)
 Expand the given binomial up to the term with x4: (1+3x)^{6}. Use your expansion to evaluate (1.3)^{6} correct to 4 decimal places. (4mks)
Ans = 4.8115  Line PQ is the diameter of a circle. Find the equation of the circle in the form ax^{2}+bx+cy^{2}+dye =0, given the coordinates P(0,2) and Q(6,2). (3mks)
 Determine the amplitude, the period and phase angle for the graph y=3sin(1/4x90º). (3mks)
amplitude 3 units
period = 360/1/4 = 1440º
phase angle = 90º  Make x the subject of the formular. (3mks)
p^{2} = bx^{2}  ax
x^{2}
p^{2} = b
a
x
a = b  p^{2}
x
= x = a
b  p^{2}  Under a transformation represented by a matrix , a triangle of area 10cm^{2} is mapped onto atriangle of area 110cm^{2}. Find x. (3mks)
A.S.F = 110/10 = 11
=> 5x^{2}  (6x) = 11
5x^{2} + 6x  11 = 0
(5x + 11) (x  1) = 0
x = 1 or 11/5  Evaluate without using Mathematical tables or a calculator. (3mks)
log5^{2}  log16^{1/2} + log 40^{2}
log 25 x 1600
4
log 10 000
log 10^{4}
4 log 10
= 4  A body starts from rest and after t seconds its velocity in ms1 was recorded as shown below;
T in (sec)
0
1
2
3
4
5
6
Velocity
0
0.29
5.4
7.7
9.7
11.4
12.7
A = 1/2h (sum of ends + 2(sum of middle)
= 1/2 x 1 (0.1 + (0.29 + 12.7) + 2(5.4 + 7.7 + 9.7 + 11.4)
= 1/2 (12.99 + 2(34.21)
= 40.695m 
 Complete the table below to 2 decimal places. (2mks)
X
0^{0}
30^{0}
60^{0}
90^{0}
120^{0}
150^{0}
180^{0}
210^{0}
240^{0}
270^{0}
300^{0}
330^{0}
360^{0}
– Cos x
– 1
0.87 – 0.5
0 0.5
0.87
1 0.87
0.5 0 0.5
0.87
1 Sin ( x – 30^{0})
0 0.0
0.5
0.87 1 0.87
0.5
0 – 0.5
0.87 1 – 0.87
– 0.5
 Draw the graphs of y = sin (x – 30º) and y = – Cos x on the same axes, for 0º ≤ x ≤ 360º. (5mks)
 Use your graph to solve the equation sin (x  300) + Cos x = 0. (3mks)
 Complete the table below to 2 decimal places. (2mks)
 Figure below is a pyramid on a rectangular base. PQ=16cm, QR = 12cm and VP = 13cm.
Length of line Qs = √162 + 122
= √256 + 144 = √400
= 20 cm
Height of the pyramid
= √13^{2}  10^{2} = √169 – 100 = √69
Find
= 8.3cm ( 1d.p)
Cos q = 10/13 = 0.7692
q = 39.72° The length of QS. (2mks)
 The height of the pyramid to 1 decimal place. (2mks)
 The angle between VQ and the base. (2mks)
 The angle between plane VQR and the base. (2mks)
h^{2} = 13^{2} 6^{2} = 169  36
h = √133 = 11.53 cm( 2d.p)
Cos ∝ = 8 = 0.6938
11.53
∝ = 46.06°  The angle between planes VQR and VPS (2mks)
θ = 2 x sin^{1} 8/11.53
= 2 x 43.93
= 87.87^{0}
 The following table shows the distribution of marks obtained by 50 students in a test.
Marks
45 – 49
50 – 54
55 – 59
60 – 64
65 – 69
70 – 74
75 – 79
No. of students
3
9
13
15
5
4
1
 The mean (5mks)
Let A = 62Marks
f
x
d = x A
fd
d^{2}
fd^{2}
45  49
50 54
55  59
60 – 64
65 – 69
70 – 74
75  79
3
9
13
15
5
4
1
47
52
57
62
67
72
77
15
10
5
0
5
10
15
45
90
65
0
25
40
15
225
100
25
0
25
100
225
675
900
325
0
125
400
225
f = 50
∑fd = ^{}120
∑fd^{2} = 2650
∑f
= 62 + 120
50
= 62  2.4
= 59.6  The variance (3mks)
 The standard deviation (2mks)
 The mean (5mks)

 Hellen’s earnings are as follows:
Basic salary sh. 38000 per month
House allowance sh. 14000 per months
Travelling allowance sh.8500 per month and
Medical allowance Ksh.3300 per month.
She is given a personal relief of Ksh. 12672 per annum
The table for payable tax is shown below
Income in K£ p.a Payable tax rate in Kshs per K£
06000 2
600112000 3
1200118000 4
18001 24000 5
2400130000 6
3000136000 7
3600142000 8
4200148000 9
Over 48000 10
Calculate Hellen’s taxable income in K£ p.a (2mks)
(38000 + 14000 + 8500 + 3300) x 12 = 38,280 ksh p.a
20  Her P.A.Y.E (5mks)
6000 x 2 =12000
6000 x 3 = 18000
6000 x 4 = 24000
6000 x 5 = 30000
6000 x 6 = 36000
6000 x 7 = 42000
2280 x 8 = 18240
Gross tax p.a = 180,240
less relief = 12672
P.A.Y.E = 167,568 p.a.
= 13,964 p.m  Hellen is deducted the following items per month
NHIF Ksh.320
Cooperative shares Ksh.2000
Loan repayment Ksh5000
Determine her net salary per month (3mks)
total ded = (13969 + 320 + 2000 + 1000)
= ksh 21 284
Net salary = 63800  21284
= ksh 42, 516
 Hellen’s taxable income in K£ p.a (2mks)
 Hellen’s earnings are as follows:
 In the figure below, O is the centre of the circle. A, B, C and D are points on the circumference of the circle. A, O, X and C are points on a straight line. DE is a tangent to the circle at D. Angle BOC= 48º and angle CAD = 36º.
 Giving reasons or otherwise, find the value of the following angles:
 Angle CBA (1mk)
90º > angle subtended by the diameter of the circumference  Angle BDE (2mks)
36º  alternate segment theorem  Angle CED (3mks)
180  (36 + 36 + 61)
180  133º
= 47º angles in a traingle add upto 180º
 Angle CBA (1mk)
 It is also given that AX = 12 cm, XC = 4 cm, DB = 14 cm and DE = 20 cm. Calculate:
 DX (2mks)
DX = X
X6 = (14  X)
AX.XC = BX.XB
12 x 4 = x(N  x)
48 = 14x  x^{2}
x^{2}  14x + 48 = 0
(x  8)(x  6) = 0
x = 8 or x = 6 cm  AE (2mks)
AE.CE = DE^{2}
20^{2} = 16(16 + a)
400 = 256 + 169
CE = a
144 = 169
a = 9 cm
 DX (2mks)
 Giving reasons or otherwise, find the value of the following angles:
 The position of two towns A and B on the earths surface are (36ºN, 49ºE) and (36ºN, 131ºW) respectively.
 Find the difference in longitude between town A and town B. (2mks)
131 + 49
= 180  Given that the radius of the earth is 6370, calculate the distance between town A and B.
 In nm. (2mks)
60 x ∝ Cos θ
= 60 x 180 Cos 36º
= 60 x 180 x 0.8090
= 8737.38 pm  In kilometers. (2mks)
∝ x 2π R Cos θ
360
180 x 2 x 3.142 x 6370 Cos 36º
360
= 16192.103 km
 In nm. (2mks)
 Another town C is 840km East of town B on the same latitude as town A and B. Find the longitude of town C. (4mks)
∝ x 2 x 3.142 x 6370 Cos 36º
360
= 840 km
∝ = 840 x 360º
628 x 6370 Cos 36º
= 9.34ºLongitude of town c
13 1º – 9.34º
= 121.66º
 Find the difference in longitude between town A and town B. (2mks)

 Using a ruler and compasses only, construct triangle ABC such that AB=4cm, BC=5cm and ∠ABC = 120º. Measure AC. (4mks)
AC = 7.8 ± 0.1  On the diagram, construct a circle which passes through the vertices of the triangle ABC. Measure the radius of the table. (4mks)
Radius = 4.5 ± 0.1 cm  Construct a perpendicular from the centre of the circle to the line BC. Measure the length of the perpendicular. (2mks)
length of perpendicular = 4.5 ± 0.1
 Using a ruler and compasses only, construct triangle ABC such that AB=4cm, BC=5cm and ∠ABC = 120º. Measure AC. (4mks)
 A metal sphere has a radius 5cm and a density of 2.4g/cm³.
 Calculate the mass of the ball in kg (4mks)
f = m/v
v = 4/3πr^{3}
= 4/3 x 22/7 x 5^{3}
= 523.5987756cm^{3}
= fv
= 523.5987756 x 2.4
= 1256.637061g
= 1.256637061 kg  The ball is dropped into a cylindrical container which is partially filled with water. The ball is fully submerged. If the cylinder has a base radius of 8cm, calculate the change in the water level. (3mks)
v = πr^{2}h = 523.5987756 cm^{3}
h = 523.5987756
π x 8 x 8
h = 2.604166667  The sphere is melted down to form a metal cylinder of same radius. Calculate the height of the cylinder formed. (3mks)
v = πr^{2}h = 523.5987756 cm^{3}
h = 523.5987756
π x 25
h = 6.667 cm
 Calculate the mass of the ball in kg (4mks)
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