INSTRUCTIONS TO CANDIDATES.
- This paper consistsof twosection I and II.
- Answer ALL questions in sectionIand only five questions from section II.
- Marks may be given for correct working even if the answer is wrong
- Non-programmable electronic calculators may be used.
Questions
SECTION I (50 MARKS)
Answer ALL Questions from this section in the spaces provided
- Use logarithms to evaluate (4mks)
5.246 x log 0.2349
0.06364½ - Make 4 the subject of the formula. (3mks)
t = 2m/n √(L-A/3k) - Express the recurring decimal below as a fraction: leaving your answer in the form of a/b where a and b are integers. (2mks)
- Translation Q is represented by the column vector and another translation R by the column vector . A point S is mapped onto a point T by Q and a point T is mapped into a point U by R. If point U is (8, - 4), determine the co-ordinates of point S. (3mks)
- Find the values of a and b. Given: (4mks)
3 + 3√5 = a + b√5
3 + √5 3 - √5 - The figure below shows a circle centre O. Chord AB subtends 300 at the centre. If the area of the minor segment is 5.25cm2, find the radius of the circle. (3mks)
- The position vectors for points A and B are 2i + 5j + 3k and 4i – 7j + 3k respectively. Express vectors AB in terms of unit vectors i, j and k. Hence find the length of AB leaving your answer in simplified surd form 3marks
- Given that the matrix has no inverse, find x. (2mks)
- In the figure below ABC is a tangent to the circle at point B.Given that BE =6.9cm, FE=7.8cm, GE=4.1cm, DC=11.2cm and ED = xcm.Determine the length BC; give your answer in four significant figures. (4mks)
- Given that matrix , Find matrix B such that: A + B (3mks)
- A quantity P varies partly as t and partly as the square of t.When t = 20, p = 45 , and when t = 24 , p = 60.
- Express p in terms of t. (2mks)
- Find p when t = 32. (2mks)
- Achang’a deposited sh. 20 000 in a saving account. Find the interest after two years. If the interest was paid at 16% per annum compound semi-annually. (3mks)
- Using a ruler and a pair of compasses only construct triangle ABC such that BC=6cm, <ABC=75o and BCA=45o. Drop a perpendicular to BC from A to meet BC at O hence find the area of triangle ABC (4 mks)
- Five men working 8 hours daily complete a piece of work in 3 days. How long will it take 12men working 5hours a day to complete the same work. (2mks)
- Find the integral values of x which satisfy 6 < 2x + 1 and 5x – 29 < - 4 . (3mks)
- In a fund-raising committee of 45 people, the ratio of men to women is 7: 2.Find the number of women required to join the existing committee so that the ratio of men to women changes to 5:4. (3mks)
SECTION II (50 MARKS )
Attempt any five questions from this section
- James’ earning are as follows:- Basic salary 38,000 p.m, House allowance Sh. 14, 000p.m Travelling allowance Sh. 8,500p.m. Medical allowance sh. 3,300
The table for the taxable income is as shown below
Income tax in k£ p.a Tax in Sh. Per pound 1 – 6000
6001 – 12000
12001 – 18000
1001 – 24 000
24001 - 30 000
30001 - 36000
36001 – 42 000
42001 – 48 000
Over 48 0002
3
4
5
6
7
8
9
10- Calculate Jame’s taxable income p.a (2mks)
- Calculate Jame’s P.A.Y.E if he is entitled to a tax relief of Sh. 18 000 p.a(4mks)
- James is also deducted the following per month:-
NHIF Sh. 320
Pension scheme Sh. 1000
Co-operative shares Sh. 2000
Loan repayment Sh. 5000
Interest on loan Sh. 500- Calculate James’ total deduction per month in Ksh. (2mks)
- Calculate his net salary per month. (2mks)
- In the figure below, ABCD is a square.Points P,Q,R and S are the midpoints of AB,BC,CD and DA respectively.
- Describe fully:
- A reflection that maps triangle QCE onto triangle SDE. (1mk)
- An enlargement that maps triangle QCE onto triangle SAE. (2 mks)
- A rotation that maps triangle QCE onto triangle PEB. (3 mks)
- The triangle ERC is reflected on the line BD. The image of ERC under the reflection is rotated clockwise through an angle of 900 about P.
Determine the images of R and C:- Under the reflection ( 2mks)
- After the two successive transformations ( 2mks)
- Describe fully:
- The figure below shows a pulley system where a conveyor belt is tied round the two wheels. The radius of the large wheel is 180cm and the distance between the centres of the wheel is 300cm and XOY = 1400
Determine- Length XV (2mks)
- Length VBW (3mks)
- Length XAY(2mks)
- The total length of the conveyor belt
- From Jane’s home, her school is 150m on a bearing of N60oE. The shopping centre is on a bearing of 150o from the school and 110o from Jane’s home. The church she attends is 180 m from her home on a bearing of 320o.. Using a scale of 1cm to represent 30m
- Show the relative positions of the four points (4mks)
- Use the drawing to determine
- The distance of the school from the shopping centre (1mks)
- The distance of her home from the shopping centre (1mks)
- The distance and bearing of the church from the shopping centre (2mks)
- The distance and bearing of the school from the church (2mks)
- P varies directly as V and inversely as the square root of R. Given that P = 180, R = 25 when V = 9.
- Find P when V = 6 and R = 26. (3 marks)
- Find the value of V when P = 360 and R = 0.64. (3 mks)
- If V increased by 16% and R decreases by 25%, find the percentage change in P. (4mks)
-
- Complete the table below. (2mks)
- Draw the graph of y = sin (x+30) and y=cos(x-15) for -30≤X≤2700on the same grid. Take 1cm to represent 30o on x-axis and 1cm to represent 0.2units on y-axis. (5 mks)
- Using your graph drawn (b) above
- Find the values of x for which cos (x-15) –sin (x+30) = 0 (2mks)
- Estimate the angle corresponding to cos(x-15) = 0.6. (1 mk)
- A tailor bought a number of suits at a cost of sh 57600 from Jimco clothing. Had he bought the same suits from Gikondi clothing, it would have cost him sh 480 less per suit. This would have enabled him to buy 4 extra suits for the same amount of money.
If x represents the number of suits bought;- Write an expression of the cost per suit bought from
- Jimco clothing (1mk)
- Gikondi clothing (1mk)
- Form an equation in x and determine the number of suits the tailor bought (4mks)
- The tailor later sold each suit for sh 720 more than he had paid for it. Determine the percentage profit. (4mks)
- Write an expression of the cost per suit bought from
-
- The first term of an arithmetic progression (AP) is 2. The sum of the first 8 terms of the AP is 256.
- Find the common difference of the AP (2mks)
- Given that the sum of the first n terms of the AP 416, find n (2mks)
- The 3rd, 5th, and 8th terms of another AP forms the first three terms of a geometric progression (GP).If the common difference of the AP is 3.
Find- The first term of GP (4mks)
- The sum of the first 9 terms of the GP to 4 s.f (2mks)
- The first term of an arithmetic progression (AP) is 2. The sum of the first 8 terms of the AP is 256.
Marking Scheme
- 5.246 x log 0.2349
0.06364 ½
= 5.246 x 1.3708
0.6364 ½
=-(5.246 x 0.6292)
0.063664 ½
No Log
5.246 0.7198
0.6292 1.7988
0.5186 0. 5186 -
0.06364 x 2.8038 1.4019
1.1167
10-1 x 1.309
= -0.1309 - t² = 4m²/n² (4m²/3k)
t² = 4m²l - 4m²A
3n²k
3n²t²k = 4m²l - 4m²A
4m²A = 4m²l - 3n²t²k
A= 4m²t - 3n²t²k
4m² - let x = 4.37272
10x = 43.72
100x = 437.2
1000x = 4372.72
1000x - 10x = 4372.72 - 43.72
990x = 4329
x 4329 = 481
990 110 -
- 3 (3 - √5) + 3√5(3 + √5)
9 -5
= 9-3√5 + 9√5 + 15
4
24 + 6√5
4
= 6 + 3/2√5
∴ a = 6 and b = 2/3 - 30/360 x 22/7 x r2 - 1/2r2 Sin 30° = 5.25
1/42r2 - 1/4r2 = 5.25
1/84r2 = 441
r = ± 21cm
r = 21cm
AB =
AB = 2i – 12j + Ok
=|AB| = √(22 + (-12)2 + 02)
= √4 + 144
= 148
= 12.17- 4(5 - x)- 6x = 0
20 - 4x - 6x = 0
10x = 20
x = 2 - FE x ED = GE x EB
7.8x = 4.1 x 6.9
7.8x = 28.29
x = 3.627
= 253.4224cm
BC2 = CD x CF
y2 = 11.2 x 22.627
= 253.4224
y = √253.4224 = 15.92cm -
-
- p ± x t and p x t2
∴ p = mt + nt2
45 = 20m + 400n
60 = 24m + 400n
60 = 24m + 576n
9 = 4m + 80n .....(i) x 2
5 = 2m + 48n ....(ii) x 4
18 = 8m + 160n
20= 8m + 192n
-2 = -32n
n = 2/32 = 1/16
and m = 5 - (48 x 1/16)1/2 = 7/2
hence p = 7/2t + 1/16t2 - p = (7/2 x 32) + (1/16 x 32 x 32) = 112 + 64 = 176
- p ± x t and p x t2
- A = P(1 + r/100)n
A=20000(1 + 8/100)4
= 27210
I = 27210 - 20000
= Sh 7, 210 -
- Men hours days
5 8 3
12 5 x
x = 5 x 8 x 3
12 x 5
= 2 days - Area Scale Factor = Determinant of the matrix
60/10 = x (x + 3)(x + 3) - 12
6= x2 + 3x - 12
(x2 - 3x) + (6x - 18) = 0
(x + 6)(x - 3) = 0
X= -6 or 3
X=3 - men = 7/9 x 45 = 35
Women = 2/9 x 45 = 10
let the additional number of women be x
35/10+x = 5/4
50 + 5x = 140
5x = 90
x = 18 women -
- IF = 38000 + 14000 + 8500 + 3300 = Ksh 63, 800pm
IF = (63800 x 12) = 38280
20 - 1st 6000x2=12000
2nd 6000x3 =18000
3rd 6000x4 =24000
4th 6000x5 =30000
5th 6000x6 =36000
6th 6000x7 =42000
7th 2280x8 =1842
Tax due =Ksh.180,240pa
Less relief 18,000
Tax payable 162,240
PAYE = 13,520 - Total deduction
=13520+320+1000+2000+5000+500
=Ksh.22,340 - 38280-22340 =Ksh.15,940
- IF = 38000 + 14000 + 8500 + 3300 = Ksh 63, 800pm
-
-
- Reflection in the line PR or ER
- Enlargement centre E
Scale factor -1 - Rotation about point R
Through 900
clockwise
-
- R - S B1
C - A B1 - R’ - Q B1
C’ - E B1 10
- R - S B1
-
-
-
-
-
- length XV = 300 cm 70°
= 281.9cm A1 - Arc length VBW =
140/360 x 2πr = 140/360 x 2x22/7 x 180
= 439.88cm - Length XAY
C = θ/360 2πr = 220/360 x 2x22/7 x 180 = 691.24 - Length of the conveyor belt
=691.2+439+(2x281.9)
=1694.88
- length XV = 300 cm 70°
-
-
-
-
-
- 7 x 30 = 210m ,
- 9.2 x 30 = 276m
- 13.7 x 30 = 411m
bearing= 310° ± 1° - 7.5 x 30 = 225m
Bearing =110° ± 1°
-
-
- Pα V/√R
P = K V/√R
180 = K = 9/√25
K= 180 x 5
9
K = 100
180 - 100V/√R
P = 100 x 6 = 117.67
√26 - P = 100V 360 = 100V
√R √R0.64
360 x 0.8 = V
100
V= 2.88 - V = %116 R = 75%
P = 100 x 1.16
√ 0.75
P = 133.945 - After change
133.945 - 100 = 33.945%
- Pα V/√R
-
-
x -30 0 30 60 90 120 150 180 210 240 270 Sin (x - 30) 0.87 0.5 0.00 0.87 -1.00 Cos (x - 15) 0.97 0.71 -0.26 0.71 -
- cos (x – 15)- sin (x – 30) = 0
cos (x – 15 ) = sin(x + 30)
x = 36 ±2 or x = 216±2 - co-ordinates of the turning point of y = cos (x – 15) on the negative section (198;-1)
- Cos (x – 15 ) = 0.6
- X-15 = 66°
- cos (x – 15)- sin (x – 30) = 0
-
-
-
- 57600
x - 57600
x + 4
- 57600
- 57600 - 57600 = 480
x x+ 4
X2+4x=480=0
X2+24x-20x-480=0
X(x+24)-20(x+24)=0
(x+24)(x-20)=0
X=20 - Price per suit
57600
20
=sh 2880
% profit 720 x 100 =25%
2880
-
-
- Sn = A/2 {2a + (n-1)d}
a = 2, n = 8 and Sn = 156d
8/2 {2 x 2 + (8-1)d}= 156
4(4 + 7d)= 156
16 + 28d = 156
⇒28d = 156 - 16
28d = 140
d=5 - n/2 {2 x 2 + (n-1)5}= 416
n/2 (4 + 5n - 5)= 416
4n + 5n2 - 5n = 416 x 2 ⇒ 4n - 5n + 5n2 = 832
5n2 - n - 832 = 0
n = n 1 ± √(1) - 4 x 5 - 832 = 1 ± 129
10 129
n = 1 ± 129 = 130 = 13
10 10
n = 1 - 129 = -128 = n = 12.8
10 10
ignore n = 12.8
∴ n = 13 - ∴ a + 2d, a + 4d , a + 7d
a + 4d = a + 7d
a + 2d a + 4d
(a + 4d)(a + 4d) = (a + 2d)(a + 7d)
a2 + 4ad + 4ad + bd2 = a2 + 2ad + 14d2
8ad + 16d2 = 9ad
16d2 = 9ad - 8ad
16d2 = 1d
9d
a = 16d
but d = 3
⇒a = 16 x3
= 48
r = a + 4d = 48 x 4 x 3 = 60 = 10
a + 2d 48 + (2 x 3) 54 9
Sn = a(rⁿ - 1)
r -1
48 (¹⁰/₉) - 1
¹⁰/₉ - 1
=683.067
- Sn = A/2 {2a + (n-1)d}
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