INSTRUCTIONS TO CANDIDATES
- This paper consists of TWO sections: Section I and Section II.
- Answer ALL the questions in Section I 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.
SECTION 1 (50 MARKS)
- Evaluate using squares, cubes and reciprocal tables (4 marks)
- Make x the subject in (3 marks)
- Ali deposited Ksh.100,000 in a financial institution that paid simple interest at the rate of 12.5% p.a. Mohamed deposited the same amount of money as Ali in another financial institution that paid compound interest. After 4 years, they had equal amounts of money. Determine the compound interest rate per annum to 1 decimal place. (3 marks)
- Simplify (3 marks)
- Expand (1 - 2x)4, hence find the value of (1.02)4 correct to 3 significant figures. (3 marks)
- If sin x =2b and cos x=2b√3, find the value of b (3 marks)
- Find the relative error in (a + b) given that a=77ml, b=23ml, c=36ml, and d=16ml. (3 marks)
(c − d) - Without using a calculator or mathematical tables, express √3 in surd form and simplify. (3 marks)
1− cos30° - The equation 3x2 − 8px + 12 = 0 has real roots. Find the value of P. (2 marks)
- A construction company employs 200 artisans and craftsmen in the ratio 1:3 every week. An artisan is paid 2 ½ times as much as a crafts man. At the end of 3 weeks the company paid ksh 1485000 to those employees. Find how much each artisan and each craftsman is paid. (a working week has six days) (3 marks)
- A dam containing 4158m3 of water is to be drained. A pump is connected to a pipe of radius 3.5cm and the machine operates for 8 hours per day. Water flows through the pipe at the rate of 1.5m per second. Find the number of days it takes to drain the dam. (4 marks)
- Two brands of coffee Arabica and Robusta costs sh.4,700 and sh.4,200 per kilogram respectively. They are mixed to produce a blend that costs shs.4,600 per kilogram. Find the ratio of the mixture. (3 marks)
- Under a transformation represented by a matrix , a triangle of area 10cm2 is mapped onto a triangle whose area is 110cm2. Find x (3 marks)
- Find the distance between the centre 0 of a circle whose equation is 2x2 + 2y2 + 6x + 10y + 7 = 0 and a point B(−4,1). (3 marks)
- Solve for x in the equation: (log2x)2+ log28 = log2x4 (4 marks)
- The figure below shows a circle inscribed in an isosceles triangle ABC. If Q, P and R are the points of contact between the triangle and the circle, O is the centre of the circle,
BO=19.5cm and BQ=18cm. Find the radius of the circle and hence the length of the minor arc PQ. (3 marks)
SECTION II (50 MARKS)
ANSWER ONLY FIVE QUESTIONS
-
- Mr. Mackey pays a tax of Kshs.5,800 per month according to the income tax
table given below. He is married and entitled to a family relief of K€ 420p.a.
Taxable income Rate (Ksh per K€ ) (K€ p.a.)
1 – 9,600
9,600 - 19,200
19,201 - 29,800
29,801 - 38,400
38,401 - 47,200
Over 47,200
2
3
5
7
9
10 - The difference between compound interest and simple interest on Kshs.P over a duration of 36 months at the rate of 15% p.a. is Kshs.52,477.50. Calculate the value of P. (4 marks)
- Mr. Mackey pays a tax of Kshs.5,800 per month according to the income tax
-
- Complete the table below for y = x3 + 4x2 − 5x − 5 (2 marks)
X −5 −4 −3 −2 −1 0 1 2 Y 19 −5 - On the grid provided, draw the graph of y = x3 + 4x2 − 5x − 5 for −5 ≤ × ≤ 2 (3 marks)
-
- Use the graph to solve the equation x3 + 4x2 − 5x − 5 = 0 (2 marks)
- By drawing a suitable straight line on the graph, solve the equation x3 + 4x2 − 5x − 5 = − 4x − 1 (3 marks)
- Complete the table below for y = x3 + 4x2 − 5x − 5 (2 marks)
- OPQ is a triangle in which OP=P and OQ=q. x is a point on OP such that OP:XP=5:2 and y is another point on PQ such that PY:YQ=1:2. Lines OY and XQ intersect at T.
- Express the following vectors in terms of P and q
- PQ (1 mark)
- OY (1 mark)
- OX (1 mark)
- If OT=kOY and QT=hQX express OT in two different ways. Hence or otherwise find the values of h and k. (6 marks)
- Determine the ratio OT:TY (1 mark)
- Express the following vectors in terms of P and q
- If (x-1 1⁄8), x and (x+3⁄2) are the first three consecutive terms of a geometric progression;
- Determine the values of x and the common ratio. (4 marks)
- Calculate the sum of the first 6 terms of this progression. (3 marks)
- Another sequence has the terms;
− 13,− 16,− 19, ……………………………− 310. Find the sum of this sequence. (3 marks)
- The figure below shows a belt passing round two pulleys of centres A and B. The radius of the pulleys is 4cm and 6cm respectively and the distance between the centres is 25cm.
Calculate the length of the belt used for the pulley system. (10 marks) - The points P(2,1), Q(4,1) R(4,3) and S(3, 3) are coordinates of a quadrilateral.
- Plot the quadrilateral PQRS on the grid provided. (1 mark)
- Find the coordinates of P1Q1R1S1 the image of PQRS under the transformation represented by the matrix
M = (2 marks) - Draw and label P1Q1R1S1 on the same grid.
- Find the coordinates of P11Q11R11S11 on the image of P1Q1R1S1 under the transformation represented by the matrix N = (2 marks)
- Draw and label P11Q11R11S11 on the same grid. (1 mark)
- Determine the matrix that maps PQRS directly onto P11Q11R11S11 . (3 marks)
- The table below shows the ages of people in years who attended a wedding ceremony.
Age in years 10 - 19 20 - 29 30 - 39 40 - 49 50 - 59 60 - 69 70 - 79 Frequency 2 4 4 8 6 3 2 - State the modal class (1 mark)
- Using an assumed mean of 44.5 calculate
- The mean age (3 marks)
- The standard deviation (3 marks)
- The median age (3 marks)
- A supermarket is stocked with plates which come from two suppliers A and B. They are bought in the ratio 3:5 respectively, 10% of plates from A are defective and 6% of the plates from B are defective.
- A plate is chosen by a buyer at randon. Find the probability that;
- It is from A (2 marks)
- It is from B and it is defective (2 marks)
- It is defective (2 marks)
- Two plates are chosen at random. Find the probability that;
- Both are defective (2 marks)
- At least one is defective (2 marks)
- A plate is chosen by a buyer at randon. Find the probability that;
MARKING SCHEME
- 0.339 + ( 1 )
(5.041 X 10−3) ……M1 Correct square & cube root
0.3309 + 3(198.4)
0.3309 + 595.2
(595.5309)−2
= 1
595.53092 M1 Correct reciprocal
= 1
35.46 ×104
= 1 = 0.2820 × 10−5…… M1
3.546 × 105
= 2.82 × 10−6……..A1 4
x2 +2 =K …………..M1
x2 = K − 2
x = ±√(k−2) ……………..A1 3- 100,000 ×12.5 × 4 = 50,000 ………M1
100
A=100,000 + 50,000 = 150,000
100,000(1 + r/100) = 150,000 ……….. M1
1+r/100 = (1.5)0.25 r = 10.7 (1d.p) …….A1 -
- 1 + 4 (−2x)+6 (−2x)2 + 4 (−2x)3 +(−2x)4
= 1 − 8x + 24x2 − 32x3 +16x4 ……. B1
1− 2x = 1.02
−2x = −0.0
x = −0.01
∴1−8(−0.01)=24(− 0.01)2 ………… M1
32(−0.01)3 + 16(−0.01)4 …..M1 ( substitution of −0.01)
1 + 0.08 + (0.0024) + 0.000032 + 0.00000016
= 1.08243216
= 1.08 (3sf) ……..A1 3\ - tan x = 2b
2b√3
= 1
√3 …….. M1
x = tan−1 (1/√3) = 30°
sin 30° = 2b ………M1
b = sin 30
2
= 0.5 = 0.25 ……A1 3
2 - Max value = 77.5 + 23.5 = 101 = 5.3158
35.5 −16.5 19 …….M1
Min value = 76.5 + 22.5 = 4.7143
36.5 −15.5
Absolute error = 5.3158 − 4.7143 …….M1
2
= 0.30075
% error = 0.30075 × 100 =6.015% √ A1
5 3 - Cos 30° = √3/2 ………..B1
= 4√3 + 6 ………A1 3 - 3x2 − 8px + 12=0
( −8p/2)2 = 3×12 ……..M1
16p2 = 36
P2 = ±√(36/16) =± 6/4
p=±1.5 ……….A1 - Let the pay for the craftsman be y
¾ (200)y+ ¼ (200) 2½y = 1485000 √ M1
18
150y +125y = 82,500 M1
y = sh.300
crafts man = sh.300……
artisan = 5/2 × 300 = 750 A1 3 - Cross-section area of the pipe = 22/7 × 3.5 × 3.5
= 38.5cm2
Vol. drained per sec
= 22/7 × 3.5 ×3.5 ×1.5√ ……. M1
100 ×100
= 0.005775m3
Vol. drained per day
= 0.005775 ×8×3600√… M1
=166.32 m3 A1
No.of days required
= 4158
166.32
= 25 days B1 4 - Arabica – x shs 4700
Robusta – 1− x → shs.4200
4700x + 4200 − 4200x=4600 √ M1
47x − 42x=46 − 42
5x=4
x = 4/5 √ A1
Arabica : Robusta = 4:1 √ ….B1 - Area scale factor = determinant = 110/10
=11√ m1
5x2 + 6=11 ……√ m1
5x2 = 5
x2 = 1
x = ±1 ……. A1 3 - x2 + y2 3x + 5y + (3/2)2 + (5/2)2 = −7/2 + 9/4 + 25/4
(x+3/2)2+(y+5/2)2 = 5………. B1
BO=√((-3/2+4) + (5/2-1) )2)…….M1
=√(25/4 + 9/4)
=√(34/4) = 2.915 units A1 3 - (log2x)2 + log28 = log2x4
let log2x be t
log28 = 3……….
∴ t2 + 3=4t
t2 − t − 3t + 3=0……… B1
t(t − 1) −3( t − 1)=0
(t−3)(t−1)=0
t=3 or t=1 ……… B1
∴log2x = 3 or log2x =1
23 = x 21 = x
8 =x B1 x = 2 B1 4 - <QOP=2sin-1(18/19.5) M1
=134.76° √ B1
Arc PQ=134.76/360 × 22/7 × 15
=17.65m ……… A1
Radius = √(19.52 ) c 182
=7.5cm √ B1 -
- Paye (p.a) = 5800x12
= kshs.69,600/=
Gross tax = 69,600 + 420 x 20
= 78,000 √ B1
Taxable income Tax (Kshs)
9600 x 2 19,200 √ M1
9600 x 3 28,000 √ M1
y x 5 30,000 √ M1
78,000
5y=30,000
y=K€ 6000 √ A1
Gross annual salary
= K€ 25,200 √ B1 - P (1+r/100)t − 1 − rt/100 =52,477.50
P(1.153 − 1 − 0.15 × 3) = 52,477.50√√ M1 M1
P = 52.477.50 √ M1
0.070875
P=740,000 √ M1
- Paye (p.a) = 5800x12
-
- Complete the table below for y = x3 + 4x2 − 5x − 5 (2 marks)
X −5 −4 −3 −2 −1 0 1 2 Y −5 15 19 13 3 −5 −5 9 - On the grid provided draw the graph of y = x3 + 4x2 − 5x − 5 for -5≤ × ≤2 (3 marks)
- Use the graph to solve the equation
x3 + 4x2 − 5x − 5 = 0 (2 marks) - By drawing a suitable straight line on the graph, solve the equation x3 + 4x2 − 5x − 5 = -4x-1 (3 marks)
- Use the graph to solve the equation
- Complete the table below for y = x3 + 4x2 − 5x − 5 (2 marks)
-
-
- PQ = − P + q B1
- OY = 2/3 P + 1/3 q B1
- QX = − q + 3/5 P B1
- OT = 2/3 pk + 1/3 qk …….(i) M1
OT= q(1− h) + 3/5 ph ……. (ii) M1
1/3 qk = (1− h)q …….. M1
1/3 k = 1− h
K=3 – 3h
3/5 ph = 2/3 pk
3/5h = 2/3 k
3/5 h = 2/3 (3 − 3h)
3/5 h = 2− 2h
3h = 10− 10h …… B1
h = 10/13 …….. B1
k = 3 − 3(10/13)
k = 9/13 …….B1 - OT:TY=9:4 …………. B1
-
-
-
-
= 46.80 √ A1 - Sn = n/2 (a + l)
Tn = a + (n-1)d
−310 = −13 + (n −1)(−3)
n=100 ……………√ B1
S100 = 100/2 {−13 + (−310)} √…………. M1
=−16,150 √ A1
-
- DC=EF
DC = EF = AX√(252 − 22) = 24.92cm …..B1 for the length
Sin φ = 2/25 = 0.8
∴φ = 4.589° ………A1
2φ = 9.178°
Reflex Angle = 180 + 9.178 = 189.178 ………..B1
Angle subtended by belt = 360°−189.178°
On small circle =170.8° ….B1
Minor arc DE = 170.8/360 × 2 × 22/7 × 6 =11.93cm….. M1 A1
Major arc CF = 189.2/360 × 2 × 22/7 × 6 = 19.82cm ….M1 A1
Length of belt = 2 × 24.92 + 11.93 + 19.82 …..M1
= 81.59cm…….. A1 10 -
P1 (3,4) Q1 (5,8) R1 (7,8) S1 (6,6) √ A1 (All coordinates)-
P11 (−2,4) Q11 (−2,8) R11 (−6,8) S11 (−6,6) √ A1 (Must be in coordinate form) -
√( M1 (matrix equ)
2a + b = −2 2c +d = 4 M1 formation of 4 equations
4a + b = −2 4c +d = 8
2a + b = −2 2c + d = 4
2a = 0 2c = 4
a = 0 c = 2
b = −2 d = 0
matrix = √ A1
-
- Modal class
40 - 49 B1 -
Age Frequency(f) Midpoint x d = x − 44.5 fd d2 Fd2 cf 10-19
20-29
30-39
40-49
50-59
60-69
70-792
4
4
8
6
3
2
εf=2914.5
24.5
34.5
44.5
54.5
64.5
74.5−30
−20
−10
0
10
20
30−60
−80
−40
0
60 M1
60
60
εfd=0900
400
100
0
100
400
9001800
1600
400
0
600 M1
1200
1800
εfd22
6
10
18
24
27
29
M1- x̄ = A + εfd
εf
= 44.5 + 0/29 √m1
= 44.5 √ A1 - Standard deviation
S2 = εfd2 = 7400 √ m1
εf 29
= 255.17 √ A1 - Median
= 39.5 + 5.625 = 45.125 A1
- x̄ = A + εfd
- Modal class
-
-
- P(A)= 3⁄8 √ B1
- P(B and D) = 5/8 × 6/100 √ m1
= 3/80 √ A1 - P(D)=P(A and D)or P(B and D)
= (3/8 × 10/100) + (5/8 × 6/100) √ m1
= 30/800 + 3/80
= 3/40 √ A1
-
- P(Two defective)=P(ADD) or P(BDD)
= (3/8 × 10/100 × 10/100) + (5/8 × 6/100 × 6/100) √ m1
= 3 + 180
800 80000
= 3 √ A1
500 - P(at least one defective) = 1− {P(AD1D1 or BD1D1 )} m1
= 1 − {3/8 × 90/100 × 90/100 + 5/8 × 94√/100 × 94√/100}
= 1 − 68480 = 11520 = 0.144 √ A1
80000 80000
- P(Two defective)=P(ADD) or P(BDD)
-
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