# Mathematics Paper 1 Questions and Answers - Asumbi Girls Highschool Pre-Mock Exams May - June 2022

INSTRUCTIONS TO CANDIDATES

1. Write your name, index number and class in the spaces provided above.
2. The paper contains two sections: Section I and Section II.
3. This paper contains 14 PRINTED pages make sure all PAGES ARE PRINTED and NON IS MISSING
4. Answer ALL the questions in Section I and ANY FIVE questions from Section II.
5. All working and answers must be written on the question paper in the spaces provided below each question.
6. Marks may be awarded for correct working even if the answer is wrong.
7. Negligent and slovenly work will be penalized.
8. Non-programmable silent electronic calculators and mathematical tables are allowed for use.

For examiners 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

SECTION I (50 marks)
Answer all the questions in this section in the spaces provided.

1. Without using Mathematical tables or a calculator, evaluate (3 marks)
2. Antony spent one quarter of his net January salary on school fees. He spent a quarter of the remainder on electricity and water bills. He then spent one ninth of what was left on transport. If he finally had sh. 3 400, calculate his net January salary. (3 marks)
3. A residential estate is to be developed on a 6 hectares piece of land. 1 500 m2 is taken up by the roads while the rest is divided into 40 equal plots. Calculate the area of each plot. (3 marks)
4. The equation of a straight line L1 is 3y + 4x – 6 = 0. Another straight line L2 is perpendicular to L1 and passes through point P (⁻3, 6). Determine the equation of L2 in the form y = mx + c, where m and c are constants. (3 marks)
5. A shopkeeper bought a number of eggs for which he paid a total of Ksh. 1000. Four eggs were broken. He sold the rest at 131/3 % profit, thereby making a cash profit of Ksh. 100. Calculate the number of eggs that he had bought at the first place. (3 marks)
6. Without using Mathematical tables a calculator evaluate  (3 marks)
7. The longest side of a right-angled triangle is (2x) cm. The other sides are (x + 3) cm and (2x – 4) cm. Find the value of x and hence the lengths of the sides of the triangle. (4 marks)
8. Below is part of sketch of a wedge ABCDEF. Complete the sketch of the solid, showing the hidden edges with broken lines. (3 marks)
9. If x is a positive integer, solve the inequality  and hence list the integral values that satisfy the inequality. (3 marks)
10. The matrix has no inverse. Determine the possible values of k. (3 marks)
11. A Kenyan Bank buys and sells foreign currencies at the exchange rates shown below
Kshs                Kshs
1 Euro                  147.86                148.00
1 US Dollar           74.22                  74.50
An American arrived in Kenya with 20000 Euros. He converted all the Euros to Kenya shillings at the Bank. He spent kshs.2,512, 000 while in Kenya and converted the remaining Kenya shillings into US Dollars at the bank. Find the amount in Dollars that he received. (3 marks)
12. Given that tan (θ + 20)º = ⁻0.7660, find θ, to the nearest degree, in the range 0º ≤ θ  ≤ 360º (3 marks)
13. The area of an island on a map of scale 1:100 000 is 200 cm2. Calculate the actual area on the ground in square kilometers. (3 marks)
14. P(2, ⁻1), Q(8, 11) and R(12, 19) are three points on a Cartesian plane. Show that P, Q and R are collinear. (3 marks)
15. A bus leaves Nairobi travelling towards Mombasa at a speed of 70 kmh⁻1. Half an hour later, a car leaves Nairobi travelling in the same direction at a speed of 90 kmh⁻1. Calculate the distance travelled by the car when it overtook the bus. (3 marks)
16. Use a ruler and a pair of compasses in this question.
1. Construct a quadrilateral PQRS in which PQ = 4 cm, QR = 6 cm, PS = 3cm, angle PQR = 135º and angle SPQ = 60º. (3 marks)
2. Measure the length of RS. (1 mark)

SECTION II
Answer five questions only in this section.

1.
1. Aisha sold 180 bags of rice in September 2017. The cost of each bag was sh. 2800.
Calculate the amount of money that he received from the sale of rice that month. (1 mark)
2.
1. In October that year, the price of a bag of rice decreased by 24% and the number of bags that she sold increased by 30%. Determine the percentage decrease in the amount of money she received from the sale of rice. (3 marks)
2. In November that year, the price of a bag of rice changed in the ratio of 7:8. Find the price of each bag in November. (2 marks)
3. The amount that he received from the sale of rice in September was sh. 1260 more than what was received in November. If the number of bags that were sold in November were t% more that those sold in September, find t. (4 marks)
2.
1. Complete the table below for the function y = 2x2 - 3x - 4 = 0 for -4 ≤ x ≤ 2  (2 marks)
 x -2 -1 0 1 2 3 2x2 2 0 2 8 - 3x - 4 2 -4 -13 y -4 5
2. On the grid below, draw the graph of  y = 2x2 - 3x - 4 = 0 for  -2 ≤  x ≤  3 (3 marks)
3. Use your graph to estimate the roots of y = 2x2 - 3x - 4 = 0 (2 marks)
4. Use your graph to solve = 4x2 - 7x = 12 (3 marks)
3. The marks scored by a certain number of students in a mathematics contest are as shown in the table below.
 Marks 45-49 50-54 55-59 60-64 65-69 70-74 75-79 No. of students 10 11 14 41 27 18 19
1. Calculate to 2d.p. the mean of the marks scored. (4 marks)
2. State the median class and hence calculate the median. (4 marks)
3. Calculate the difference between the mean and the median. (1 mark)
4. State the modal class. (1 mark)
4. The figure below shows a solid frustum of right pyramid with a rectangular base EFGH measuring 24cm by 7cm. The frustum was obtained by cutting off a small pyramid along plane ABCD that is parallel to base EFGH. Plane ABCD measures 16.8cm by 4.9cm, and is exactly seven tenths way up the vertical height of the original pyramid.

Given that the original pyramid had a slant edge of 32.5cm, find:
1. The altitude (perpendicular height) of the frustum. (4 marks)
2. The volume of the frustum (3 marks)
3. The surface area of the original pyramid. (3 marks)
5. Triangle ABC has vertices A(-6,5), B(0,1) and C(-2,-3). Triangle MNP is the image of triangle ABC under an enlargement. The vertices of triangle MNP are M(-4,6), N(-1,4) and P(-2,2). A reflection then maps triangle MNP onto triangle XYZ whose vertices are X(7,-5), Y(5,-2) and Z(3,-3).
1. Plot the three triangles on the grid below. (3 marks)
2. Determine the centre and the scale factor of enlargement that maps ABC onto MNP. (3 marks)
3. Find the equation of the mirror line of the reflection. (2 marks)
4. Given that triangle XYZ has an area of Qcm2, state the area of triangle ABC in terms of Q. (2 marks)
6. In the figure below, AB = 11cm, BC = 8cm, AD = 3cm, AC = 5cm and ‹DAC is a right angle.

Calculate, correct to one decimal place:
1. The length DC (2 marks)
2. The size of ‹ADC (2 marks)
3. The size of ‹ACB (3 marks)
4. The area of the quadrilateral ABCD (3 marks)
7. Three warships P, Q and R are at sea such that ship Q is 400km on a bearing of 0300 from ship P. Ship R is 750km from ship Q and on a bearing of 1200 from ship Q. An enemy warship S is sighted 1000km due south of ship Q.
1. Taking a scale of 1cm to represent 100km locate the relative positions of ships P, Q, R and S. (4 Marks)
2. Find the compass bearing of
1. P from S (1 mark)
2. S from R (1 mark)
3. Use the scale drawing to determine the distance of;
1. S from P (1 mark)
2. R from S (1 mark)
4. Find the bearing of;
1. Q from R (1 mark)
2. P from R (1 mark)
8. A particle moving along a straight line passes through a fixed point P. Its displacement S metres from P after a period at t seconds is given S=t3-5t2+3.
Find;
1. The particle’s displacement from P at t=4 (2 marks)
2. The particle’s velocity at t = 4. (3 marks)
3. The possible Value(s) of t when the particle is momentarily at rest. (3 marks)
4. The acceleration of the particle at t = 3. (2 marks)

## MARKING SCHEME

 SECTION I 1 = (-8) x 4 + (156 ÷ -26)         -3 + 16 + 6= -38   19 = ⁻2 M1 M1 A1 Numerator completely simplified  Denominator completely simplified 3 2 School fees = 1/4Remainder = 1 - 1/4 = 3/4Electricity and water bills = 1/4 x 3/4  = 3/16Remainder = 1 - (1/4 + 3/16) = 1 - 7/16 = 9/16Transport =  1/9 x 9/16 = 1/16Remainder = 1 -  (1/4 + 3/16 + 1/16) = 8/16 = 1/2Let the net salary be SThen 1/2S = 3 400S = Ksh. 6 800 B1 M1 A1 Remaining fraction Equating to 3 400 3 3 Total area = (6 x 10 000)m2                      = 60 000 m2           Plots area = 60 000 – 1 500                   = 58 500 m2Area of 1 plot =  58 500                                40                          = 1 462.5 m2 B1 M1 A1 Conversion of hectares to m2 Division by 40 3 4 L1 : 3y + 4x – 6 =0      3y = ⁻4x + 6y = -4/3x + 6Gradient m1 = -4/3Gradient m2 = 3/4 = y - 6 = 3    x + 3   44(y – 6) = 3 (x + 3)4y – 24 = 3x + 94y = 3x + 33y = 3/4x + 33/4 B1 M1 A1 Gradient of L2 Or equivalent equations 3 5 Let the total number of eggs bought be x.Then C.P. per egg = 1000/xS.P. per egg =   = 340/300(1000/x) = 3400/3xNumber of good eggs = x – 4Total S.P. = (3400/3x)(x – 4) = 11003 400x – 13 600 = 1 100x100x = 13 600x = 136 eggs M1 M1 A1 Expression for the S.P. per egg Equating total S.P. to 1 100 3 6 M1 M1 A1 Removal of all the negative indices Removal of all the roots 3 7 (x + 3)2 + 2x – 4)2 = (2x)2x2 + 6x + 9 + 4x2 – 16x +16 = 4x2x2 – 10x + 25 = 0(x – 5) (x – 5) = 0x – 5 = 0x = 52x – 4 = 10 – 4 = 6 cmx + 3 = 5 + 3 = 8 cm2x = 2(5) = 10 cm M1 M1 A1 B1 Applying Pythagoras theorem Complete factorisation Solving the quadratic equation For all the three sides 4 8 B1  B1 B1 Lines AF and ED drawn equal and parallel to BC, 0.1cm For any 2 pairs of lines drawn equal and parallel: AB and FC, AE and DF, BE and CD.   Completion of the solid with the broken lines shown. Note: The lines must not be extended 3 9 B1 B1 B1 Solving for x Solving for x  All the 3 values correct 3 10 Determinant of M = 0Thus 2k2 – 1(k + 3) = 02k2 – k – 3 = 02k2 + 2k – 3k – 3 = 02k (k + 1) – 3 (k + 1) = 0(2k – 3) (k + 1) = 0Either 2k – 3 = 0 2k = 3k =  3/2ork = ⁻1 M1 M1 A1 Equating the determinant to zero Complete factorization. For both values of k 3 11 20,000×147.86 =Ksh 2,957,2002597200 74.50 US\$5975.8389 M1 M1 A1 3 12 For 370 For the pair of 1430 and 1230 For the pair of 3230 and 3030 3 13 L.F.S. =        1                              100 000A.S.F. = (L.S.F.)2A.S.F. = (1/100 000)2 = 1/100 0002Actual area =  (200 x 100 000 x 100 000) cm2Actual area = (200 x 100 000 x 100 000) km2                               10 000 000 000= 200 km2 B1 M1 A1 For A.S.F.For conversion of cm2 to km2 3 14 B1 B1 B1 For both conditions  NB: Follow through the candidate’s work. 3 15 Distance covered by the bus in 30 minutes = 70 x 1/2 = 35 kmRelative speed of the car over the bus =  90 – 70 = 20 kmh⁻1Time taken for the car to catch up with the bus = 90 - 70 = 20 kmh-1t = 1.75 hoursDistance travelled by the car = 90 x 1.75D = 157.5 km. B1 M1 A1 For 20 kmh⁻1 seen For multiplication C.A.O. 3 16 RS = 7.0 ±  0.1 cm B1 B1 B1 B1 Correct location of point R, i.e., construction of PQ = 4.0 ±  0.1 cm, 135 ± 10 and QR = 6.0 ± 0.1 cm. Correct location of point S, i.e., construction of i.e., 60 ± 10 at P and PS = 3.0 ±  0.1 cm. Correct quadrilateral PQRS completed. 4

17.

(a) 180 x 2800 = 504000

(b)(i) October’s sales:
= (180 x 1.3) x (2800 x 100- 24)
100
= 234 x 2128
= sh. 492, 952
504000 - 497952 x 100% = 1.2%
504000
(ii) New price =2128 x 7/8
= sh. 1,862
(c)  November’s bags = 502740 - 1260
1862
=270
t% = 270 - 180 x 100% = 50%
180
t = 50

B1

M1

A1

B1

M1

A1

M1

A1

M1

A1

10

18

(a)

 -2 -1 0 1 2 3 8 2 0 2 8 18 2 -1 -4 -7 -10 -13 10 1 -4 -5 -2 5
(b)

(c) x = -0.85,x = 2.35

(d) 4x2 - 7x = 12
2x2 - 7/2x - 6 = 0
y = 2x2 - 3x - 4
y = 1/2x + 2
x = -1.1,x = 2.8

B1
B1

B1

B0

S1

P1

C1

B1

B1
B1

B1

B1

all correct values

6-8 correct values

Less than 6 values

Scale

plotting

curve

for Line

Each value one mark

For Both values

10

19

(a)

 marks x f fx cf 45-49 47 10 470 10 50-54 52 11 572 21 55-59 57 14 798 35 60-64 62 41 2542 76 65-69 67 27 1809 103 70-74 72 18 1296 121 75-79 67 19 1273 140 8760
Mean= 8760 = 62.57

140
(b) median class = 60 – 64
Median = 59.5 + 149/2 - 35 x 5
41
= 59.5 + 4.268
= 63.77
(c) 63.77 – 62.57 = 1.2
(d) 60 - 64

B1

B1

M1

A1

B1

M1

M1

A1

B1

B1

For Fx

For cf

10

20

(a)   The diagonal of the base =√242 + 72 = 25
The perpendicular height of the original pyramid
√32.52 - 12.52
=30
Perpendicular height of the frustum= 7/8x30
=21cm
(b)   The altitude of the cut-off pyramid is 30-21= 9cm
Volume  of frustum
= (1/3 x 24 x 7 x 30) - (1/3 x 16.8 x 4.9 x 9)
= 1680-246.96
= 1433.04cm3
(c) Perpendicular height of face VHG
=√32.52 x 122
=30.2
Perpendicular height of face VHG
=√32.52 x 3.52=
32.3
SA of original pyramid
= (24 x 7) + 2(1/2 x 24 x 30.2) + 2(1/2 x 7 x 32.3)
=1118.9cm2

M1

M1

M1

A1

M1

M1

A1

M1M1

A1

10

21

(a)

(b)   Q(-2,7) - (centre of enlargement)
Scale factor = 1/2

(c) y = x - 1

(d)   ASF=(LSF)2
2   = 4
Area of triangle ABC = 4Qcm2

S1

B1

B1

B1

B1

B1

B1

B1

M1

A1

Scale

Each  for each triangle

For centre of enlargement

Scale factor of enlargement

mirror line Drawn

eqn of the mirror line

10

22

(a) DC = √32 + 52
√34
= 5.8cm

(b)   Tan α= 5/6 = 1.667
α      = Tan-1  (1.677)
α      = 59.00

(c) c2 = a2 + b2 - 2abcosø
cos = 82 + 52 - 112
2 x 8 x 5
cos  ø=-0.4
ø = cos-1(-0.4)
ø= 113.60

(d)   Area = (1/2 x 3 x 5) + (1/2 x 5 x 8 x sin113.6)
= 7.5+18.3
= 25.8cm2

M1

A1

M1

A1

M1

M1

A1

M1M1

A1

10

23

(a)

(b) (i) 3430 ±1

(ii) 2260 ±1

c) (i)6.8x100=680km

(ii)9x100km=900km

d) (i) 3000

(ii)2720

B1

B1

B1

B1

B1

B1

B1

B1

B1

B1

for each point located

10

24

(a) S = (4)3 – 5(4)2 + 4(4) + 3
= 3m

(b) v = ds/dt = 3t2 - 10t + 4
=3(4)2 – 10(4) + 4
=12m/s

(c)  3t2-10t+4=0
t = 10 ± √100 - 48
6
t = 2.87 and t=0.46

(d) a = dv/dt = 6t - 10
= 6(3) - 10 = 8m/s2

M1

A1

M1

M1

A1

M1

M1

A1

M1

A1

(both values)

10

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