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

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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)
    1 daaddad
  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 6 adada (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)
    8 adadada
  9. If x is a positive integer, solve the inequality 9 dsfsfsfs and hence list the integral values that satisfy the inequality. (3 marks)
  10. The matrix 10 auhygdadhas 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
                                   Buying              selling
                                    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)
      b asdasdada
    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.
    20 auydgad
    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)
      b asdasdada
    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.
    22 adadadad
    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

1 daaddad
= (-8) x 4 + (156 ÷ -26)
         -3 + 16 + 6
= -38
   19
 
=2

M1

M1

A1

Numerator completely simplified 

Denominator completely simplified

 

 

3

 

2

School fees = 1/4
Remainder = 1 - 1/4 = 3/4
Electricity and water bills = 1/4 x 3/4  = 3/16
Remainder = 1 - (1/4 + 3/16) = 1 - 7/16 = 9/16
Transport =  1/9 x 9/16 = 1/16
Remainder = 1 -  (1/4 + 3/16 + 1/16) = 8/16 = 1/2
Let the net salary be S
Then 1/2S = 3 400
S = 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 m2
Area 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 + 6
y = -4/3x + 6
Gradient m1 = -4/3
Gradient m2 = 3/4
 = y - 6 = 3
    x + 3   4
4(y – 6) = 3 (x + 3)
4y – 24 = 3x + 9
4y = 3x + 33
y = 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/x
S.P. per egg =   = 340/300(1000/x) = 3400/3x
Number of good eggs = x – 4
Total S.P. = (3400/3x)(x – 4) = 1100
3 400x – 13 600 = 1 100x
100x = 13 600
x = 136 eggs

M1

M1

A1

Expression for the S.P. per egg

Equating total S.P. to 1 100

 

 

3

 

6

6 dadada

62 auhada

M1

M1

A1

Removal of all the negative indices

Removal of all the roots

 

 

3

 

7

(x + 3)2 + 2x – 4)2 = (2x)2
x2 + 6x + 9 + 4x2 – 16x +16 = 4x2
x2 – 10x + 25 = 0
(x – 5) (x – 5) = 0
x – 5 = 0
x = 5
2x – 4 = 10 – 4 = 6 cm
x + 3 = 5 + 3 = 8 cm
2x = 2(5) = 10 cm

M1

M1

A1

B1

Applying Pythagoras theorem

Complete factorisation

Solving the quadratic equation

For all the three sides

 

 

4

 

8

8 dddadaa

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

9 adadada

B1

B1

B1

Solving for x

Solving for x

 All the 3 values correct

 

 

3

 

10

10 adadad
Determinant of M = 0
Thus 2k2 – 1(k + 3) = 0
2k2 – k – 3 = 0
2k2 + 2k – 3k – 3 = 0
2k (k + 1) – 3 (k + 1) = 0
(2k – 3) (k + 1) = 0
Either 2k – 3 = 0
 2k = 3
k =  3/2
or
k = ⁻1

M1

M1

A1

Equating the determinant to zero

Complete factorization.

For both values of k

 

 

3

 

11

20,000×147.86

=Ksh 2,957,200
2597200
 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               
               100 000
A.S.F. = (L.S.F.)2
A.S.F. = (1/100 000)2 = 1/100 0002
Actual area =  (200 x 100 000 x 100 000) cm2
Actual 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

14 ssfsfs

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 km
Relative speed of the car over the bus =  90 – 70 = 20 kmh1
Time taken for the car to catch up with the bus = 90 - 70 = 20 kmh-1
t = 1.75 hours
Distance travelled by the car = 90 x 1.75
D = 157.5 km.

B1

M1

A1

For 20 kmh1 seen

For multiplication

C.A.O.

 

 

3

 

16

16 auygdyuada
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
Percentage decrease in amount received:
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)
18 aygdad
(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)
21 ssdsds
(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)23 sdadad 

(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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