Mathematics Paper 1 Questions and Answers - KCSE 2022 Mock Exams Set 1

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INSTRUCTIONS TO CANDIDATES

  • Write your name and index number in the spaces provided at the top of this page.
  • Write your school name, sign and write the date of the examination in the spaces provided above.
  • This paper consists of Two sections: Section I and Section II
  • Answer ALL questions in Section I and any 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.
  • Non-Programmable silent electronic calculators and KNEC Mathematical Tables may be used.
  • Candidates should check the question paper to ensure that all the pages are printed as indicated and no questions are missing.
  • Candidates should answer the questions in English.

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

 

 

 

 

 

 

   

Total

 

Grand Total

 

17

18

19

20

21

22

23

24

 

QUESTIONS

SECTION I (50 Marks)
Answer all the questions in this section

  1. Without using mathematical tables or a calculator, evaluate: (3 marks)
    1 adadad
  2. Find the exact value of 2.3 ̇4 ̇-1.9 ̇1 ̇ (4 marks)
  3. The HCF and GCD of three numbers are 7920 and 12 respectively. Two of the numbers are 48 and 264. Using factor notation, find the third number if one of its factors is 32. (3 marks)
  4. The graph below (not drawn to scale) is a plot for the function y=ax2+bx+k where a, b and k are constants.
    4 adada
    Determine the values of a, b and k. (3 marks)
  5. Three camps, P, Q and R shared 875 bags of beans donated by a charitable organization such that the ratio of P to Q was 2: 3 and that of R to Q was 5: 4. Find the number of bags of beans received by camp Q. (3 marks)
  6. Find the number of revolutions made by cylindrical roller of diameter 1.02 m and thickness 1.3 m if it rolls over a surface area of 291.72 m2. Use π = 22/7 (3 marks)
  7. Dr. June needs to import a car from Japan that costs US dollars (USD) 5 000 outside Kenya. He intends to buy the car through an agent who deals in Japanese Yen (JPY). The agent charges a 20% commission on the price of the car and a further 80 325 JPY for shipping the car to Kenya. Find the amount in Kenya shillings that Dr. June will need to send to the agent to get the car given that: (3 marks)
    1 USD = Ksh 120
    1 USD = 135 JPY
  8. The figure below shows a vertical mast AB. The angles of elevation of the top of the mast from two points C and D on the level ground are 400 and 320 respectively. The distance CD is 50 metres.
    8 szadada
    Calculate the distance BC, of the tower correct to 1 decimal place (3 marks)
  9. The figure below shows a solid with the given dimensions.
    9 adadad
    Calculate the mass of the solid to in kilograms given that its density is 3.5 g/cm3. (3 marks)
  10. Solve for x in the equation (3 marks)
    9x+1+32x=810
  11. A lampshade is in the shape of a frustum of a cone such that its slanting height is 30 cm while the top and bottom radii of the lampshade are 10 cm and 17.5 cm respectively. Calculate the surface area of the material used to make the lampshade. (4 marks)
  12.                            
    1. Find the range of values of x which satisfies the following inequalities simultaneously (2 marks)
      4x-9≤x+6
      4+x≥8-3x
    2. Represent the range of values of x in (a) above on a number line. (1 mark)
  13. The graph below is a histogram showing the marks scored by students in a Mathematics contest.
    13 sfsfsfs
    Prepare a frequency distribution table for the above data, hence determine the total number of students who did the contest. (3 marks)
  14. Use logarithms only to evaluate (4 marks)
    14 ssfsfs
  15. A sales lady earns a basic salary of Ksh 12 400 per month. In addition, she gets a 2% commission on the first Ksh 80 000 worth of good and a further 3% commission on goods worth over Ksh 80 000. In a certain month, she earned a total of Ksh 20 000. Calculate the value of the goods sold that month. (3 marks)
  16. The figure below shows a triangular prism whose cross-section is an equilateral triangle
    16 adadad
    1. On the prism, show the positions of the lines of symmetry. (1 mark)
    2. Draw the net of the prism. (1 mark)

SECTION II (50 Marks)
Answer any five questions in this section

  1. A straight line L1 has a gradient -1/2 and passes through the point P(-1,3). Another line L2 passes through the points Q(1,-3) and R(4,5)
    Find:
    1.                        
      1. The equation of Lin the form y=mx+c, where m and c are constants (2 marks)
      2. Hence find k given that S(0,k) (1 mark)
    2.                          
      1. The gradient of L2 (1 mark)
      2. The equation of L2 in the form ax+by=c, where a,b and c are integral values. (2 marks)
    3. The equation of a line L3 passing through a point T(0,5) and perpendicular to L2. (3 marks)
    4. Calculate the acute angle that L3 makes with the x-axis. (1 mark)
  2. Water flows through a cylindrical pipe of diameter 7 cm at a rate of 15 m per minute.
    1. Calculate the capacity of water delivered by the pipe in one minute in litres. Use π=22/7. (3 marks)
    2. A storage tank that has a circular base and depth 12 m is filled with water from this pipe and at the same rate of flow. Water begins flowing into the empty storage tank at 6.30 a.m. and is full at 1310 hours. Calculate the area of the cross-section of this tank in square metres. (4 marks)
    3. A school consumes the capacity of this tank in one month. The cost of water is Ksh. 100 for every 1 000 litres and a standing charge of Ksh 1 950. Calculate the cost of the school’s water bill for one month. (3 marks)
  3.                      
    1. Find A-1 given that 19 adadada (1 mark)
    2. An ICT firm bought 8 printers and 12 copiers for a total of Ksh 294 000. Had the firm bought 1 more printer and 3 more copiers, it would have spent Ksh 43 500 more.
      1. Form two equations to represent the information above. (2 marks)
      2. Hence, using A-1 in (a) above, calculate the cost of each item. (4 marks)
      3. A two-digit number is such the that difference between tens and ones digit is 1. If the digits are reversed, the sum of the two numbers is 165. Find the original number (3 marks)
  4. Two equal circles with centres P and Q and radius 8 cm intersect at points A and B as shown below.
    20 adadada
    Given that the distance between their centres is 12 cm, find, correct to 4 significant figures;
    1. The length of chord AB (2 marks)
    2. The area of the shaded region . Use π = 3.142 (8 marks)
  5. In the figure below, P, Q, R, S and T lie on the circumference of a circle centre O. Line UPV is a tangent to the circle at P. Chord ST of the circle is produced to intersect with the tangent at U. Angles UPT, RST and ORQ are 28º, 100º and 50º respectively.
    21 ssfsfs
    1. Determine the sizes of the following angles;
      1. RTP (3 marks)
      2. QTP (3 marks)
    2. Given that PQ = 6 cm, calculate correct to 1 decimal place, the radius of the circle. (4 marks)
  6.                                
    1. Two trains, A and B are such that they are 40 m long and 160 m long respectively. Their speeds are 60 km/hr and 40 km/hr respectively. The two trains are 100 m apart and moving towards the same direction in a pair of parallel tracks. Calculate the time in seconds it takes train B to completely overtake train A. (4 marks)
    2. The figure below (not drawn to scale) shows a velocity time graph for a robot in a robotic challenge.
      22 adada
      1. If the distance covered by the robot in the first 15 seconds was 180 metres, calculate the value of m (3 marks)
      2. Describe the movement of the robot between the 15th and 45th seconds. (1 mark)
      3. Calculate the deceleration of the particle in the last 20 seconds. (2 mark)
  7. A farmer wanted to make a trough for cows to drink water. He had a metal sheet of dimensions 240 cm by 120 cm and 1 cm thick. The density of the metal sheet is 2.5 g/cm3. A square of sides x cm is removed from each corner of the rectangle and the remaining part folded to form an open cuboid.
    1. Sketch the sheet after removing the squares from the four corners, showing all the dimensions. (2 marks)
    2. Calculate
      1. the value of x, to the nearest whole number, that maximizes the volume of the cuboid (5 marks)
      2. hence calculate the maximum volume of the box . (3 marks)
  8.                            
    1. Complete the table below for the curve y=x3-5x2+2x+9 for -2≤x≤5. (2 marks)
       x  -2  -1  0  1  2  3  4 5
       y      9          
    2. On the grid provided, draw the graph of y=x3-5x2+2x+9 for -2≤x≤5. (3 marks)
      24 adadad
    3. Use the graph in (b) above to find the roots to the following equations:
      1. x3-5x2+2x+9=0 (2 marks)
      2. x3-5x2+6x=-5 (3 marks)

 MARKING SCHEME

NO.

WORKING

REMARKS

1.       

1 adada

Removal of decimal places

Expressing as product of prime factors

 

Total

 

2.       

2.34 = 2.343434343....... = r
100r = 234.34343434343...
r = 2.34343434...
99r = 232
r = 232
      99

Also 

1.91 = 1.919191 .. q
100q = 191.919191...
q = 1.919191 ...
99q = 190
q = 190
       99

Hence

232 - 190
  99    99
= 42 = 14
   99    33

Expressing 2.34 as a fraction

Expressing 1.91 as a fraction

Difference between the 2 fractions

 

Total

 

3.       

HCF -> 7920 = 24 x 32 x 5 x 11
GCD -> 22 x 3
48 = 24 x 3
264 = 23 x 3 x 11

Other number -> 22 x 32 x 5 = 180

Expressing HCF, GCD and the 2 given numbers in power form

 

Total

 

4.       

From the graph,

x = -3 -> x + 3 = 0 and x = 2 -> x - 2 = 0
(x + 3) (x - 2) = 0
x(x - 2) + 3(x - 2) = 0
x2 - 2x + 3x - 6
x2 + x - 6 = 0
Hence 
a = 1, b = 1 and k = -6

Factorization by grouping

 

Total

 

 5. P:Q = 2:3 ... x 4 = 8:12
R : Q = 5: 4 ... x 3 = 15:12
Hence P:Q:R = 8:12:15
Q -->          12      x 875
        8 + 12 + 15
Q --> 12 x 875 = 300 bags
         35
 
   Total  3
 6. curved surface area of cylinder
C.S.A = 22/7 x 1.02 x 1.3
Number of revolutions
=         291.72     
22/7 x 1.02 x 1.3
= 70
 
   Total  3
 7. cost --> 120/100 x 5000 = USD 6000
Shipping --> 80325 = JPY USD
                      135
Total --> 6000 + 595 = 6595
TOTAL COST --. 6595 X 120
= Ksh 791, 400
Total cost in USD
Conversion of USD to Ksh
   Total  3
 8.

Let BC = x 
tan 40º = AB --> AB = x tan 40º
                 x
Also,
tan 32º =    AB   --> AB = (x + 50)tan 32º
              x + 50
x tan 40º = (x + 50) tan 32º
0.8391x = (x + 50)0.6249
0.8391x = -0.6249x = 31.245
0.2142x = 31.245
x   =   31.245   = 145.9 m
         0.2142

 

Expressing AB in terms of
tan 32º and tan 40º

Equating AB to AB

145.9 seen

   Total  3
 9. v = {1/2 x 8(13 + 20)} x 12
V = 1584
mass = 3.5 x 1584
                1000
= 5.544 kg
 
   Total  3
 10.  32x + 2 + 32x = 810
32 x 32x + 32x = 810
9(32x) + 32x = 810
10(32x) = 810 --> 32x = 81
32x = 34
2x = 4 --> x = 2
Expressing right hand side in terms of base 3
   Total  3
 11.  Let the slant length of the smaller cone be L
      L     = 10 
   L + 30  17.5
17.5L = 10L + 300
L = 300/7.5 = 40
C.S.A  = 3.142{(17.5 x 70) - (10 x 40)}
C.S.A = 2 592.15 cm2
 
   Total  4
 12. 4x - x ≤ 6 + 9
3x ≤ 15 -> x ≤ 5

x + 3x ≥ 8 - 4
4x ≥ 4 -> x ≥ 1
1 ≤ x ≤ 5
12 adada
 

For x ≤ 5 and x ≥ 1

Compound inequality shown

Number line drawn 
   Total  3
 13.

Frequency distribution table

Marks

f

10 – 14

2

15 – 19

5

20 – 24

7

25 – 29

12

30 – 34

4

∑f = 2 + 5 + 7 + 12 + 4 = 30

 

All classes/class boundaries

 

All frequencies

 

Total frequency30 seen

   Total  3
 14.

Logarithms

Number

Std Form

Logarithm

7.089
124.5

7.089 x 100
124.5 x 102

0.8506
2.0952

   

2.9458

86.73

86.73 x 10

1.9382

   

1.0076 ÷ 3

2.167

2.167 x 100

0.3359

 

All logs

Correct +/- of logs

Correct multiplication by 2 and division by 3

Accuracy

   Total  4
 15. Commission -> 20000 - 12400 = 7600
2/100 x 80 000 = 1600
7600 - 1600 = 6000
6 000 = 3/100 x A 
A = 6 000 x 100/3 = 200 000
Total value
200 000 + 80 000 = 280 000
 

Amount from commission

Expression for excess of 80 000

   Total  3
 16.

a) Lines of symmetry

16 a adad

b) Net

16 b adad

For all the 4 lines of symmetry drawn

 

Correct net drawn

 

Correct measurements transferred from the solid

  Total  2
17.
  1.                            
    1. Equation of l1
      3=-1/2 (-1)+c
      6=1+2c
      2c=5→c=5/2
      l_1→y=-1/2 x+5/2
    2. If S(0,k)→S – y-intercept
      k=5/2=2.5
  2.                      
    1. Gradient of l2
      m2=(5-(-3))/(4-1)=(5+3)/3=8/3
    2. Equation of l2
      8/3=(y-5)/(x-4)
      8x-32=3y-15
      8x-3y=17
  3. Equation of l3
    m3×8/3=-1→m3=-3/8
    (-3)/8=(y-5)/x
    -3x=8y-40
    3x+8y=40
  4. Acute angle of l3 and x-axis
    tan⁡θ=3/
    θ=tan-1(3/8)
    θ=20.560
 

Substituting (-1,3) in y=mx+c

y=-1/2 x+5/2 seen

2.5 or equivalent seen

8/3 seen

  Total  10
18.
  1. Amount of water delivered in 1 minute
    V=22/7×3.5×3.5×15×100
    V=57 750

    Capacity
    (57 750)/(1 000)=57.75 litres
  2. Area of base of tank
    Time difference
    1310 hrs
    0630 hrs
    6 hrs 40 minutes→6×60+40=400 minutes
    1 minute→57.75 litres
    400 minutes=400×57.75
    =23 100 litres
    1 litres=1 000 cm3
    23 100 litres=23 100×1 000
    23 100×1 000=Base Area×12
    Base Area=(23 100×1 000)/(12×10 000)
    = 192.5 m2
  3. Monthly water bill
    1 000 litres→Ksh 100
    23 100 litres=(23 100×100)/(1 000)
    =Ksh 2 310
    Bill
    2 310+1 950=Ksh 4 260

Time difference

 

Capacity in 400 minutes

 

Expression for base area

  Total  10
19.
  1.                            

    19 a saddada
  2.                        
    1. Equations
      8p+12c=294 000
      (8+1)p+(12+3)c=294 000+43 500
      9p+12c=337 500
    2. 19 b adada
  3. Let the number be ab
    a-b=1
    (10a+b)+(10b+a)=165
    11a+11b=165→a+b=15

    a-b=1
    a+b=15
    2a=16→a=8

    8-b=1→b=8-1=7
    Hence the number is 87

Accept if all elements as fractions

 

Matrix equation

 

Premultiplying by A^(-1)

 

 

Both values 

 

Forming 2 equations in a and b

Solution for a and b using any method

  Total  10
20.

 20 auygdada

  1. Consider ΔAPT
    PT=1/2 PQ=6 cm
    AT=√(82-62 )=5.291 cm
    AB=2×5.291=10.58 cm
  2. Shaded area
    Let ∠APT=θ in ΔAPT
    cos⁡ θ=6/8→θ=cos-1 (6/8)= 41.41º
    ∠APB=2× 41.41º = 82.82º

    Area of segment
    A=82.82/360×3.142×82-1/2×8×8×sin⁡ 82.82º
    A=46.261-31.749
    A=14.512

    Both segments
    14.512×2=29.024
    Area of APBQ
    A=2×31.749=63.498

    Shaded area=63.498-29.024
    =34.74 cm2

Angle APT

 

Area of sector, area of ΔAPB

 

Area of both segments
Area of APBQ

Shaded Area

  Total 10
21.
  1.                              
    1. ∠RTP
      ∠RQT= 180º - 100º= 80º – opposite angles of cyclic quadrilateral QRST are supplementary
      ∠TQP=∠UPT= 28º – angle between a chord and a tangent is equal to the angle subtended by the same chord on the circumference of the alternate segment.
      ∠RQP= 80º + 28º=108º
      Hence
      ∠RTP= 180º- 108º= 72º – opposite angles of cyclic quadrilateral QRTP are supplementary
    2. Join O to Q and consider ∠RQO
      ∠RQO= 50º – base angles of isosceles ΔRQO
      ∠ROQ= 180º -2× 50º = 80º  – sum of angles in ΔRQO is 180º
      ∠RTQ=1/2× 80º = 40º – angle at the centre is twice angle at the circumference

      Hence
      ∠QTP= 72º - 40º = 32º
  2. Consider
    ∠QPT= 180º - 28º + 32º = 120º
    ∠TQO= 80º - 50º = 30º
    In ΔQPT
    6/sin 32º  =QT/sin⁡ 120º
    QT=(6 sin⁡ 120º )/sin⁡ 32º =9.806
    Let M be the midpoint of QT
    MQ=1/2×9.806=4.903
    Consider ΔOQM
    cos⁡ 30º  =  OQ/4.903
    OQ (radius)=4.903/cos⁡ 30º  =5.7 cm

∠RQT

∠RQP

∠RTP

∠ROQ

∠RTQ 

∠QTP

Sine Rule applied

Half of QT

Attempt to get radius

  Total 10
22.
  1. Time
    Total length = 40+100+160=300 m
    Relative speed=60-40=20 km/h
    Time=0.3/20×3600
    =54 seconds
  2.                                
    1. Value of m
      180=1/2×15(4+m)
      180×2=15(4+m)
      (180×2)/15=4+m
      4+m=24→m=20 m/s
    2. No acceleration
    3. Deceleration
      a=(0-20)/(60-45)
      a=(-20)/15=-1 1/3  m/s2
      Hence, a deceleration of 1 1/3 m/s2

Total length
Relative speed

 

Equation distance to area of trapezium

Collecting like terms

  Total 10
23.
  1. Sketch
    23 adada
  2. Value of x for maximum volume
    23 b adada
    V=(240-2x)(120-2x)x
    V={240(120-2x)-2x(120-2x)}x
    V=(4x2-720x+28800)x
    V= 4x3-720x2+28800x

    dV/dx=12x2-1440x+28800

    For maximum volume
    12x2-1440x+28800=0→x2-120x+2400=0
    x=(-(-120)±√((-120)2-4×2400))/2
    x=(120±√4800)/2
    x=(120±69.28)/2

    Either
    x=(120+69.28)/2=94.64≅95

    And
    x=(120-69.28)/2=25.36≈25

    Hence x=25
  3. Mass of empty box
    External dimensions
    (240-2×25) cm by (120-2×25) cm by 25 cm
    190 cm by 70 cm by 25 cm

    Internal dimensions
    188 cm by 68 cm by 24 cm
    V=(190×70×25)-(188×68×24)
    V=332500-306816=25684
    Mass=25684

Correct sketch

 

Dimensions shown on the sketch

 

Expression for volume

 

Equating volume to 0 at maximum volume

 

Both values of x

 

Value of x

 

Internal and external dimensions

  Total 10
24.
  1. Table

    x

    -2

    -1

    0

    1

    2

    3

    4

    5

    y

    -23

    1

    9

    7

    1

    -3

    1

    19

  2. Graph
    24 asdada
  3. Roots
    1. x3-5x2+2x+9=0
      y=x3-5x2+2x+9
      0=x3-5x2+2x+9-
      y=0
      x=-1.2 or x=2.2 or x=3.8 – all ±0.2
    2. x3-5x2+6x=-5
      y=x3-5x2+2x+9
      0=x3-5x2+6x+5-
      y=-4x+4

      x=-0.6±0.2
 

All the y values (B1 for at least 5 y values )

 

Linear scales used on both axes – accommodates all table values

 

All points plotted within the graph paper

 

Smooth curve drawn

 

y=0 shown or implied in the roots

 

All the values of x

 

attempt to get

 y=-4x+4

Line y=-4x+4 drawn 

value of x

  Total 10

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