Mathematics Paper 1 Questions and Answers - Form 4 Term 1 Opener Exams 2021

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FORM 4 TERM 1
OPENER EXAMS

MATHEMATICS PAPER 1
TIME: 2½ HRS.

INSTRUCTION TO STUDENTS:

  • This paper consists of two Sections; Section I and Section II.
  • Answer ALL the questions in Section I and only five questions from Section II.
  • Show all the steps in your calculation, giving your answer at each stage in the spaces provided below each question.
  • Marks may be given for correct working even if the answer is wrong.
  • KNEC Mathematical tables may be used, except where stated otherwise.
  • Candidates should answer the questions in English.

SECTION I (50 marks)

  1. Without using a calculator, evaluate. (3 marks)
    -8+(-5) x(-8)-(-6)
    -3+(-8) ÷ 2 x 4
  2. Evaluate without using a calculator (2 marks)
    MathF4OPP1Q2
  3. In fourteen years time, a mother will be twice as old as her son. Four years ago, the sum of their ages was 30 years. Find how old the mother was, when the son was born.( 4 marks)
  4. A Kenya bank buys and sells foreign currencies as shown below
      Buying
    (In KShs) 
    Selling
    (In Kshs) 
    1 Hong Kong dollar  9.74  9.77
    1 South African rand  12.03  12.11
    A tourist arrived in Kenya with 105000 Hong Kong dollars and changed the whole amountto Kenyan shillings. While in Kenya, she spent Kshs. 403,897 and changed the balance to South Africa rand before leaving for South Africa. Calculate the amount, in South African rand that she received. (3 marks)
  5.    
    1. Using a ruler and a pair of compasses only, construct a quadrilateral PQRS in which PQ= 5cm, PS = 3cm, QR = 4cm,<PQR = 135°and <SPQ is a right angle. (2 marks)
    2. The quadrilateral PQRS represents a plot of land drawn to a scale of 1:4000. Determine the actual length of RS in meters. (2 marks)
  6. The ratio of goats to cows in a farm is 2:5 while the ratio of sheeps to cows is 3:4. If there are 15 sheep, how many animals are there in farm?(2 marks)
  7. Mr.Maina who deals in electronics sells a radio to a customer at Ksh. 1440 after giving a discount of 10% but find that he makes a 20% profit. Find the profit Mr. Maina would make if he does not give a discount. (3 marks)
  8. Use the reciprocal and square table to evaluate to four significant figure, the expression.(3 marks)
    MathF4OPP1Q8
  9. Simplify the following expression completely. (3 marks)
    MathF4OPP1Q9
  10. Given that sin (x + 60)° = Cos (2x)°, find tan (x + 60)°. (3 marks)
  11. The figure below shows triangle PQR in which PR = 12 cm, T is a point on PR such that TR = 4cm. line ST is Parallel to QR
    MathF4OPP1Q11
    If the area of triangle PQR is 336 cm2, find the area of the quadrilateral QRTS. (3 marks)
  12. A square brass plate is 2 mm thick and has a mass of 1.05 kg. The density of the brass is 8.4 g/cm3. Calculate the length of the plate in centimeters (3 marks)
  13. The diagram below represent a solid made up of a hemisphere mounted on a cone. The common radius is 6 cm and the height of the solid is 15cm.
    MathF4OPP1Q13
    Calculate the external surface of the solid (4 marks)
  14. Solve the simultaneous inequalities given below and list all the integral values of x (3 marks)
    3-x/2x+1/32x+1/-3
  15. A construction company employs technicians and artisans. On a certain 3 technicians and 2 artisans were hired and paid a total of Ksh 9,000.
    On another day the firm hired 4 technicians and one artisan and paid a total of Ksh 9500. Calculate the cost of hiring two technicians and 5 artisans in a day
  16. The figure below is not drawn to scale.
    MathF4OPP1Q15
    Find correct to 1 decimal place;
    1. Length PQ. (2 marks).
    2. Angle ABC (2 marks)

SECTION II(50 marks)
CHOOSE ANY FIVE QUESTIONS IN THIS SECTION

  1. Two lines L1: 2y - 3x - 6 = 0 and L2: 3y+x-20 = 0 intersect at point A.
    1. Find the coordinates of A. (3 marks)
    2. A third line L3 is perpendicular to L2 at point A. Find the equation of L3 in the form y = mx +c, where m and c are constants. (3 marks)
    3. Another L4 is parallel to L1 and passes through (-1, 3). Find the x and y intercepts of L(4 marks)
  2. Town B is 180km on a bearing 0500 from town A. Another town C is on a bearing of 110° from town A and on a bearing of 150° from town B. A fourth town D is 240 km on a bearing of 320° from A. Using scale drawing, such that 1cm rep 30km,
    1. Show the relative position of the towns (4 marks)
    2. Using the diagram, find
      1. Distance AC (2 marks)
      2. Distance CD (2 marks)
      3. Compass bearing of C from D (2 marks)
  3. The table shows the marks obtained by 40 candidates in an examination
    Marks   5-14 15-29 30-34 35-44  45-49 
    Frequency   2  12  7  15  x

    1. Find the value of X (1 mark)
    2. Calculate the mean mark (2 marks)
    3. On the grid provided below draw a histogram to represent the data (4 marks)
    4. drawing a straight line on the graph above determine the median mark (3 marks)
  4. The figure below shows a triangle ABC inscribed in a circle. AC = 10cm, BC = 7cm and AB = 10cm.
    MathF4OPP1Q20
    1. Find the size of angle BAC. (3 marks)
    2. Find the radius of the circle. (2 marks)
    3. Hence calculate the area of the shaded region. (5 marks)
  5. The figure below represents a speed time graph for a cheetah which covered 825m in 40 seconds.
    MathF4OPP1Q21
    1. State the speed of the cheetah when recording of its motion started (1 mark)
    2. Calculate the maximum speed attained by the cheetah (3 marks)
    3. Calculate the acceleration of the cheetah in:
      1. The first 10 seconds (2 marks)
      2. The last 20 seconds (1 mark)
    4. Calculate the average speed of the cheetah in first 20 seconds (3 marks)
  6. Given the quadratic function y=3x2 + 4x - 1
    1. Complete the table below for values of x ranging - 4≤ x ≤3. (2 marks)
       X -4 -3 -2 -1 0 1 2 3
       Y                
    2. Using the grid provided draw the graph of y = 3x2 + 4x —2 for -4≤ x ≤ 3 (3 marks)
    3. Using the graph, find the solution to the equations.
      1. 3x2+4x-2=0 (2 marks)
      2. 3x2+7x+2=0 (3 marks)
  7. The diagram represents a solid frustum with base radius 21cm and top radius 14cm. The frustum is 22.5cm high and is made of a metal whose density is 3 g/cm3.
    (Take π=22/7)
    MathF4OPP1Q23
    1. Calculate
      1. The volume of the metal in the frustum. (5 marks)
      2. The mass of the frustum in kg. (2 marks)
    2. The frustum is melted down and recast into a solid cube. In the process 20% of the metal is lost. Calculate to 2 decimal places the length of each side of the cube.(3 marks)
  8. A saleswoman is paid a commission of 2% on goods sold worth over Ksh 100,000. She is also paid a monthly salary of Ksh 12,000.In a certain month, she sold 360 handbags at Ksh 500 each.
    1. Calculate the saleswoman’s earnings that month. (3 marks)
    2. The following month, the saleswoman’s monthly salary was increased by 10%. Her total earnings that month were Ksh 17,600.
      Calculate:
      1. The total amount of money received from the sales of handbags that month.(5 marks)
      2. The number of handbags sold that month. (2 marks)


MARKING SCHEME

SECTION I (50 marks)

  1. Without using a calculator, evaluate. (3 marks)
    -8+(-5) x(-8)-(-6)
    -3+(-8) ÷ 2 x 4

    -8 -5 x -8 +6
    -8 + 40 +6
    =38   

    -3 – 8 ÷ 2 x 4
    -3 – 4 x 4 = -19
    = 38/-19
    = -2


  2. Evaluate without using a calculator (2 marks)
    MathF4OPP1Q2
    17/7 - 11/6 = 25/42 X 6/5 = 5/7
    2/3 X 9/4 = 3/2 - 8/7 = 15/14
    5/7 X 14/5 = 2

  3. In fourteen years time, a mother will be twice as old as her son. Four years ago, the sum of their ages was 30 years. Find how old the mother was, when the son was born.( 4 marks)
    Let mother's years be x and son's be y
    x+14= 2(y +14)
    x+14= 2y + 28
    x-2y = 14

    (x-4)+(y-4) = 30
    x +y = 38

       (x- 2y= 14)
      -(x+ y= 38)
    = -3y=-24
    =3y=24
    y=8( son's age currently)
    x+y=38
    x=30( mother's age currently)
    At son's birth, mother's age = 30-8=22 years

  4. A Kenya bank buys and sells foreign currencies as shown below
      Buying
    (In KShs) 
    Selling
    (In Kshs) 
    1 Hong Kong dollar  9.74  9.77
    1 South African rand  12.03  12.11
    A tourist arrived in Kenya with 105000 Hong Kong dollars and changed the whole amount to Kenyan shillings. While in Kenya, she spent Kshs. 403,897 and changed the balance to South Africa rand before leaving for South Africa. Calculate the amount, in South African rand that she received. (3 marks)
    105000 x 9.74 = 1,022,700
    (1022700 - 403879)/51100

  5.    
    1. Using a ruler and a pair of compasses only, construct a quadrilateral PQRS in which PQ= 5cm, PS = 3cm, QR = 4cm,<PQR = 135°and <SPQ is a right angle. (2 marks)
    2. The quadrilateral PQRS represents a plot of land drawn to a scale of 1:4000. Determine the actual length of RS in meters. (2 marks)
      MathF4OPT1p1a5

  6. The ratio of goats to cows in a farm is 2:5 while the ratio of sheeps to cows is 3:4. If there are 15 sheep, how many animals are there in farm?(2 marks)
    Goats(G), Cows(C), Sheep(S)
    G:C:S
    2:5
        4 :3
    G :C  : S
    8 :20:15
    sheep = 15
    15+20+8= 43 Animals

  7. Mr.Maina who deals in electronics sells a radio to a customer at Ksh. 1440 after giving a discount of 10% but find that he makes a 20% profit. Find the profit Mr. Maina would make if he does not give a discount. (3 marks)

    100% = x
    90% = 1440

    1440 x 100
          90
    =1600

    120% =1600

    100% =x

    1600 x 100
          120
    =1333.3
    1600 – 1333.3 = 266.70

  8. Use the reciprocal and square table to evaluate to four significant figure, the expression.(3 marks)
    MathF4OPP1Q8
    1/0.03654 - 4.1512
    4.1512 = 17.231
    1/0.03654 = 1/3.654 x 10-2= 1/3.654 x 1/10-2
    0.237 x 102 = 27.37
    27.37 - 17.231=10.139

  9. Simplify the following expression completely. (3 marks)
    MathF4OPP1Q9
    12a2 - 3b2 = 3(2a +b)(2a-b)
    2a2-ab-b2 = 2a(a-b)+b(a-b)
    3(2a +b)(2a-b)
    2a(a-b)+b(a-b)
     =3(2a-b)
         a-b

    = 3(2a-b)/a-b

  10. Given that sin (x + 60)° = Cos (2x)°, find tan (x + 60)°. (3 marks)
    Sin x+ 60º = Cos 2x
    x+ 60 + 2x =90
    3x = 30
    x=10
    Tan (x+60)= Tan 70
    Tan 70 = 2.748, 2.7475

  11. The figure below shows triangle PQR in which PR = 12 cm, T is a point on PR such that TR = 4cm. line ST is Parallel to QR
    MathF4OPP1Q11
    If the area of triangle PQR is 336 cm2, find the area of the quadrilateral QRTS. (3 marks)
    12-4 = 8
    12/8= 3/2
    a.s.f = 9/12
    4/9 x 336 = 1491/3
    Area QRST = 336- 149 1/3 = 186 2/3
  12. A square brass plate is 2 mm thick and has a mass of 1.05 kg. The density of the brass is 8.4 g/cm3. Calculate the length of the plate in centimeters (3 marks)
    Volume = mass/density 
    1050cm3/8.4 = 125 cm3
    ∴ L x L x 0.2 cm = 125 cm3

    L2 = (125 cm)/0.2 = 625
    L = √(625 ) = 25 cm

  13. The diagram below represent a solid made up of a hemisphere mounted on a cone. The common radius is 6 cm and the height of the solid is 15cm.
    MathF4OPP1Q13
    Calculate the external surface of the solid (4 marks)
    2 x 3.142 x 36 = 226.224
    3.142 x 610.82 = 204.0
    Total = 430.224cm

  14. Solve the simultaneous inequalities given below and list all the integral values of x (3 marks)
    3-x/2x+1/32x+1/-3
    9-3x>2x +2
    7/5 >x
    x+1>-2x -1
    x > -2/3
    -2/3 < x < 12/5
    3x > -2

  15. A construction company employs technicians and artisans. On a certain 3 technicians and 2 artisans were hired and paid a total of Ksh 9,000.
    On another day the firm hired 4 technicians and one artisan and paid a total of Ksh 9500. Calculate the cost of hiring two technicians and 5 artisans in a day

    3t+2a=9000..... (i)
    4t+a =9500...... (ii)
    Multiply (ii)x2
    8t+2a =19000-
    3t+2a =9000
    5t = 10000
    t = 2000√
    substituting in (i) above
    3 x 2000 + 2a = 9000
    a = 3000/2
    a = 1500√
    t = 2000
    2 x 2000 + 5 x 1500=11500

  16. The figure below is not drawn to scale.
    MathF4OPP1Q15
    Find correct to 1 decimal place;
    1. Length PQ. (2 marks).
       7.62 + 4.82 – 2 x 7.6 x 4.8 cos 80
      = 57.76 + 23.04 – 12.67
      = 68.13 
      R = 68.13
      = 8.3

    2. Angle ABC (2 marks)
      Sin B/ 4.8 = Sin 80/8.254
      SinB = 0.5727
      B = sin -10.5727
      = 34.90
      .

 

SECTION II(50 marks)
CHOOSE ANY FIVE QUESTIONS IN THIS SECTION

  1. Two lines L1: 2y - 3x - 6 = 0 and L2: 3y+x-20 = 0 intersect at point A.
    1. Find the coordinates of A. (3 marks)
      2y – 3x = 6
      3y + x = 20
      2y - 3x = 6
      9y + 3x = 60
      11y = 60
      Y= 6
      X = 20 -18
      = 2
      Co-ordinates of A are (2,6)

    2. A third line L3 is perpendicular to L2 at point A. Find the equation of L3 in the form y = mx +c, where m and c are constants. (3 marks)
       L2: 3y = -x + 20
      y = - 1/3 x + 20 
      Gradient of perpendicular = 3
      (y-6)/(x-2) =3
      Y= 3X- 6 + 6
      Y = 3X

    3. Another L4 is parallel to L1 and passes through (-1, 3). Find the x and y intercepts of L(4 marks)
      Gradient of L4= gradient of L1
      =3/2
      (y-3)/(x-1) = 3/4
      2y -6 = 3X + 3
      2y - 3X = 9
      When X = 0 y = 4.5
      When y = 0 x = -3

  2. Town B is 180km on a bearing 0500 from town A. Another town C is on a bearing of 110° from town A and on a bearing of 150° from town B. A fourth town D is 240 km on a bearing of 320° from A. Using scale drawing, such that 1cm rep 30km,
    1. Show the relative position of the towns (4 marks)
    2. Using the diagram, find
      1. Distance AC (2 marks)
      2. Distance CD (2 marks)
      3. Compass bearing of C from D (2 marks)
        (representation, not to scale)
        MathF4OPT1p1a18
  3. The table shows the marks obtained by 40 candidates in an examination
    Marks   5-14 15-29 30-34 35-44  45-49 
    Frequency   2  12  7  15  x

    1. Find the value of X (1 mark)
      4
    2. Calculate the mean mark (2 marks)
      Σfx/Σf = 1287.5/40 = 32.1875
    3. On the grid provided below draw a histogram to represent the data (4 marks)
      CB  4.5 - 14.5  14.5 -29.5  29.5 - 34.5  34.5 - 44.5  44.5 - 49.5
       Fd  0.2  0.8  1.4  1.5  0.8
       Fx  19  264  224  595.5  188

      MathF4OPT1p1a19
    4. drawing a straight line on the graph above determine the median mark (3 marks)
      total area A = 10 x 0.2 = 2 
      B= 15 x 0.8 = 12
      C = 1.4 x y = 6 
      D = 1.5 x 10 = 15
      E = 0.8 x 5 = 4
      2 + 12 + 7 + 15 + 4 40/2= 20

      Point to draw the lie is 29.5 + 6/7 =30.36

  4. The figure below shows a triangle ABC inscribed in a circle. AC = 10cm, BC = 7cm and AB = 10cm.
    MathF4OPP1Q20
    1. Find the size of angle BAC. (3 marks)
      a2 =b2+ c2 – 2bc cos A
      72 = 102 + 82 - (2 x 10 x 8) cos A
      49=164- 60 cos A
      -116 =160 cos A
      Cos A =116/160
      Cos A = 0.725
      Cos1 0.725 = 43.53115
      <BAC = 43.53

    2. Find the radius of the circle. (2 marks)
      a/(sin sin A ) = 2R
      7/(sin 43.53) = 2R
      R =5.082cm

    3. Hence calculate the area of the shaded region. (5 marks)
      Sin 43.53 = 3.5/r
      r = 3.5/(sin sin 43.53 )= 5.082 cm
      Area of △OCB =1/2 ab sin θ
      = 1/2 x 5.082 sin 87.06
      = 12.896cm2
      Area of sector ACB
      =θ/360 πr2
      =87.06/360 x 22/7 x 5.082 = 19.630
      Shaded region
      (19.630 — 12.896) = 6.734 cm2


  5. The figure below represents a speed time graph for a cheetah which covered 825m in 40 seconds.
    MathF4OPP1Q21
    1. State the speed of the cheetah when recording of its motion started (1 mark)
      15m/s
    2. Calculate the maximum speed attained by the cheetah (3 marks)
      1/2 (15+h)×10+1/2 (10+30)h=825
      75+5h+20h=825
      25h=750

    3. Calculate the acceleration of the cheetah in:
      1. The first 10 seconds (2 marks)
        h=30m/s(i) = (30-15)/10
        = 1.5m/s2
      2. The last 20 seconds (1 mark)
        0-30/20 = -1.5m/s2
    4. Calculate the average speed of the cheetah in first 20 seconds (3 marks)
      [1/5 (15+30)×10+10×30]÷20
      = (225+300)÷20
      =26.25 m/s

  6. Given the quadratic function y=3x2 + 4x - 1
    1. Complete the table below for values of x ranging - 4≤ x ≤3. (2 marks)
       X -4 -3 -2 -1 0 1 2 3
       Y  30 13  2  -3  -2  5  18  37 
    2. Using the grid provided draw the graph of y = 3x2 + 4x —2 for -4≤ x ≤ 3 (3 marks)
    3. Using the graph, find the solution to the equations.
      1. 3x2+4x-2=0 (2 marks)
      2. 3x2+7x+2=0 (3 marks)
        MathF4OPT1p1a21
  7. The diagram represents a solid frustum with base radius 21cm and top radius 14cm. The frustum is 22.5cm high and is made of a metal whose density is 3 g/cm3.
    (Take π=22/7)
    MathF4OPP1Q23
    1. Calculate
      1. The volume of the metal in the frustum. (5 marks)
        Volume
        Ratio x/14 =(22.5+x)/21
        21X=14(22.5 +X)

        21= 14(22.5 + x)
        21x - 14x = 315
        x = 45cm
        Volume of whole cone
        =1/3 x 22/7 x 21 x 21 x 67.5 =31185 cm3
        Volume of small cone

        =1/3 x 22/7 x14 x 14 x 45 = 9240 cm3 
        Volume of frustum
        31185- 9240 = 21 945 cm3

      2. The mass of the frustum in kg. (2 marks)
        Mass of frustum
        Mass=21945x3g/cm3 =65835g 
        Mass in kg= (65.835 g)/1000g = 65835kg

    2. The frustum is melted down and recast into a solid cube. In the process 20% of the metal is lost. Calculate to 2 decimal places the length of each side of the cube.(3 marks)
      20% of 65.835kg = 13.167 kg
      65.835 - 13.167 = 52.668 kg
      Volume of material remaining =
      56682/3= 17556cm3
      Length of cube = ∛(17 556 )= 25.99 cm

  8. A saleswoman is paid a commission of 2% on goods sold worth over Ksh 100,000. She is also paid a monthly salary of Ksh 12,000.In a certain month, she sold 360 handbags at Ksh 500 each.
    1. Calculate the saleswoman’s earnings that month. (3 marks)
      360x500=180,000
      180,000-100,000= 80,000

      80,000x2/100 = 1600
      12000+1600=13600

    2. The following month, the saleswoman’s monthly salary was increased by 10%. Her total earnings that month were Ksh 17,600.
      Calculate:
      1. The total amount of money received from the sales of handbags that month.(5 marks)
        12000x110/100=13200 - basic salary
        17600-13200 =4400
        4400x100/2 = 220,000
        220,000+100,000= 320,000

      2. The number of handbags sold that month. (2 marks)
        320,000/500 = 640
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