Mathematics Paper 1 Questions and Answers - Kassu Joint Mock Examination 2021

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INSTRUCTIONS TO CANDIDATES

  • Write your name and Admission number in the spaces provided at the top of this page.
  • This paper consists of two sections: Section I and Section II.
  • Answer ALL questions from section I and ANY FIVE from section II
  • All answers and workings must be written on the question paper in the spaces provided below each question.
  • Show all the steps in your calculation, giving your answer at each stage in the spaces
    below each question.
  • Non – Programmable silent electronic calculators and KNEC mathematical tables may be used, except where stated otherwise.

SECTION A (50 marks)

  1. Without using a calculator or tables, evaluate: (3 marks)
    1
  2. Solve the equation for x. 52x+1+52x-750 =0 (3 marks)
  3. Simplify 8mn-6m+8n2-6n (3 marks)
                      8n-6
  4. Use squares, square roots and reciprocals tables to evaluate the following giving your answer to 2 decimal places. (4 marks)
        1    +     2   
    √20.52   (6.23)2
  5. Susan made a loss of 20% by selling a blender at sh. 2,400. What profit would she have made had she sold it at sh. 3300? (3 marks)
  6. Solve for x and y using substitution method:
    1/3(x + y) – 2 = 0
    1/4(x - y) = 1 (3 marks)
  7. The number of sides of two regular polygons differs by one. If the sum of the interior angles of these polygons is in the ratio 2:3, calculate the number of sides of each polygon and name them. (3 marks)
  8. Solve for x in the following equation: Sin(½ x -10º)=Cos2x (3 marks)
  9. A vehicle moves at an initial speed of 20m/s with a constant acceleration of 2m/s2 for five seconds before breaks are applied. If the car comes to rest under constant deceleration 4 seconds, determine the total distance travelled during the 9 seconds (3 marks)
  10. Simplify completely the expression (4 marks)
    2
  11. A point P divides the line AB shown below internally in the ratio 2:3. By construction, find the position P and measure AB.
    A B (3 marks)
    3
  12. In the figure below, O is the centre of the circle and reflects angle AOC =1420. Find angle ABC. (3 marks)
    4
  13. A tourist arrived in Kenya with 10,000 US dollars which he converted to Ksh on arrival. He spent Kshs.428,500 and converted the remaining amount to Sterling pounds. How much did he receive in Sterling pounds? The currency exchange rate of the day was as follows; (3 marks)
    Currency Buying Selling
    1 Sterling pound 135.50 135.97
    1US dollar  72.23 72.65  
  14. Adam harvested 200 bags of wheat from 2 ha of his farm. How many bags of wheat would he harvest from 16 ha if he maintained the rate? (3 marks)
  15. Complete the solid below whose length is 7cm (3 marks)
    5
  16. Write down three inequalities which fully describe the unshaded region R in the figure below (3 marks)
    6

SECTION B (50 marks)

  1. Three points P, Q and S are pm the vertices of a triangular plain field. P is 400m from Q on a bearing of 3000 and R of 550m directly south of P.
    1. Using a scale of 1 cm to represent 100m on the ground, draw a diagram to show the position of the points. (3 marks)
    2. Use the scale drawing to determine;
      1. The distance and bearing of Q from R. (2 marks)
      2. The bearing and distance of point S from P given that point S is directly 600m East of R. (3 marks)
      3. The bearing and distance of Q from S. (2 marks)
  2. A bus travelling at a speed of 80km/hr left Mombasa at 8.00am for Nairobi. Two hours later, a car travelling at a speed of 100km/hr left Nairobi for Mombasa.
    1. Given that the distance between both cities is 500km, find the time of the day when the two vehicles met. (6 marks)
    2. After meeting, the speed of both vehicles dropped to 60km/hr due to traffic jam. At what time did each vehicle arrive at its destination? (4 marks)
  3. The figure below represents an histogram of heights against age brackets of members of a village.
    7
    Using the figure above,
    1. Develop a frequency distribution table (3marks)
    2. Using the table in (a) above find;
      1. The mean. (3marks)
      2. The median class (1mark)
      3. The median (3marks)
  4. The diagram below shows a container base made of a frustum of a square pyramid. The top is a solid frustum of a cone.
    8

    AB = BC = 12 cm
    EF = 6
    Height of 10 and 6 cm respectively.

    1. Calculate the surface area of the bottom solid. (5 marks)
    2. Calculate the surface area of the top side. (4 marks)
    3. Calculate the total area. (1 mark)
  5.      
    1. Complete the table below for the function
      y=-x2+10x (2marks)
      X -1 0 1 2 3 4 5 6 8 10
      Y     9     24       0
    2. On the grid provided draw the graph of y=-x2+10x (3marks)
      9
    3. Using the graph above solves the equations:
      1. 10x-x2=0 (2marks)
      2. x2-7x-8=0 (3marks)
  6. Two lines L1=2y-3x-6 and L2=3y+x-20=0 intersect at point A.
    1. Find the coordinates of A (3marks)
    2. A third line L3 is perpendicular to L2 at point A. Find the equation of L3 in form of y=mx+c, where m and c are constants. (3marks)
    3. Another line L4 is parallel to l1 and passes through (-1,3). Find the x-intercept and the y-intercept of L4. (4marks)
  7.    
    1. PQRS is a quadrilateral with vertices P(1,4), Q(2,1), R(2,3) and S(6,4). On the grid provided, plot the quadrilateral. (1 mark)
      9
    2. Draw P’Q’R’S’ the image of PQRS under a positive quarter turn about the origin and write down its co-ordinates. (3 marks)
    3. Draw P”Q”R”S” the image of P’Q’R’S’ under an enlargement scale factor -1 and center (0,0) and write down its co-ordinates. (3 marks)
    4. Determine the matrix of a single transformation that maps PQRS onto P”Q”R”S (3 marks)
  8. A curve whose equation is 3y=9-18x+27/2 x2- 3x3 turns at points P and R.
    1. Find the coordinates of P and R (5 marks)
    2. Determine the nature of points P and R (3 marks)
    3. Sketch the curve (2 marks)

 

 

 

 



MARKING SCHEME

SECTION A (50 marks)

  1. Without using a calculator or tables, evaluate: (3 marks)
    1
    1
    (203/135 ÷ -429/56)-2 
    (203/135 ÷ 56/-429)-2 
    (135x - 429)2
       203 x 56
  2. Solve the equation for x. 52x+1+52x-750 =0 (3 marks)
    52x+1 + 52x = 750
    52x.51+52x = 75
    52x(5 + 1)= 750
    52x(6) = 750
    52x=53
    2x = 3
    x = 1.5
  3. Simplify 8mn-6m+8n2-6n (3 marks)
                      8n-6
    2m(4n - 3)+2n(4n - 3)(4n - 3)(2m + 2n)
                2(4n - 3)                  2(4n - 3)
    =2(m + n)
            2
    = m + n
  4. Use squares, square roots and reciprocals tables to evaluate the following giving your answer to 2 decimal places. (4 marks)
        1    +     2   
    √20.52   (6.23)2
    √20.56 = 4.5299
    (6.23)2 = 38.81
    1(rec 4.530) + 2(rec 38.81)
    0.2208 + 2 x 0.02576
    = 0.2208 + 0.05152
    = 0.27232
  5. Susan made a loss of 20% by selling a blender at sh. 2,400. What profit would she have made had she sold it at sh. 3300? (3 marks)
    80% = 2400
    100% = ?
    100% x 2400
           80
    = 3000
  6. Solve for x and y using substitution method:
    1/3(x + y) – 2 = 0
    1/4(x - y) = 1 (3 marks)
    x + y - 6 = 0
    x - y - 4 = 0
    x = y + 4
    (y + 4)+ y = 6
    2y = 2
    y = 1 
    x = 1 + 4
    = 5
  7. The number of sides of two regular polygons differs by one. If the sum of the interior angles of these polygons is in the ratio 2:3, calculate the number of sides of each polygon and name them. (3 marks)
        180 (n - 2)  2/3
     180(n + 1 - 2)
    3(n - 2) = 2(n - 1)
    3n - 6 = 2n - 2
    3(n - 2) = 6 - 2
    n = 4 - Quadrilateral
    n + 1 = 5 - Pentagon
  8. Solve for x in the following equation: Sin(½ x -10º)=Cos2x (3 marks)
    (½x - 10)+ 2x = 90
    2.5x = 100
    x = 40
  9. A vehicle moves at an initial speed of 20m/s with a constant acceleration of 2m/s2 for five seconds before breaks are applied. If the car comes to rest under constant deceleration 4 seconds, determine the total distance travelled during the 9 seconds (3 marks)
    2
    D=(30 + 20)5 + ½ x 30 x 4
              2
    = 125 + 60
    = 185m
  10. Simplify completely the expression (4 marks)
    2
            (25x2 - 9y2)
           225                      
     1  (25x2 + 30xy + 9y2)
    225
           (5x - 3y)(5x + 3y)     
    25x2 + 15xy + 15xy + 9y2
          (5x - 3y)(5x + 3y)    
    5x(5x - 3y)+3y(5x + 3y)
     (5x - 3y)(5x + 3y)
    (5x - 3y)(5x + 3y
    5x - 3y
       5x + 3y
  11. A point P divides the line AB shown below internally in the ratio 2:3. By construction, find the position P and measure AB.
    A B (3 marks)
    3
    • for a line forming <with AB
    • for parallel line
    • for point P
  12. In the figure below, O is the centre of the circle and reflects angle AOC =1420. Find angle ABC. (3 marks)
    4
    <ABC= 180º - ½(142)
    = 180º - 71
    = 109º
    Alt<ABC = 360 - 142
                           3
    =218
        2
    =109º
        2
  13. A tourist arrived in Kenya with 10,000 US dollars which he converted to Ksh on arrival. He spent Kshs.428,500 and converted the remaining amount to Sterling pounds. How much did he receive in Sterling pounds? The currency exchange rate of the day was as follows; (3 marks)
    Currency Buying Selling
    1 Sterling pound 135.50 135.97
    1US dollar  72.23 72.65  
    1usd = 72.23
    (10000 x 72.23)
    sh.722300
    spend 428500
    (722300 - 428500)
    =sh 293800
    1 sterling pound = sh135.97
    sh293800
    =293800 = 2160.8 sterling pounds
       135.97
  14. Adam harvested 200 bags of wheat from 2 ha of his farm. How many bags of wheat would he harvest from 16 ha if he maintained the rate? (3 marks)
    16 x 200
     2
    = 1600bags
  15. Complete the solid below whose length is 7cm (3 marks)
    5
    showing hidden side
    complete solid
  16. Write down three inequalities which fully describe the unshaded region R in the figure below (3 marks)
    6
    L1
    x/4 + y/4 = 1 
    x/3(12) + y/4(12) =1(12)
    4x + 3y = 12
    8 + 0 = 12
    8 < 12
    4x + 3y ≤ 12

    L2
    x/4y/4 = 1 (6,0)
    6/4 - ¼ 
    5/4 > 1

    L3
    x/4y/4 ≥1
    or
    -x + y ≥ -4
    or
    x - y ≥ -4
    y ≥ -3

SECTION B (50 marks)

  1. Three points P, Q and S are pm the vertices of a triangular plain field. P is 400m from Q on a bearing of 3000 and R of 550m directly south of P.
    1. Using a scale of 1 cm to represent 100m on the ground, draw a diagram to show the position of the points. (3 marks)
    2. Use the scale drawing to determine;
      1. The distance and bearing of Q from R. (2 marks)
      2. The bearing and distance of point S from P given that point S is directly 600m East of R. (3 marks)
      3. The bearing and distance of Q from S. (2 marks)
        6
  2. A bus travelling at a speed of 80km/hr left Mombasa at 8.00am for Nairobi. Two hours later, a car travelling at a speed of 100km/hr left Nairobi for Mombasa.
    1. Given that the distance between both cities is 500km, find the time of the day when the two vehicles met. (6 marks)
      7
      D = t x s
      = 2 x 80
      = 160km
      500 - 160 = 340km
      A.S = 80 + 100
      = 180km/hr
      t = d/s 

      T =340
           180
      17/9 hrs
      = 18/9hrs →1hr . 531/3
      10:00:00
      1   53  20
      11:53:20 am
    2. After meeting, the speed of both vehicles dropped to 60km/hr due to traffic jam. At what time did each vehicle arrive at its destination? (4 marks)
      d = t x s
      17/9 x 100
      Distance covered at the time of meeting
      18/9 x 80 = 1511/9km
      Bus = 160 + 1511/9 = 3111/9km
      Car = 1511/9km
      remaining distance
      Bus = 500 - 3111/= 1888/9km
      Car = 500 - 1511/9 = 3488/9km

      Time taken by
      Bus =1888/9 = 3hr 9mins
                60
      Car = 3488/9 = 5hrs 49mins         
                 60
      Arrival time
      Bus 11:53
             3  09
        15.20hrs
      Car 11:53
            5   49
      17:42 hrs
  3. The figure below represents an histogram of heights against age brackets of members of a village.
    7
    Using the figure above,
    1. Develop a frequency distribution table (3marks)
      f x xf cf
      0-15
      15-20
      20-55
      55-65
      120
      50
      560
      60  
      850 
      7.5
      17.5
      47.5
      60

      900
      875
      26600
      3600  
      31975
      120
      170
      730
      790

    2. Using the table in (a) above find;
      1. The mean. (3marks)
        x =31975
               790
        = 40.475
      2. The median class (1mark)
        790 = 395
         2
      3. The median (3marks)
        20 +(395 - 170)35
                    730
        = 30.788
  4. The diagram below shows a container base made of a frustum of a square pyramid. The top is a solid frustum of a cone.
    8

    AB = BC = 12 cm
    EF = 6
    Height of 10 and 6 cm respectively.

    1. Calculate the surface area of the bottom solid. (5 marks)
      9
      AC2 = 144 + 144
      AC2 = 288
      AC = √288
      OC = √288
                 2
      (VC2) = 162(√288)2
                                2
      = 256 + 72
      VC = √328
      H12 = (√328)2 - 62
      = 328 - 36
      (H1) = √292
      1 = 
      H1     H
        h  1 =
      √292    16
      h
      13/8 (√292)
      A = [½ x 12 x √292 -½ x 6 x 3/8 √292]4
      =[√292(6 - 9/8)]4
      =[83.304cm2]4
      = 333.216cm22 = 12 x 12 = 144cmcm
      A2 = 12 x 12 = 144cm2 
      A3 = (6 x 6) - 22/7 x 3 x 3
      = 36 - 282/7
      = 7.7142cm2
      Total = 333.216 + 144 + 7.7142
      = 484.93cm
    2. Calculate the surface area of the top side. (4 marks)
      A1 = 22/7 x 2 x 2
      = 12.5714cm2 
       h/Hr/R
        h  2/3
      6+h
      3h = 12 + 2h
      h = 12
      L2 = 122 + 22 
      L = 12.166
      12.1662/3
          L
      2L = 36.4966
      L = 18.248
      A2 = πRL - πrl
      =π x 30.412
      =95.581
      total area = 12.57N + 95.581
      =108.152
    3. Calculate the total area. (1 mark)
      484.93 + 108.15
      = 593.082cm2
  5.      
    1. Complete the table below for the function
      y=-x2+10x (2marks)
      X -1 0 1 2 3 4 5 6 8 10
      Y -11  0 9 16 21 24 25 24 16 0
    2. On the grid provided draw the graph of y=-x2+10x (3marks)
      10
    3. Using the graph above solves the equations:
      1. 10x-x2=0 (2marks)
        x = 0 or x = 0
      2. x2-7x-8=0 (3marks)
        y = -x2 + 10x + 0
        +
        0 = x2 - 7x - 8
        y = 3x + 8 
        x = 0, y = -8, (0,8)
        x = 2, y = -2, (2,-2)
        x = -1 or x = 8
  6. Two lines L1=2y-3x-6 and L2=3y+x-20=0 intersect at point A.
    1. Find the coordinates of A (3marks)
      (2y - 3x = 6) 3
      (3y + x = 20) 2
      6y - 9x = 18
      6y + 2x = 40
      -11x = -22
      x = 2
      2y - 3(2) = 6
      2y = 12
      y = 6
      A(2,6)
    2. A third line L3 is perpendicular to L2 at point A. Find the equation of L3 in form of y=mx+c, where m and c are constants. (3marks)
      3y = -x + 20
      y = -1/3 x + 20/3 
      m = 3
      y - 63/1
      x - 2
      y - 6 = 3x - 6
      y = 3x + 0
    3. Another line L4 is parallel to l1 and passes through (-1,3). Find the x-intercept and the y-intercept of L4. (4marks)
      2y = 3x + 6
      y = 3/2 x + 3
      y - 33/2
      x + 1
      2(y - 3) = 3(x + 1)
      2y - 6 = 3x + 3
      2y = 3x + 9
      y = 3/2 x + 9/2 
      x intercept y = 0
      2/33/2x = -9/2x/3
      x = -3
      y intercept  x =0
      y = 9/2
      y = 4.5
  7.    
    1. PQRS is a quadrilateral with vertices P(1,4), Q(2,1), R(2,3) and S(6,4). On the grid provided, plot the quadrilateral. (1 mark)
      11
    2. Draw P’Q’R’S’ the image of PQRS under a positive quarter turn about the origin and write down its co-ordinates. (3 marks)
      P'(-4,1) Q'(-1,2) R'(-3,2) S'(4,-6)
    3. Draw P”Q”R”S” the image of P’Q’R’S’ under an enlargement scale factor -1 and center (0,0) and write down its co-ordinates. (3 marks)
      P"(-4,1) Q"(-1,2) R"(-3,2) S"(4,-2)
    4. Determine the matrix of a single transformation that maps PQRS onto P”Q”R”S (3 marks)
      The transformation of +90º about the origin
      for constructions
  8. A curve whose equation is 3y=9-18x+27/2 x2- 3x3 turns at points P and R.
    1. Find the coordinates of P and R (5 marks)
      [3y = 9 - 18x + 22/2x2 - 3x3]x 1/3
      y = 3 - 6x + 4.5x2 - x3
      dy/dx = -6 + 9x - 3x2
      at st.p dy/dx = 0
      -6 + 9x -3x2 = 0
      x2 - 3x + 2 = 0
      x(x-1) -2(x - 1) = 0
      x = 1, y = 3 - 6 + 4.5 + 1
      y = ½
      when x = 2
      y = 3 - 12 + 18 - 8
      y = 1
      p1(1,½) P2(2,1)
    2. Determine the nature of points P and R (3 marks)
      y" = 9 - 6x
      at x = 1
      y = 9 - 6 = 3
      p1(1,½) min point
      y" = 9 - 6x at x = 2
      y" = 9 - 6(2) = -3
      P2(2,1) max point
    3. Sketch the curve (2 marks)
      12


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