Mathematics Paper 1 Questions and Answers - Kassu Jet Joint Exams 2020/2021

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INSTRUCTIONS TO CANDIDATES

  • 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)

Answer all questions in this section

  1. Without using a calculator or tables, evaluate: (3 marks)
    mathsp1kassu20q1
  2. Solve the equation for x. 5(2x+1)+52x − 750 = 0 (3 marks)
  3. Simplify  (3 marks)
    mathsp1kassu20q3
  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(1/2x − 10°) = Cos2x (3 marks)
  9. A vehicle moves at an initial speed of 20 m/s with a constant acceleration of 2 m/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)
    mathsp1kassu20q10
  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. (3 marks)
    mathsp1kassu20q11
  12. In the figure below, O is the centre of the circle and reflects angle AOC = 142°. Find angle ABC. (3 marks)
    mathsp1kassu20q12
  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)
    mathsp1kassu20q15
  16. Write down three inequalities which fully describe the unshaded region R in the figure below (3 marks)
    mathsp1kassu20q16

SECTION B (50 marks)

Answer any five questions from this section

  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.
    mathsp1kassu20q19
    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.
    mathsp1kassu20q20
    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 = −x+ 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=-x+ 10x (3marks)
      graphpaperq21b
    3. Using the graph above solves the equations:
      1. 10x − x= 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)
      graphpaperq21b
    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 + 27/2x2 − 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

  1.  
    mathsp1kassu20ans1
  2. 52x+1 + 52x = 750
    52x.51 + 52x = 750
    52x(5+1) = 750
    52x(6) = 750
    52x = 125
    52x = 53
    2x = 3
    x = 1.5

  3. 2m(4n − 3) + 2n(4n−3) = (4n − 3)(2m + 2n)
             2(4n − 3)                       2(4n − 3)
    = 2(m + n)
              2
    = m+n
  4. √20.56 = 4.5299
    (6.23)2 = 38.81
    1(Reciprocal 4.530) + 2 (Reciprocal 38.81)
    0.2208 + 2 × 0.02576
    = 0.2208 + 0.05152
    = 0.27232
  5. 80% = 2400
    100% = ?
    100 × 2400
          80
    = 3,000
    3300 − 3000
    = 300
  6. x + y − 6 = 0
    x − y − 4 = 0
    x = y + 4
    (y + 4) + y = 6
    2y = 2
    y = 1
    x = 1 + 4
    = 5
  7.    180 (n − 2)   = 2
    180 (n + 1 −2)    3
    3 (n−2) = 2(n−1)
    3n − 6 = 2n − 2
    3n − 2n = 6 − 2
    n = 4 - Quadrilateral
    n + 1 = 5 - Pentagon
  8. (1/2x − 10) + 2x = 90
    2.5x = 100
     x = 40
  9.  
    mathsp1kassu20ans9
    v − u = a
        t
    v − 20 = 2
        5
    v − 20 = 10
    v = 30
    D = (30 + 20)5 + 1/2 × 30 × 4
                 2
    = 125 + 60
    = 185 m
  10.  
    mathsp1kassu20ans10
  11.  
    mathsp1kassu20ans11
  12.  
    mathsp1kassu20ans12
    ∠ABC = 180° − 1/2(142) 
    = 180° − 71
    = 109°
    Alt ∠ ABC = 360 − 142
                             2
    = 218/2
    = 109°
  13. 1 US dollar = 72.23
    (10000 × 72.23)
    Sh. 722300
    Spend 428500
    (Sh. 722300 − 428500)
    = Sh. 293, 800
    1 sterling pound = sh. 135.97
    ?                       = sh. 293, 800
    293, 800 = 2160.8 sterling pounds
     135.97
  14. Bags     Farm
    200        2ha
    ?            16ha
    16/2 × 200 
    = 1600 bags
  15.  
    mathsp1kassu20ans15
  16.  
    mathsp1kassu20ans16
  17.  
    mathsp1kassu20ans17
  18.  
    1.  
      mathsp1kassu20ans18
      D = T × S
         = 2 × 80
          = 160 km
      500 − 160 = 340 km
      A.S = 80 + 100
            = 180 km/h
      T = D/S
      T = 340/180
         = 17/9
         = 1 8/9 hrs → 1 hr 53mins 20 secs
      10:00:00
      +1:53:20
      11: 53:20 a.m
    2. D = T × S
         = 17/9× 100
      Distance covered at the time of meeeting
      18/9 × 80 = 151 1/9 Km
      Bus = 160 + 151 1/9 = 311 1/9 Km
      Car = 151 1/9 Km
      Remaining distance
      Bus = 500 − 311 1/9 Km = 188 8/9 Km
      Car = 500 − 151 1/9 Km = 348 8/9 Km
      Time taken by
      Bus = 188 8/9 ÷ 60 
            = 3 hr 9 mins
      Car = 348 8/9 ÷ 60 = 5 hrs 49 min
      Arrival time
      Bus 11:53
              3:09
             1502 hrs
      Car 1153
              549
            1742hrs
  19.  
    1.  x  f  x  xf  cf
       0 - 15  120  7.5  900  120
       15 - 20  50   17.5  875  170
       20 - 55  560  47.5  26600  730
       55 - 65  60  60  3600  790
         850   31975  
    2.  
      1. x̄ = 31975/790
            = 40.475
      2. 790/2 = 395th
        20 - 55

      3. 20 + (395 − 170)35
                      730
        = 30.788
  20.  
    mathsp1kassu20ans20
  21.    
    1.  
       X  −1  0  1  2  3  4  5  6  8  10
       Y  −11  0  9  16  21  24  25  24  16  0
    2.  
      mathsp1kassu20ans21
    3.  
      1. x = 0 or x = 10
      2. 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
  22.  
    1. (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. 3y = −x + 20
      y = −1/3x + 20/3
      m = 3
      y − 6 = 3
      x − 2    1
      y − 6 = 3x − 6
      y = 3x + 0
    3. 2y = 3x + 6
      y = 3/2x + 3
      y − 3   = 3
      x + 1      2
      2(y−3) = 3 (x + 1)
      2y − 6 = 3x + 3
      2y = 3x + 9
      y = 3/2x + 9/2
      x - intercept y = 0
      2/3 × 3/2x = −9/2 × 2/3 
      x = − 3
      y - intercept x = 0
      y = 9/2
      y = 4.5
  23.  
    1.  
      mathsp1kassu20ans23
    2. P' (−4, 1) Q'(−1,2) R'(−3,2) S'(−4,6)
    3. P" (4, −1) Q"(1, −2) R"(3, −2) S"(4, −2)
    4. The transformation of +90° about the origin
      ✓1 for constructions
  24.  
    1. [3y = 9 − 18 x + 27/2x2 − 3x3]1/3
      y = 3 − 6x + 4.5x2 − x3
      dy/dx = − 6 + 9x − 3x2
      at s.t.p dy/dx = 0
      −6 + 9x − 3x2 = 0
      x2 − 3x + 2 = 0
      x(x−1) − 2(x−1) = 0
      x = 1 or x = 2
      when x = 1, y = 3 − 6 + 4.5 + 1
      y = 1/2
      when x =2, y = 3 − 12 + 18 − 8
      y = 1
      P1 (1, 1/2) P2 (2,1)
    2. y" = 9 − 6x
      at x = 1
      y = 9 − 6 = 3
      P1 (1, 1/2) min point
      y" = 9 − 6x at x = 2
      y" = 9 − 6(2) = − 3
      P2 (2,1) max point
    3.  
      mathsp1kassu20ans24

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