KCSE 2014 Mathematics Alternative A Paper 1 Questions with Marking Scheme

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QUESTIONS

SECTION I (50 marks)

Answer all questions in this section in the spaces provided.

  1. Ntutu had cows, sheep and goats in his farm. The number of cows was 32 and the number of sheep was 12 times the number of cows. The number of goats was 1344 more than the number of sheep. If he sold ¾ of the goats. Find the number of the goats that remained.    (4mark)
  2. Use the prime factors of 1764 and 2744 to evaluate                      (3 marks)
    factors KCSE 2014
  3. The mass of a solid cone of radius 14 cm and height 18 cm is 4.62 kg. Find its density in g/cm3. (3 marks)
  4. The figure below represents a triangular prism ABCDEF. X is a point on BC.
    prism shape KCSE 2014
    1. Draw a net on the prism (2 marks)
    2. Find the distance DX. (1 mark)
  5. A business man makes a profit of 20% when he sells a carpet for Ksh 36 000. In a trade fair he sold one such carpet for Ksh 33 600. Calculate the percentage profit made on the sale of the carpet during the trade fair. (3 marks)
  6. Simplify   (3 marks)
    square roots KCSE 2014
  7. The area of a sector of a circle, radius 2.1 cm, is 2.31 cm2. The arc of the sector subtends an angle θ, at the centre of the circle. Find the value of in radians correct to 2 decimal places. (2 marks)
  8. Expand and simplify  (x + 2y)2 - (2y - 3)2   (2 marks)
  9. A plane leaves airstrip L and flies o a bearing of 0400 to airstrip M, 500 km away. The plane then flies on a bearing of 3160 to airstrip N. The bearing of N from L is 3500. By scale drawing, determine the distance between airstrips M and N. (4 marks)
  10. The sum of interior angles of a regular polygon is 18000. Find the size of each exterior angle. (3 marks)
  11. A cow is 4 years 8 months older than a heifer. The product of their ages is 8 years. Determine the age of the cow and that of the heifer. (4 marks)
  12. Solve 4 ≤ 3x - 2 < 9 + x hence list the integral numbers that satisfies the inequality.(3 marks)
  13. The figure below shows a rectangular container of dimensions 40 cm by 25 cm by 90 cm. A cylindrical pipe of radius 7.5 cm was fitted in the container as shown.
    cuboid KCSE 2014
    Water is poured into the container in the space outside the pipe such that the water level is 80% the height of the container. Calculate the amount of water, in litres, in the container correct to 3 significant figures. (4 marks)
  14. A minor arc of a circle substends an angle of 1050 at the centre of the circle. if the radius of the circle is 8.4 cm, find the length of the major arc. ( Take π = 22/7) (3 marks)
  15. Twenty five machines working at the rate of 9 hours per day can complete a job in 16 days. A contractor intends to complete the job in 10 days using the similar machines working at the rate of 12 hours per day. Find the number of machines the contractor requires to complete the job.    (3 marks)
  16. Points A (-2, 2) and B (-3, 7) are mapped onto A' (4, -10) and B' (0, 10) by an enlargement. Find the scale factor of the enlargement. (3 marks)



SECTION II (50 marks)
Answer any five questions in this section in the spaces provided.

  1. A line L passes through points (-2,3), and (-1, 6) and is perpendicular to line P at ( -1, 6).
    1. Find the equation of L. (2 marks)
    2. Find the equation of P in the form ax + by - c, where a, b and c are constants.   (2 marks)
    3. Given that another line Q is parallel to L and passes through point (1, 2) find the x and y intercepts of Q.  (3 marks)
    4. Find the point of intersection of lines P and Q. (3marks)
  2. The lengths, in cm, of pencils used by pupils in a standard one class in a certain day were recorded as follows
    3 7 9 9 20 14 10 6 8 13
    14 3 17 13 8 12 5 15 14 15
    7 12 11 6 10 19 9 14 6 9
    10 16 13 9 12 11 10 7 10 11

    1. Using a class width of 3, and starting with the shortest lengths of the pencils, make a frequency distribution table for the data. (2 marks)
    2. Calculate ;
      1. The mean length of the pencils. (3 marks)
      2. The percentage of pencils thst were longer than 8 cm but shorter than 15 cm. (2 marks)
    3. On the grid provided, draw a frequency polygon for the data.  (3 marks)
  3. The figure below represents a speed time graph for a cheetah which covered 825 m in 40 seconds.
    distance time graph kcse 2014
    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; (2marks)
      2. The last 20 seconds. (1 mark)
    4. Calculate the the average speed of the cheetah in the first 20 seconds. (3 marks)
  4. The figure below shows  a right pyramid VABCDE. The base ABCDE is a regular pentagon.
    AO = 15 cm and VO = 36 cm.
    right pyramid kcse 2014
    Calculate:
    1. The the area of the base correct to 2 decimal places;  (3 marks)
    2. The length of AV;  (1 mark)
    3. The surface area of the pyramid correct to 2 decimal places; (4 marks)
    4. The volume of the pyramid correct to 4 significant figures. (2 marks)
  5. Complete the table below for the function y = x2 - 3x + 6 in the range  -2 ≤ x ≤ 8.   (2 marks)
    trapezium rule KCSE 2014
    1. Use the trapezium rule with 10 strips to estimate the area bounded by the curve, y = x2 - 3x + 6, the lines x = -2, x = 8 and the x -axis. (3 marks)
    2. Use the mid-ordinate rule with 5 strips to estimate the area bounded by the curve, y = x2 - 3x + 6, the lines x = -2, x = 8 and the x -axis. (2 marks)
    3. By integration, determine the actual area bounded by the curve y = x2 - 3x + 6, the lines x = -2, x = 8 and the x -axis. (3 marks)

    1. Using a pair of compasses and a ruler only, construct;
      1. Triangle ABC in which AB = 5 cm. <BAC = 300 and angle ABC = 1050; (3 marks)
      2. A circle that passes through the vertices of the triangle ABC. Measure the radius. (3 marks)
      3. The height of triangle ABC with AB  as the base. Measure the height. (2 marks)
    2. Determine the area of the circle that lies outside the triangle correct to 2 decimal places. (2 marks)
  6. The figure below represents a piece of land in the shape of the quardilateral in which AB = 240 m, BC = 70 m, CD = 200 m, <BCD = 1500 and < ABC = 900.
    angles KCSE 2014
    Calculate;
    1. The size of <BAC correct to 2 decimal places; (2 marks)
    2. The length of AD correct to one decimal place. (4 marks)
    3. The area of the piece of land, in hectares, correct to 2 decimal places. (4 marks)
  7. The equation of a curve is given by y = x3-4x2-3x .
    1. Find the value of y when x = -1. (1 mark)
    2. Determine the stationary points of the curve. (5 marks)
    3. Find the equation of the normal to the curve at x  = 1.  (4 marks) 

MARKING SCHEME

  1. Cows = 32
    Sheep = 32 × 12
    = 384
    Goats = 384 + 1344
    = 1728
    Number of goats that remained
    = 1 × 1728
       4
    = 432
  2.                     
    2 JATFGDA
    =2 × 3 × 7
          2 × 7
    = 3
  3. Volume = 1 × 22 × (14)2 × 18
                   3    7
    = 3696 cm3
    Density = 4.62 × 1000
                         3696
    = 1.25 g/cm3
  4.      
    4 HJAGYTDAGD
    DX = 5.3 ± 0.1
    √ measurements and angles
    √ complete net (labelled)
  5. C.P. for carpet
    = 36000 × 100
               120
    = 30000
    % profit made during trade fair
    = 33600 - 30000 × 100
              30000
    = 12%
  6.                
    6 JAUYDGA
    = 27 × 25
           9
    = 75
    √ manipulation of all indices or equivalent
    √ simplification
  7. θ × π × 2.1 × 2.1 = 2.31

    θ  = 2.31 × 2
          2.1 × 2.1
    = 1.05º
  8. (x + 2y)2 - (2y - 3)2
    = (x2 + 4xy + 4y2) - (4y2 - 12y + 9)
    = x2 + 4xy + 12y - 9
  9.                 
    9 HHGAGTDA
    Distance MN = 6.8 × 100
    = 680 km
    √ location of M
    √ location of N
  10. (2n - 4) ×  90 = 1800
    180n = 2160
    n = 12
    size of each exterior angle
    = 360 = 30º
        12
  11. let age of cow be x years
    therefore: x(x - 42/3) = 8
    3x2 - 14x - 24 = 0
    (3x + 4) (x - 6) = 0
    x = 6 or - 4
                   3
    Age of cow = 6 years
    Age of heifer = 11/3 years
  12. 4 ≤ 3x - 2 < 9 + x
    4 ≤ 3x - 2      3x - 2 < 9 + x
    6 ≤ 3x            2x < 11
    x ≥ 2           x < 51/2
    therefore: 2 ≤ x < 51/2
    Integral values
    2, 3, 4, 5
  13. Volume of water in container
    = 80 × 90(40 × 25 -  π × 7.52)
      100
    = 59276.54975
    59276.54975
         1000
    = 59.3
  14. Angle for major arc = 360 - 105
    = 255º
    Length of arc = 255 × 2 × 8.4 × 22
                           360                    7
    = 37.4 cm 
  15. Amount of work = 25 × 16 × 9
    Machines required
    = 25 × 16 × 9
          12 × 10
    = 30 M1
  16.                  
    16 hgfvytafda
    = 4
  17.                        
    1. Equation of L
      gradient = 6 - 3
                     -1 - 2
      = 3
      equation = y - 6 = 3
                      x + 1
      = y - 3x = 9
    2. equation of P
      = y - 6 =- 1
         x + 1     3
      3y + x = 17
    3. equation of Q
      = y - 2 = 3
         x - 1
      y = 3x - 1
      x intercept
      when y = 0 = x = 1
                                 3
      y intercept
      when x = 0 =  y =-1
    4. Intersection of lines P and Q
      3y + x = 17..(i)
      y - 3x =- 1.. (ii)
      3y + x = 17
      3y - 9x =- 3
      10x = 20 = x = 2
      subset 3y + 2 = 17 = y = 5
      therefore: point of intersection (2, 5)
  18.                            
    1.                      
      class  3-5  6-8  9-11  12-14  15-17  18-20 
      frequency  3  8 13   10  2
    2.                                        
      1. mean length = Σfx
                               Σf
        = 4 × 3 + 7 × 8 + 10 × 13 + 13 × 10 + 16 × 4 + 19 × 2
                                                40
        = 10.75
      2. = 23 × 100
           40
        = 57.5%
    3.                                        
      18 ajgdya
  19.                                  
    1. 15 m/s
    2. maximum speed
      1 × (15 + h) × 10 + 1 (10 + 30)h = 825
      2                             2
      75 + 5h + 20h = 825
      25h = 750
      h = 30 m/s
    3.                        
      1. = 30 - 15
               10
        = 1.5 m/s2
      2. = 0 - 30 =-1.5 m/s2
             20
    4. [1 (15 + 30) × 10 + 10 × 30] ÷ 20
       2
      = (225 + 300) ÷ 20
      = 26.25 m/s
  20.                        
    1. base area
      = 1 × 15 × 15 sin 72 × 5
         2
      = 534.97
    2. Length AV
      =b jagtda
    3. Area of triangular faces:
         AB     =   15   
      sin 72      sin 54
      AB = 15 sin 72
                sin 54
      = 17.63
      therefore: area
      c jafgytdfa
      = 334.89
      Total area = 334.89 × 5 + 534.97
      = 2209.42
    4. volume of pyramid
      = × 534.97 × 36
         3
      = 6419.63 cm2
      = 6420 (4 s.f.)
  21.              
    1.                            
      -2  -1  5 6 7 8
      y 16  10  10  16 24 34 46
    2. Area using trapezium rule
      = 1/2 × 1[16 + 46 + 2(10 + 6 + 4 + 4 + 6 + 10 + 16 + 24 + 34)]
      = 1/2 [62 + 2(114)]
      = 145
    3. Area using mid-ordinate rule
      = 2 × (10 + 4 + 6 + 16 + 34)
      = 140
    4. Area using integration method
      21 ajgydta
  22.                          
    1.                        
      1.                    
        22 ahvdfyta
      2. radius = 3.5 ± 0.1
      3. height construction
        height = 3.4 ± 0.1
    2. area of circle outside triangle
      = π × 3.52 - 1/2 × 3.4 × 5
      = 29.98
  23.                  
    1.  tan θ = 70 
                  240
      = 0.2917
      θ = 16.26º
    2. AC = √702 + 2402
      = 250 m
      angle ACD = 150º - (90º - 16.26º)
      = 76.26º
      AD2 = 2002 + 2502 - 2 × 200 × 250 cos 76.26
      AD = √40000 + 62500 - 100000 cos 76.26º
      = 280.6
    3. Area of plot
      = 1/2 × 240 × 70 + 1/2 × 250 × 200 sin 76.26º
      = 8400 + 24284.59
      = 32684.59 m2
      = 32684.59
          10000
      = 3.27 ha
  24.                  
    1. Value of y when x =-1
      y =-1 - 4 + 3 =-2
    2. Stationary points
      dy = 3x2 - 8x - 3
      dx
      for stationary points
      3x2 - 8x - 3 = 0
      (3x + 1)(x - 3) = 0
      x =- 1 or x = 3
             3
      when x =- 1 , y = 14
                      3         27
      when x = 3 , y =-18
      therefore:  stationary points
      - 1 ,  14
        3    27
      and (3, -18)
    3. Equation of normal to curve:
      gradient of tangent at x = 1
      dy = 3 - 8 - 3 =- 8
      dx
      gradient of normal
      = 1
         8
      therefore: equation of normal at x = 1
      y + 6 = 1
      x - 1     8
      y + 6 = 1 x - 1
                  8      8
      y = 1 x - 6 1
            8         8

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