KCSE 2014 Mathematics Alternative A Paper 2 Questions with Marking Scheme

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QUESTIONS

SECTION I (50 marks)
Answer all questions in this section in the spaces provided.

  1. The length of two similar iron bars were given as 12.5 m and 9.23 m. Calculate the maximum possible difference in length between the two bars. (3 marks)
  2. The first term of an arithmetic sequence is -7 and the common difference is 3.
    1. List the first six terms of the sequence; (1 mark)
    2. Determine the sum of the first 50 terms of the sequence. (2 marks)
  3. In the figure below, BOD is the diameter of the circle centre O. Angle ABD =300 and angle AXD = 700.
    Angles in a circle KCSE 2014
    Determine the size of ;
    1. Reflex angle BOC; (2 marks)
    2. Angle ACO. (1 mark)
  4. Three quantities L, M and N are such that L, varies directly as M and inversely as the square of N. Given that L = 2 when M = 12 and N = 6, determine the equation connecting the three quantities. (3 marks)
  5. The table below shows the frequency distribution of marks of students scored in a test.
    Marks  Frequency 
     1-10
     11-20  4
     21-30  11
     31-40  5
     41-50  3
    Determine the median mark correct to 2 s.f. (4 marks)
  6. Determine the amplitude and period of the function, y = 2 cos (3x - 45)0. (2 marks)
  7. In a transformation, an object with n area of 5 cm2 is mapped onto an image whose area is 30 cm2. Given the matrix of transformation is matrix KCSE 2014, find the value of x. (3 marks)
  8. Expand (3 - x )7 up to the term containing x4. Hence find the aproximate value of (2.8)7. (3 marks)
  9. Solve for the equation;
    2 log 15 - log x = log 5 + log (x - 4). (4 marks)
  10. The figure below represents a cuboid PQRSTUVW.
    PQ = 60 cm, QR = 11 cm, RW = 10 cm.

    cuboid KCSE 2014
    Calculate the angle between line PW and plane PQRS, correct to 2 decimal places. (3 marks)
  11. Solve the simultaneous equations; (4 marks)
    3x - y = 9
    x2 - xy = 4
  12. Muga bought a plot of land for Ksh 280 000. After 4 years, the value of plot was Ksh 495 000. Determine the rate of appreciation, per annum, correct to one decimal place. (3 marks)
  13. The shortest distance between two points A (400N, 200W) and B(00S, 200W) on the surface of the earth IS 8008 km. Given that the radius of the earth is 6370 km, determine the position of B.
    (Take π = 22/7 )(3 marks)
  14. Vectors r and s are such that r = 7i + 2j - k and s = - i + j - k. Find |r + s|. (3 marks)
  15. The gradient of a curve is given by gradient of curve KCSE 2014. The curve passes through the point (1, 0). Find the equation of the curve. (3 marks)
  16. The graph below shows the rate of cooling of a liquid with respect to time.
    average
    Determine the average rate of cooling of the liquid between the second and the eleventh minutes. (3 marks)



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

  1.  A paint dealer mixes three types of paint A, B and C, in the ratios A:B = 3:4 and B:C = 1:2. The mixture is to contain 168 litres of C.
    1. Find the ratio A:B:C. (2 mark)
    2. Find the required number of litres of B. (2 marks)
    3. The cost of A is Ksh 160, type B is Ksh 205 and type C is Ksh 100.
      1. Calculate the cost per litre of the mixture. (2 marks)
      2. Find the percentage profit if the selling price of the mixture is Ksh 182 per litre. (2 marks)
      3. Find the selling price of a litre of the mixture if if the dealer makes a 25% profit. (2 marks)
  2. In the figure below OS is the radius of the circle centre O. Chords SQ and TU are extended to meet at P and QR is perpendicular to QS at R. QS = 61 cm, PU = 50 cm, UT = 40 cm and PQ = 30 cm.
    angles in a circle kcse 2014
    1. Calculate the length of:
      1. QS; (2 marks)
      2. QR. (3 marks)
    2. Calculate correct to 1 decimal place:
      1. The size of angle ROS; (2 marks)
      2. The length of the minor arc QS. (3 marks)
  3. The table below shows income tax rates for a certain year.
    income rates kcse 2014
    A tax relief of Ksh 1162 per month was allowed. In a certain month, of that year, an employee's taxable income of that year in the fifth band was Ksh 2018.
    1. Calculate:
      1. The employees total taxable income in that month; (2 marks)
      2. The tax payable by the employee in that month. (5 marks)
    2. The employee's income included a house allowance of Ksh 15 000 per month.
      The employee contributed 5% of the basic salary to a co-operative society.
      Calculate the employee's net pay for that month. (3 marks)
  4. The dimensions of a rectangular floor of a proposed building are such that:
    - the length is greater than the width but at most twice the width;
    -the sum of the width and the length is, more than 8 metres but less than 20 metres. If x represents the width and y the length.
    1. Write the inequalities to represent the above information. (4 marks)
      1. Represent the inequalities in part (a) above on the grid provided. (4 marks)
      2. Using integral values of x and y, find the possible maximum area of the floor. (2 marks)
  5. Each morning Gataro does one of the following exercises:
    Cycling, jogging and weightlifting.
    He chooses the exercise to do by rolling a fair die. The faces of the die are numbered 1, 1, 2, 3, 4 and 5.
    If the score is 2, 3 or 5, he goes for cycling.
    If the score is 1, he goes jogging.
    If the score is 4, he goes weightlifting.
    1. Find the probability that:
      1. On a given morning, he goes for cycling or weightlifting; (2 marks)
      2. On two consecutive mornings he goes for jogging. (2 mark)
    2. In the afternoon, Gtaro plays either football or hockey but never both games. The probability that Gataro plays hockey in the afternoon is:
      1/3 if the cycled
      2/5 if he jogged and
      1/2 if he did weightlifting in the morning.
      Complete the tree diagram below by writing the appropriate probability on each branch. (2 marks)
      probability KCSE 2014
    3. Find the probability that on any given day:
      1. Gataro plays football; (2 marks)
      2. Gataro neither jogs nor plays football. (2 marks)
  6. In the figure below OA = a and OB = b. M is the mid point of OA and AN:NB = 2:1.

    Vectors KCSE 2014
    1. Express in terms of a and b:
      1. BA; (1 mark)
      2. BN; (1 mark)
      3. ON. (2 marks)
    2. Given that BX = hBX and OX = kON determine the values of h and k. (6 marks)
  7. Figure ABCD is a scale drawing representing a square plot of side 80 metres.
    Locus of a point KCSE 2014
    1. On the scale drawing, construct:
      1. The locus of a point P, such that it is equidistant from AD and BC. (2 marks)
      2. The locus of point Q such that <AQB = 600. (3 marks)
    2.  
      1. Mark on the drawing the point Q1, the intersection of the locus of Q and lines AD.
        Determine the length of BQ1, in metres. (1 marks)
      2. Calculate, correct to the nearest m2, the area of the region bounded by the locus of P, the locus of Q and the line BQ1. (4 marks)
  8. In an experiment involving two variables t and r, the following results were obtained.
    equation of line KCSE 2014
    1. On the grid provided, draw the line of best fit for the data. (4 marks)
    2. The variables r and t are connected by the equation r = at + k where a and k are constants.
      Determine:
      1. The values of a and k; (3 marks)
      2. The equation of the line of the best fit. (1 mark)
      3. The value of t when r = 0. (2 marks)

MARKING SCHEME

  1. Limits: 12.5 ± 0.05 m and 9.23 ± 0.005 m
    Maximum difference
    = 12.55 - 9.225
    = 3.325 m B1
  2.                    
    1. First 6 terms
      -7, -4, -1, 2, 5, 8
    2. Sum of 1st 50 terms
      S50 = 50 {2 × -7 + 49 × 3}
                2
      = 3325
  3.                
    1. angles BAC = 70º - 30º = 40º
      Reflex angles BOC = 360º - 80º
      = 280º
    2. angles ACO = 40º- 30º = 10º
  4. L = kM
          N2
    2 = k × 12
             36
    k = 6
    therefore: equation L = 6M
                                       N2 
  5.                       
    marks  frequency  c.f 
    1-10 
    11-20 
    21-30  11  17 
    31-40  22 
    41-50  25
    Median
    = 20.5 + 12.5 - 6 × 10
                        11
    = 20.5 + 5.91
    = 26.41
    = 26
  6.  Amplitude = 2
    Period = 360 = 120º
                   3
  7. Area scale factor = 30 = 6
                                 5
    4x - 2x + 2 = 6
    2x = 4
    x = 2
  8. (3 - x)7 = 37 - 7 (3)6 x + 21(3)5 x2 - 35(3)4 x+35 (3)3 x4 + ...
    = 2187 - 5103x + 5103x2 - 2835x3 + 945x4
    (2.8)7 = (3 - 0.2)7
    = 2187 - 5103(0.2) + 5103(0.2)2
    - 2835(0.2)3 + 945(0.2)4
    = 1349.352
  9. Log 152 = log 5 (x - 4)
           x
    152 = 5 x - 4
     x
    x2 - 4x - 45 = 0
    (x - 9) (x + 5) = 0
    x = 9 or -5
    x = 9
  10. PR = √602 + 112 = 61
    Tan θ = 10
                61
    θ = 9.31º
  11. 3x - y = 9 ................ × x
    x2 - xy = 4
    3x2 - xy = 9x
    x2 - xy = 4        
    2x2 = 9x - 4
    2x2 - 9x + 4 = 0
    (2x - 1) (x - 4) = 0
    x = 1 or x = 4
          2
    y = 3(1/2 ) - 9 or 3(4) - 9
    = - 71/2 or 3
  12. (1 + r )4 = 495000
          100     280000
    1 + r = 1.153
       100
    r = 15.3
  13. 8008 = 40 + θ × 2 × 22 × 6370
                  360             7
    40 + θ = 8008 × 360 × 7 = 72
                   2 × 22 × 6370
    θ = 72º - 40º
    = 32º
    Position of B(32º S, 20ºW) 
  14.                              
    14 ahgftydfa
    = 7
  15.                        
    15 ytfaytfda
  16. Temperature at the 2nd minute = 60º
    Temperature at the 11th minute = 18º
    Average rate of cooling
    16 ajgfdtya
  17.                              
    1.  A = 3 B, C = 2B
             4
      = A: B: C = 3 B: B: 2B
                        4
      = 3: 4: 8
    2. (168 × 4)litres
         8
      = 84 L
    3.                 
      1. 3 × 160 + 4 × 205 + 8 × 100
                  3 + 4 + 8 
        =Ksh 140
      2. 182 - 140 × 100%
           140
        = 30%
      3. Ksh 140 × 125
                        100
        =Ksh 175
  18.            
    1.                                                          
      1. (50 + 40) (50) = 30 (30 + x)
        4500 = 900 + 30 x
        30 x = 3600
        QS = x = 120 cm
      2. RS = 1/2QS
        =1/2(120) = 60cm
        QR = √612 - 602
        =11 cm
    2.                  
      1. Sin θ = 60
                    61
        θ = 79.6º
      2. Angle at the centre
        = 2 × 79.6
        = 159.2º
        Length of minor arc QS
        = 159.2 × 2π × 61
             360
        = 169.5 cm
  19.          
    1.                                               
      1. 38392 + 2108
        = Ksh 41000
      2. 10164 × 0.1 + 9576 × 0.15 + 9576 × 0.2 + 9576 × 0.25 + 2108 × 0.3
        = 1016.4 + 1436.4 + 1915.2 + 2394 + 632.4
        = Ksh 7394.4
        monthly income tax
        = 7394.4 - 1162
        = Ksh 6232.4
    2. Amount saved in coop society
      = 5 × (41000 - 15000)
       100
      = Ksh1300
      Nett pay
      41000 - (6232.4 + 1300)
      = Ksh 33467.6 
  20.                  
    1. y > x
      y ≤ 2x
      x + y < 20
      x + y > 8
    2.                  
      1.                      
        19 uaguyda
      2. Maximum area:
        9  × 10
        = 90 m2 
  21.                      
    1.                 
      1. 3 + 1
        6    6
        = 2
           3
      2. 2 × 2
        6    6
        = 1
           9
    2.                          
      21 jgtaytda
    3.                    
      1. P(Gataro plays football)
        = 1 × 2 + 1 × 3 + 1 × 1
           2     3    3    5     6    2
        = 37
           60
      2. P(neither jogs nor plays football)
        = 1 × 1 + 1 × 1
           2     3    6    2
        = 1
           4
  22.        
    1.                               
      1. BA = a - b
      2. BN = 1BA = 1(a - b)
                3         3
      3. ON = b + 1(a - b)
                       3
        =1a + 2b
          3      3
    2. BX = hBM = h(1/2a - b)
      OX = kON = k(1/3a +2/3b)
      also
      OX = OB + BX
      = b + h(1/2a  - b)
      k(1a + 2b) = b + h (1a - b)
         3      3                  2
      1ka = 1ha
      3        2
      1k = 1h = k = 3h.............(i)
      3      2            2
      2kb = b - hb
      3
      2k = 1 - h .....................(ii)
      3
      Substituting k = 3 h in (ii)
                              2
      2 (3h) = 1 - h = h = 1
      3  2                         2
      Substituting h = 1 in (i)
                               2
      k = 3 (1) = 3
            2  2      4 
  23.            
    1.                                              
      1.                
        23 uagytdga
      2. locus of P
        construction of 30º
        identification of centre
        drawing of arc
        Identifying region
        for radius and angle of sector
    2.                  
      1. 9.2 × 10 = 92 m
      2. area of region bounded by locus of P, locus of Q and line BQ1
        angle = 60º radius = 46 m
        = π × 462 × 60 
                         360
        = 1107.94
        ≈ 1108 m2
  24.          
    1.                      
      24 uaygdtyfa  
    2.                              
      1.  value of a
        = - 0.7
              3.5
        = -0.2
        value of k = 1.7
      2. equation: r = -0.2t + 1.7
      3. value of t when r = 0
        therefore: 0 =-0.2t + 1.7
        0.2 t = 1.7
        t = 1.7 = 8.5
              0.2

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