MATHEMATICS PAPER 2 Marking Scheme - 2019 KCSE Prediction Answers Set 2

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SECTION A:50 MARKS
Answer all questions in this section

  1. Make q the subject of the formula. (3 mks)
    r=pq + 1/2tq2
    pp2 ans 1
  2. Solve for x in
    4cos2x = 3sinx + 3 for 0 ≤ xo ≤ 180o  (3 mks)
    pp2 ans 2
  3. The figure below shows two circles centre O1 and O2 which are 7.5cm apart and of radii 4cm and 2.5cm respectively.
    prediction 2 mathspp2q3
    Calculate the value of KL to 2 decimal places. (3 mks)
    pp2 ans 3
  4. Without using tables or calculators simplify (3mks)
    prediction 2 mathspp2q4
  5. Express as a single logarithm. (3mks)
    prediction 2 mathspp2q5
    pp2 ans 5
  6. The mass of some rabbits in a certain agriculture class were recorded as follows.
     Mass  1.0-1.9  2.0-2.9  3.0-3.9  4.0-4.9  5.0-5.9  6.0-6.9
     No. of rabbit  7  16  14  8  13  2
    Calculate the quartile deviation of the data to 3 s.f (3 mks)
    pp2 ans 6
  7. Given triangle AOB, O(0,0) A(2,0), B(0,4) and its image O'(√3/2,0) A'(√3,1), B'(-2,2√3). Find the matrix that maps OAB into O'A'B and describe the transformation represented by this matrix fully. (3mks)
    pp2 ans 7
  8. Ketepa tea worth Ksh.40 per kg is mixed with Sasini tea worth Ksh. 60 per kg in the ration 3:1 in what ratio should this mixture be mixed with Kericho tea worth Ksh. 50 per kg to produce a mixture worth Ksh. 47 per kg. (3mks)
    pp2 ans 8
  9. Calculate the area enclosed by the curve y = (x-1)2 + 4 and the line y = -x + 7 (3 mks)
  10. Expand in ascending process of x upto and including the term in x3 (3 mks)
    (3 - x - x2)4
    pp2 ans 10
  11. In the triangular based pyramid shown below, find the angle between planes VAC and VCB to 1 decimal place if the base is an equilateral triangle of side 10cm and edges VA = VB = VC = 16cm each. (4mks)
    prediction 2 mathspp2q11
    • Area AVB (½ × x × h) + ½(16 −x)
      = 76cm2 
      AN = 9.5 cm and CN = 3
      < ANB = 63.5o
  12. Calculate the area bounded by the curve y=x2 + 5 and lines y = 0 and x−3 = 0. Using mid-ordinate rule with 6 strips giving your answer to 4 decimal place. (4 mks)
     X  0.25  0.75  4.25  1.75  2.25  2.75
     Y  5.0625  5.5625  6.5625  8.0625  10.0625  12.5625
    h(y1 + y2 + y3 + .....+ yn)
    0.5 (5.0625 + 5.5625 + 6.5625 + 8.0625 + 10.0625 + 12.5625)
    ½(47.875) = 23.9375
    pp2 ans 12
  13. The matrix M is (41 32 ). Find the image of line y=2x + 1 under M. (3mks)
    pp2 ans 13
  14. From a metal rod of length 2.00m, 15 students each cut a piece 5cm long. If all the students had an error of ± 0.2cm in their cuts, determine the minimum and maximum length of the rod that remained. (3 mks)
    • (1.995 ≤ 2.00 ≤ 2.005) m
      15 × 5 = 75
      (4.8 ≤ 5 ≤ 5.2) m
      2.00m=200cm
      max cut 5.2 × 15 = 78 cm
      Max rod = 2.005 − 15(0.048) = 1.285 m
      min cut 4.8 × 15 = 72 cm
      min rod = 1.995 − 15(0.052) = 1.215 m
      max that remained 200 − 72 = 128 cm
      min that remained = 200 − 78 = 122 cm
  15. Given that where k is a constant and t is increased by 44% and b reduced by 88%, find the ratio by which x is changed. (3mks)
    pp2 ans 15
  16. The numbers 2, x and y form a geometric sequence. The numbers x, 30 and y form an arithmetic sequence. Find the possible value of x and y. ( 3mks)
    pp2 ans 16

SECTION B (50 MARKS)
Answer any five questions from this section

  1. A youth fund employee in Kiambu county lives in government premise in Thika. He pays a normal rent of sh. 1066 and is entitled to tax relief of sh. 1056 p.m. He also has an insurance scheme for which he pays sh. 5,300 pm and claims relief at 18% of the premium to a maximum of sh. 2000. His medical allowance is sh. 3855 p.m and his P.A.Y.E is sh. 5244.70pm. Use the tax table below to calculate the questions that follow.
     Income K£ p.a  Rate (%) per K£
     1 - 5808  10%
     5808 - 11280  15%
     11281 - 16752  20%
     16753 - 22224  25%
     22225 and above   30%
    1. His income in K£ p.a (4 mks)
      pp2 ans 17
      pp2 ans 17a
    2. His basic pay in Ksh p.m to the nearest Ksh. (3 mks)
      pp2 ans 17b
    3. If other deductions added upto sh. 3850 p.m, find his net pay p.m (3mks)
      pp2 ans 17c
  2.  
    1. A box contains n beads of which k are white and the rest are green. What is the probability of obtaining one of each colour in two successive drawing from the box.
      1. With replacement. (3mks)
      2. Without replacement. (3 mks)
    2. The figure below shows an equilateral triangle PQR of side 4cm. The shaded areas are sectors of circle centres P, Q and R with radii 1cm. If a point is selected at random from the triangle, find the probability that it lies in the unshaded region. (giving your answers in terms of pie and surd form) (4 mks)
      prediction 2 mathspp2q17b
  3.  
    1. Sketch the curve
      y = x(x+1)(x-2)  (5 mks)
      pp2 ans 19a
    2. A carpenter cuts a rectangular sheet of cardboard such that the sum of the length and width is 6m. Find the maximum area of the cardboard he cuts. (5mks)
      pp2 ans 19b
  4. A certain uniform supplier is required to supply two types of shirts. One for girls labelled G and the other for boys labelled B. The total number of shirts must not be more than 400. He has to supply more of the type G that type B. However the number of type G shirts must not be more than 300 and the number of type B shirts must not be less than 80. By taking x to be the number of type G shirts, and y the number of type B shirts.
    1. Write down in terms of x and y all the inequalities representing the information above. (4mks)
      • x + y ≤ 400
        x ≥ y
        x ≤ 300y ≥ 80 
    2. On the grid draw the inequalities and shade the unwanted regions. (3mks)
    3. Given that type G cost shs. 500 per shirt and type B costs sh. 300 per shirt.
      1. Use the graph (b) above to determine the number of shirts of each type that should be made to maximize the profit. (1 mk)
        • 210 shirts for girls
        • 109 shirts for boys
      2. Calculate the maximum possible profit. (2mks)
        • 210 × 500 + 190 × 300
          = 162,000
  5.  
    1. Using a ruler and a pair of compasses only, construct a triangle ABC such that AB = 8cm, and angle ABC = CAB = 75o. (2mks)
    2. Locate the locus of P such that angle APB = 52.5o and area of triangle APB = 20cm2. Locate the two possible loci as P1 and P2. Measure the distance P1 and P2 (4mks)
    3. On the opposite side of AB construct a rectangle ABXY such that its area is 48cm2. Shade a point r inside the rectangle ABXY such that angle ARB > 9 and <XRY > 90o. Locate and shade the region in which R lies. (4mks)
  6.  
    prediction 2 mathspp2q22
    In the triangle OAB above, OA = a and OB = b. Given that OK = 2KA and M is the midpoint of BK.
    1. Find in terms of a and b.
      1. BA = a − b
      2. BK = −b + 2/3d
      3. OM = OK + KM
        • 2/3d + 1/2KB
    2. Given further that OL = sOM and BL = pBA express OL in two ways hence or otherwise find the value of scalars S and P.
    3. P shows that points O, M and L are collinear.
  7. An aircraft leaves town p(30oS, 17oE) and moves directly northwards to Q(60oN,17oE). It then moved at an average speed of 300knots for 8 hours westwards to town R.
    Determine
    1. The distance PQ in nautical miles. (3 mks)
      PQ = 60 nm
      =60 × 90 = 5400nm
      or PQ = 90 × 2πR = 90 × 2 × 3.14 × 6370 = 5400.58 nm 
    2. The position of town R. (3 mks)
      pp2 ans 23b
    3. The local time at R if local time at Q is 3:12pm (2mks)
      • longitude difference = 80o
        80 × 4/360 = 5 hrs 20 min
        3.12 pm - 5 hrs 20 mins = 9.20 am
    4. The total distance moved from P to R in kilometers. (take 1nm = 1.853km) (2 mks)
      • Total difference = PQ + QR
        = (5400 + 2400) nm × 1.853
        =14,453.4 km
  8. The table below shows values obtained form an experiment.
     x  1  2  3  4  5  6
     y  2.70  5.70  11.15  22.62  45.2  90.51
    It is thought that y and x are connected by the formula y = A(b + x)
    1. By drawing a suitable linear graph, find the values of the constants A and b.
      • Log y = blogA + xlogA log Y = XlogA + blogA
    2. From the graph, estimate
      1. The value of x when y=24 (2mks)
        • log y4 = 1.38 = 1.4
          x=4
          blogA
          b × 0.3 = 0.2 
          b = 0.2/0.3 = 2/3 b =0.6667 = 0.7
      2. The value of y when x = 5.6 (1 mk)
        • log y = 1.875
          y = 71.989
          =74.99
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