Mathematics Paper 2 Questions and Answers - Kapsabet Mock Exams 2021/2022

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INSTRUCTIONS TO THE CANDIDATES

  • This paper contains two sections; Section I and Section II.
  • Answer all the questions in section I and only five questions from Section II.
  • All workings and answers must be written on the question paper in the spaces provided below each question.
  • Non programmable silent electronic calculators and KNEC Mathematical tables may be used EXCEPT where stated otherwise.
  • Show all the steps in your calculations, giving your answers at each stage in the spaces below each question.

SECTION I (50MKS)

  1. Simplify by rationalising the denominator
    √ 2 + √3 (3mks)
     √6 - √3
  2. Find the value of x in the equation log10 (2x-1) + log103 = log10 (8x-1). (3mks)
  3. Find the compound interest on sh. 200,000 for 2 years at 14% pa. Compounded semi-annually. (3mks)
  4. The ratio of 12th to 10th term in a geometric series is 9:1. Find the common ratio. (3mks)
  5.      
    1. Expand (2 – ¼x)5 (2mks)
    2. Use your expansion to find the value of (1.96)5 correct to 3 decimal places (2mks)
  6. Chord WX and YZ intersect externally at Q. The secant WQ = 11cm and QX = 6cm while ZQ = 4cm.
    1
    1. Calculate the length of chord YZ. (2mks)
    2. Using the answer in (a) above, find the length of the tangent SQ. (2mks)
  7. Given that 2is a singular matrix, find the possible values of y. (3mks)
  8. The masses to the nearest kg of 50 adults were recorded as follows:
    Mass (kg) Frequency (f)
    45 – 50
    51 – 56
    57 – 62
    63 – 68
    69 – 74
    75 – 80
    2
    10
    11
    20
    6
    1
    Calculate the quartile deviation. (3mks)
  9. P varies as the cube of Q and inversely as the square root of R. If Q is increased by 20% and R decreased by 36%, find the percentage change in P. (3mks)
  10. Solve 8 cos2 x – 2 cos x – 1 = 0 (3mks)
  11. Make χ the subject of the formula: (3mks)
    A = √3 + 2χ
            5 - 4χ
  12. The position vectors of A and B are given as a= 2i-3j+4k and b= -2i-j+2k respectively. Find to 2decimal places, the length of the vector(AB). (3mks) 
  13. Find the centre and the radius of a circle whose equation is x2-6x+y2-10y+30=0 (3mks)
  14. A point (x, y) is mapped onto (13, 13) by two transformations M followed by T where 3Find the point (x y) (3mks)
  15. Given that 2 ≤ A ≤ 4 and 0.1 ≤ B ≤ 0.2. Find the minimum value of   AB  (3mks)
                                                                                                              A - B 
  16. In a transformation, an object with area 9cm2 is mapped onto an image whose area is 54cm2. Given that the matrix of transformation is 4find the value of x (3mks) 

    SECTION II (50MKS)
  17. The table below shows the rates of taxation in a certain year.
    Income in K£ pa Rate in Ksh per K£
    1 – 3900
    3901 – 7800
    7801 – 11700
    11701 – 15600
    15601 – 19500
    Above 19500
    2
    3
    4
    5
    7
    9
    In that period, Juma was earning a basic salary of sh. 21,000 per month. In addition, he was entitled to a house allowance of sh. 9000 p.m. and a personal relief of ksh.105 p.m He also has an insurance scheme for which he pays a monthly premium of sh. 2000. He is entitled to a relief on premium at 15% of the premium paid.
    1. Calculate how much income tax Juma paid per month. (7mks)
    2. Juma’s other deductions per month were cooperative society contributions of sh. 2000 and a loan repayment of sh. 2500. Calculate his net salary per month. (3mks)
  18. Wainaina has two dairy farm A and B. Farm A produces milk with 3 ½ percent fat and farm B produces milk with 4 ¾ percent fat. Determine;
    1. The total mass of milk fat in 50kg of milk from farm A and 30kg from farm B. (3mks)
    2. The percentage of fat in a mixture of 50kg of milk from A and 30kg of milk from farm B. (2mks)
    3. Determine the range of values of mass of milk from farm B that must be used in a 50kg mixture so that the mixture may have at least 4 percent fat. (5mks)
  19. A cupboard has 7 white cups and 5 brown ones all identical in size and shape. There was a blackout in the town and Mrs. Kamau had to select three cups, one after the other without replacing the previous one.
    1. Draw a tree diagram for the information. (2mks)
    2. Calculate the probability that she chooses.
      1. Two white cups and one brown cup. (2mks)
      2. Two brown cups and one white cup. (2mks)
      3. At least one white cup. (2mks)
      4. Three cups of the same colour. (2mks)
  20.    
    1. Complete the table below, giving the values correct to 2 decimal places (2mks)
      Xº  15º 30º 45º 60º 75º 90º 105º 120º 135º 150º 165º 180º
      Cos 2Xº 1.00  0.87   0.00 -0.5   -1.00   -0.5 0.00 0.50 0.87 1.00
      Sin(Xº+30º) 0.50 0.71 0.87 0.97 1.00   0.87 0.71 0.50   0.00   -0.50
    2. Using the grid provided draw on the same axes the graph of y=cos 2Xº and y=sin(Xº+30º) for 0º≤X≤180º. (4mks)
      5
    3. Find the period of the curve y=cos 2x0 (1mk)
    4. Using the graph, estimate the solutions to the equations;
      1. sin(Xº+30º) = cos 2Xº (1mk)
      2. Cos 2Xº=0.5 (1mk)
  21. The For a sample of 100 bulbs, the time taken for each bulb to burn was recorded. The table below shows the result of the measurements.
    Time(in hours) 15-19 20-24 25-29 30-34 35-39 40-44 45-49 50-54 55-59 60-64 65-69 70-74
    Number of bulbs 6 10 9 5 7 11 15 13 8 7 5 4
    1. Using an assumed mean of 42, calculate
      1. the actual mean of distribution (4mks)
      2. the standard deviation of the distribution (3mks)
    2. Calculate the quartile deviation (3mks)
  22.       
    1. Using a ruler and a pair of compasses only, construct a parallelogram ABCD such that AB=9 cm, AD=7 cm and angle BAD=60º. (3mks)
    2. On the same diagram, construct:
      1. The locus of a point P such that P is equidistant from AB and AD; (1mk)
      2. The locus of a point Q such that Q is equidistant from B and C; (1mk)
      3. The locus of a point T such that T is equidistant from AB and DC; (1mk).
    3.    
      1. Shade the region R bounded by the locus of P, the locus of Q and the locus of T. (1mk)
      2. Find the area of the region shaded in (d)(i) above. (3mks)
  23. The points A (1,4), B(-2,0) and C (4,-2) of a triangle are mapped onto A1(7,4), B1(x,y) and C1 (10,16) by a transformation N =6Find
    1. Matrix N of the transformation (4mks)
    2. Coordinates of B1 (2mks)
    3. AIIBIICII are the image of A1B1C1 under transformation represented by matrix M =7Write down the co-ordinates of AIIBIICII(2mks)
    4. A transformation N followed by M can be represented by a single transformation K.
      Determine K (2mks)
  24. The roof of a ware house is in the shape of a triangular prism as shown below
    8
    Calculate
    1. The angle between faces RSTU and PQRS (3mks)
    2. The space occupied by the roof (3mks)
    3. The angle between the plane QTR and PQRS (4mks)


MARKING SCHEME

  1. (√2 +√3) (√6 + √3)
    (√6 - √3) (√6 + √3)
    √12 + √6 + √18 + 3
            6 – 3
    2√3 + √6 + 3√2 + 3
                3 
  2. 3(2x-1) = 8x-1
    6x-3 = 8x-1
    -2x = 2
    x = -1

  3. 9
    = 200,000 (1.3107960)
    = Sh. 262159.20
    I = 262159.20 – 200000 = Sh.62,159
  4. 12th term = ar11
    10th term = ar9
    ar11 =
    ar9     1
    r11-9 = 9
    r2 = 9
    r = ± 3
    r = 3 or -3
  5.      
    1. (2 – ¼ x)5 = 25 + (24)(5)(-¼x)2 +
      10(22)(-¼x)3+ 5(2)(-¼x)4 + (-¼x)5
      =32 – 20x + 5x25/8x3 + 5/128x41/1024 x5
    2. 1.965 = 32 - 20(0.16) + 5(0.16)25/8 (0.16)3 +5/128(0.16)41/1024(0.16)5
      =28.925
  6.      
    1. QW x QX = QY x QZ
      11 x 6 = 4(a+4)
      4a +16 = 66
      4a = 50
      a = 25
    2. QS2 = QY x QZ
      = 4(4+12.5)
      QS = √66
      = 8.124
  7. x(x-1) – 3x(x+1) = 0
    x2 – x – 3x2 – 3x = 0
    -2x2 -4x = 0
    -2x (x + 2) = 0
    x = 0 or x = -2
  8.       x          f     cf
    45 – 50     2      2
    51 – 56    10    12
    57 – 62    11    23
    63 – 68    20    33
    69 – 74     6     39
    75 – 80     1    40
    ¼ x 50 =12.5th = [56.5 + 12.5 – 12] 6
                                                  11
    = 56.77kg
    ¾ x 50 = 37th ;
    62.5 + [37.5 – 23] 6
                      20
    = 66.85kg
    Quartile deviation = ½ (66.85 – 56.77)
    = 5.04
  9. P = KQ3
          √R
    P1 = K (1.2Q)3
            √0.64R
    = 1.728KQ3
          0.8√R
    = 2.16 KQ3
           √3
    2.16 – 1 x 100
         1
    = 116%
  10. Let cos x be y
    8y2 – 2y – 1 = 0
    (4y + 1) (2y – 1) = 0
    y = - ¼ or ½
    cos x = ¼ ⇒ x = 75.52
    Angle in 2nd and 3rd quadrant
    . : . x = 104.48, 255.52
    Cosx = ½ ⇒ x = 60º
    Angle in 1st and 4th quadrant.
    x = 60º, 300º
    .:. x = 104.48, 255.52º, 60º 300º
  11. A² = 3 + 2χ     A² (5 - 4χ) = 3 + 2χ
           5 - 4χ
    5A² - 4A²χ = 3 + 2χ
    5A² - 3 = 2χ + 4A²χ
    5A² - 3 = χ(2 + 4A²)
    χ= 5A² - 3  
         4A² + 2

  12. 10
    /AB/= √16 + 4 + 4
    = √24
    /AB/=4.90
  13. x2-6x+9+y2-10y+25= -30+9+25
    (x-3)2+(y-5)2=4
    R=2
    C (3,5)

  14. 11
    3x+y=17
    2x+4y=10
    X=5.8
    Y=-0.4
    (-0.4,5.8)
  15. 2×0.1
    4-0.1
    =0.2
      3.9
    =2/39
  16. Area scale factor =54/9= 6
    4x – 2(x – 1) = 6
    x = 2
  17.    
    1. Taxable income = 21,000 + 9000
      p.a = sh. 30,000
      30000 x 12 = K₤ 18,000 p.a
              12
      2 x 3900 = 7,800
      3 x 3900 = 11,700
      4 x 3900 = 15,600
      5 x 3900 = 19,500
      7 x 2400 = 16,800
                        71,400
      15/100 x 2000 = 300
      Total relief p.a = (300 + 1056) 12
      = sh. 16,272
      Tax paid 71400 – 16272 = sh. 55, 128
      P.A.Y.E 55128 = sh 4594
                     12
    2. Total deductions = 4594 +2000 + 2000 + 2500 = sh. 11,094 per month
      Net salary = 30,000 – 11,094
      = sh. 18,906
  18.    
    1. 7/200 x 50 + 19/400 x 30
      1.75 + 1.425
      = 3.175
    2. 3.175 x 100
        80
      = 3.96875%
    3. let the masses be x
      [19/400 x + 7/200(50 – x)]100 = 4
                    50
      [1.25x + 1.75]100= 4
               50
      1.25 x + 175 = 200
      1.25x = 25
      x = 25 
          1.25
      x = 20
      x > 20
  19.      
    1.    
      18
    2.    
      1. (7/12 x 6/11 x 5/10) + (7/12 x 5/11 x 6/10) + (5/12 x 7/11 x 6/10)
        = 21/44
      2. (7/125/11 x 4/10) + (5/127/11 x 4/10) + (5/12 x 4/11 x 7/10)
        = 7/22
      3. (5/12 x 4/11 x 7/10) + (5/127/11 x 4/10) + (7/125/11 x 4/10) + (5/127/11 x 6/10) + (7/12 x 5/10 x 6/10) + (7/12 x 6/115/10) + (7/12 x 5/11 x 6/10) + (5/127/11 x 6/10)
        = 427/440
  20.      
    17

  21. Class Mid point X f t =X - A 
          C 
    ft  t2 ft2
    15-19
    20-24
    25-29
    30-34
    35-39
    40-44
    45-49
    50-54
    55-59
    60-64
    65-69
    70-74
    17
    22
    27
    32
    37
    42
    47
    52
    57
    62
    67
    74
    6
    10
    9
    5
    7
    11
    15
    13
    8
    7
    5
    4
    -5
    -4
    -3
    -2
    -1
    0
    1
    2
    3
    4
    5
    6
    -30
    -40
    -27
    -10
    -7
    0
    15
    26
    24
    28
    25
    24
    25
    16
    9
    4
    1
    0
    1
    4
    9
    16
    25
    36
    150
    160
    81
    20
    7
    0
    15
    52
    72
    112
    125
    144
          ∑f=100 ∑ft=28   ∑ft2=873
    1.    
      1. x = 42 + (28/100 x 5) = 43.4
      2. 873/100 - (28/100)2 = 0.7856
    2. [49.5+ (75-63) 5.29.5] x ½
                    13
      = 12.31
  22.      
    16

  23. 12
    A+4b = 7 ……….(i) x 4
    4a + 16b ………..10 (ii)
    4a + 16b = 28
    4a - 2b = 10
     18b =18
    b = 1
    a = 9-4 = 3
    c+4d =4 ………. (iii) x 4
    4c – 2d = 16 ………..(iv)
    4c + 16d = 16
    4c-2d=16
       20d=0
    d =0
    c =4
    13
  24.    

    1. 14
      Required angle is θ
      82=102+122-2x10x12cosθ
      64=244-240 cosθ
      -180= -240 cosθ
      cosθ =180
                 240
      θ= cos -1 180 = 41.41θ
                    240
    2. Space =volume
      = ½ x 6 x10sin 41.41ºx24
      = 60sin41.41ºx24
      =952.5m3

    3. 15
      TA= 10 sinθ= 10xsin41.41
      Tan∞=10xsin41.41
                       24
      tan∞=0.2756
      = 15.41º

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