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

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MATHEMATICS
PAPER 2

INSTRUCTIONS TO CANDIDATES

  • Write your name and Admission number in the spaces provided at the top of this page.
  • 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

  1. Use logarithms tables to evaluate (4mks)
    ³√36.72 x (0.46)²
        185.4
  2. T is a transformation represented by the matrix  KapPreM Mathp2q2under T, a square of area 10cm2 is mapped onto a square of area 110cm2. Find the value of x (3mks)
  3. Given that 2Cos(2x-30º)=-6/5 find x where 180º < x < 360º (3mks)
  4. Make A the subject of the formula (3mks)
    T= 2m/n √L-A/3k
  5. A quantity P is partly constant and partly varies inversely as square of t. P = 6 when t = 6 and p = 18 when t = 3. Find t when p = 11 (3mks)
  6.              
    1. Expand (5 + x/2)⁶ up to the term in x³. (2mks)
    2. Use your expansion to estimate the value of (11/6)⁶ . Correct to one decimal place. (2mks)
  7. Solve for x in the equation. (3 Mks)
    Log8 (x + 6) – Log8 (x – 3) = 2/3
  8. Solve for x and y in the simultaneous equation below. (3 mks)
    xy + 6 = 0
    x – 2y = 7
  9. The size of each interior angle of a regular polygon is five times the size of the exterior angle. Find the number of sides of the polygon. (3mks)
  10. If      1         2 + 2√5  = a + b√c find the value of a, b and c (3 mks)
        3- √5          3 + √5
  11. The data below shows marks scored by 8 form four students in Molo district mathematics contest 44, 32, 67, 52, 28, 39, 46, 64.Calculate the mean absolute deviation. (3 Mks)
  12. Steve deposited ksh.50, 000 in a financial institution in which interest is compounded quarterly. If at the end of second year he received a total amount of ksh79, 692.40. Calculate the rate of interest p.a (3 Mks)
  13. The points with coordinates (5,5) and (-3,-1) are the ends of a diameter of a circle
    Centre A
    Determine:
    1. The coordinates of A (1mk)
    2. The equation of the circle, expressing it in form x2 + y2 + ax + by + c = 0 Where a, b, and c are constants (2mks)
  14. Pipe x can fill an empty tank in 3 hours while pipe y can fill the same tank in 6 hours. When the tank is full, it can be emptied by pipe z in 8 hours. pipe x and y are opened at the same time when the tank is empty. If one hour later pipe z is also opened, find the total time taken to fill the tank. (3marks)
  15. Fatima bought maize and beans from Kami. She mixed the maize and beans in the ratio 3: 2 she bought the maize at sh.90 per kg and the beans at sh.150 per kg. If she was to make a profit of 30% what would be the selling price of 1kg of the mixture. (3mks)
  16. Given  KapPreM Mathp2q16aand KapPreM Mathp2q16bfind A‾1 B-1 (3mks)

SECTION B

  1.              
    1. A figure whose co-ordinates are A(-2, -2), B(-4, -1), C(-4, -3) and D (-2, -3) undergoes successive transformations ERS; where E, R and S are transformations represented by the matrices, On the grid provided, show the figure ABCD and its image under the successive transformations ERS. (6mks)
    2. Find the matrix representing the single transformation mapping the image found in (a) above back the object figure ABCD. (2mks)
      Marks 10-19 20-29 30-39 40-49 50-59 60-69 70-79 80-89 90-99
      Frequency  2  6  10  16  24  20  12 8 2
    3. Triangle PQR has vertices at P(2, 2), Q(4, 1) and R(6, 4). On the same grid, show the image of triangle PQR under a shear with line y = 2 invariant and point R(6, 4) is mapped onto R1(2, 4). (2mks)
  2. The following are marks by form four students in a mathematics test.
    Using an assumed mean of 54.5, calculate the
    1. Mean mark (4mks)
    2. Variance (4mks)
    3. Standard deviation (2mks)
  3. The diagram below represents a Cuboid ABCDEFGH in which FG =4.5cm, GH =8cm HC=6m
    KapPreM Mathp2q19
    1. Calculate the length FC (2mrks)
    2.          
      1. The size of the angle between the lines FC and FH (2mks)
      2. Size of the angle between the line AB and FH. (2mks)
    3. The size of the angle between the planes ABHE and the plane FGHE.(2mks)
    4. The total surface area of the cuboid (closed) (2mks)
  4.              
    1. Complete the table below, giving all your values correct to 2 d. p. for the functions y = cos x and y = 2cos (x + 30)º
      60º 120º 180º 240º 300º 360º 420º 480º 540º
      Cos x 1.00     -1.00   0.50        
      2Cos(x+30) 1.73   -1.73   0.00          
    2. For the function y = 2cos (x + 30)0
      State:
      1. The period (1mk)
      2. Phase angle (1mk)
    3. On the same axes draw the waves of the functionsy = cos x and y = 2cos (x + 30)º for 0º ≤ x ≤ 540.Use the scale 1cm rep 30º horizontally and 2cm rep 1 unit vertically. (4mks)
    4. Use your graph above to solve the inequality 2cos (x + 30o) ≤ cos x (2mks)
  5. A teacher had 5 red, 6 black and 9 blue pens in a box. The pens were all identical except for the colour.
    1. If one pen is picked from the box, what is the probability that it is
      1. Red. (1mk)
      2. Not black. (1mk)
    2. The teacher asked a student to pick two pens from the box, one at a time, without replacement. Find the probability that
      1. Both pens are of the same colour. (3mks)
      2. They are of different colours. (2mks)
    3. If the first student was allowed to take away two blue pens and another student was asked to pick two pens without replacement. What is the probability that the second student picked pens of same colour? (3mks)
  6. In the figure below, PQR is the tangent to the circle at Q. TS is a diameter and TSR and QUV are straight lines. QS is parallel to TV. Angle SQR = 35° and TQV = 60°.
    KapPreM Mathp2q22
    1. Find the following angles, giving reasons for each answer.
      1. QTS. (2mks)
      2. QRS. (2mks)
      3. QVT. (2mks)
      4. UTV. (2mks)
      5. QUT. (2mks)
  7. Use ruler and a pair of compasses only in this question
    1. Construct triangle ABC such that AB = 6cm, AC=BC and angle ACB = 135° 4mks
    2. On one side only construct the locus of P such that:
      1. < APB = 67.5° 1mk
      2. area of triangle , APB = 9cm² 3mks
    3.         
      1. Locate P1 and P2 the two possible positions of P which satisfy the two conditions above 1mk
      2. Measure the distance between P1 and P2. 1mk
  8. An arithmetic progression has the first term a and the common difference d.
    1. Write down the third, ninth and twenty – fifth terms of the progression. (3 Mks)
    2. The progression is increasing and the third, ninth and twenty-fifth terms form the first three Consecutive terms of a geometric progression. If the sum of the seventh term and twice the sixth term of the arithmetic progression is 78.
      Calculate
      1. The first term and the common difference (5 Mks)
      2. The sum of the first nine terms of the arithmetic progression (2 Mks)

MARKING SCHEME

  1.        
    No Log
    36.72            
    (0.46)2 ⇒ 2 (‾1.6628)






    3.474 x 10-1
    =0.3474
     KapPreM Mathsp2qa1
  2. A.S.F = det
    5x2 + 6 = 110/10
    5x2 + 6 -11 = 0
    5x2 = 5
    x = ± 1

  3. 2 cos(2x – 30) = -6/5
    Cos(2x -30) = - 0.6
    Cos –ve in 3rd quad
    2x – 30 = 53.13
    2x – 30 = 233.13
    2x = 263.10
    x = 131.57°
    2x-30=540-53.13
    X=258.44
    2x - 30°= 360 + 233.13°
    2x = 623.13
    x = 311.57
    x=131.57,258.44,311.57

  4. nT/2m = √ L-A/3k  

      n²T²    L - A  
      4m²           3K
    4m2 (L – A) = 3k (n2 T2)
    4Lm2 – 4Am2 = 3kn2T2
    A =   4LM²   - 3kn²T²
                4m²

  5. p=c + k/
    6= c + k/36
    18=c + k/9
    216 = 36c + k ----- (i)
    162 = 9 c + k --------(ii)
      54 = 27c
    54/27 = c = 2 k = 144
    when p = 11
    11 = 2 + 144/
    9t2 = 144
    t = 4
  6.            
    1. Coefficients: 1     6      15     20
      Terms: 56, 55 (x/2), 54(x/2)2, 53 (x/2)3
      15625, 3125x/2, 625x²/4, 125x³/8
      ( 5 + x/2)⁶ = 15625 + 3125/3x + 9375x³/4 + 625x³/2 + ...
      x= 1 Therefore, (11/2)6 =  15625 + 3125/3 + 9375/4 + 625/2
    2. = 15625 + 1041.667 + 2343.75 + 312.5
      =19322.9
  7. Log8 (x+6/x-3) = Log88²/₃
    x+6/x-3 = 4
    x + 6 = 4x – 12
    3x = 18
    x = 6

  8. 8 x = 7 + 2y
    ( 7 + 2y) y + 6 = 0
    7y + 2y2 + 6= 0
    2y2 + 7y + 6 = 0
    2y2 + 4y + 3y + 60 = 0
    2y(y + 2) +3 (y + 2) = 0
    (2y + 3) ( y + 2) = 0
    When y =-2, x = 3
    y =-3/21, x = 4

  9. X+5X=180
    6X=180
    X=30
    360/30
    12 sides 
  10.   1( 3 + √5) - ( 2 + 2√5)(3- √5)
                 (3 - √5)(3 + √5)
    3 + √5 - [6-2√5 + 6√5 -10]
                 9-5
    3+√5 - 4√5 + 4
             4
    7 - 3√5  7   - 3 √5
         4           4       4
    hence a = 7/4, b= -3/4 and c =5

  11. Mean = 44 + 32 + 67 + 52 +28 + 39 + 46 + 64
    372/8 = 46.5
    x 44 32 67 52 28 39 46 64
    d  -2.5 -14.5 20.5 5.5 -18.5 -7.5 -0.5 17.5
    (d) 2.5  14.5 20.5 5.5 18.5 7.5 0.5 17.5
    Mean absolute deviation
    Σxd 87   = 10.875
      N       8
  12. 50,000/50,000 ( 1 + r/100)⁸ =  79692.40/50,000
    ⁸√(1 + r/1000) = ⁸√1.593848
    1 + r/100 = 1.06
    r/100 = 0.06
    R = 6% per interest period
    Per annum = 6 x 4 = 24%

  13.            
    1. Co – ordinate of
      A ( 5-3/2  5-1/2)
      A (2/2   4/2)
      A ( 1,2)

    2. r = √(5-1)2 + (5-2)2
      =√16+9
      =5 units
      ( x – 1)2 + ( y – 2) 2 = 52
      x2 – 2x+ 1 + y2- 4y + 4 = 25
      x2 + y2 – 2x - 4y + 5 = 25
      x2 y2 – 2x – 4y – 20 = 0

  14. x fills 1/3 of tank in 1 hour
    y fills 1/6 of tank in 1 hour
    z empties 1/8 of tank in 1 hour
    x and y open will fill 1/3 + 1/6 = 1/2 of tank in 1 hour
    remaining fraction = 1 - 1/2 = 1/2
    x, y and z both will fill 1/3 + 1/6 - 1/8 = 3/8 of tank in 1 hour
    time taken by x, y and z to fill ½ of remaining tank
    = 1/2 ÷ 3/8=1 1/3 hours
    Total time = 1 + 1 1/3 = 2 1/3 hours
  15. M:B
    3:2
    3 x 90 x 150 x 2 = cost price
         3+ 5
    CP= 270 + 300
                  8
    S.P = 130/100 + 71.25
    = Ksh 92.625
    =Ksh 92.60
  16.    
    KapPreM Mathp2q16aKapPreM Mathp2q16b
    Det21-20 =1 Det 12-11=1
    KapPreM Mathp2qa15
  17.        
    1.     
      KapPreM Mathp2qa17a
    2. S1 R1 E1 (A”’B ”’C ”’D ”’) = ABCD
      KapPreM Mathp2qa17b
    3. ΔPQR – For object figure plotted correctly
      PQR – For image figure plotted correctly
      KapPreM Mathp2qa17c
  18.          
    marks f Mp(x) d(x-A) fd d2 fd2
    10-19
    20-29
    30-39
    40-49
    50-59
    60-69
    70-79
    80-89
    90-99
     2
    6
    10
    16
    24
    20
    12
    8
     14.5
    24.5
    34.5
    44.5
    54.5
    64.5
    74.5
    84.5
    94.5
     -40
    -30
    -20
    -10
    0
    10
    20
    30
    40
     -80
    -180
    -200
    -160
    0
    200
    240
    240
    80 
    1600
    900
    400
    100
    0
    100
    400
    900
    1600 
    3200
    5400
    4000
    1600
    0
    2000
    4800
    7200
    3200
            140    31400

    1. Mean = A + ∑fd/∑f
      = 54.5 + 140/100 = 55.9

    2. Variance = 31400/100 – (140/100)2
      = 312.04
    3. standard deviation = √variance
      = √312.04 = 17.66
  19.             
    1. FC = √62 + 4.52 + 82
      = 10.97
    2.        
      1. Sin θ = 6
        0.97
        θ = 33.16

      2.    
        KapPreM Mathp2qa19bii
        Tan θ = 8 ÷ 4.5
        = 1.778
        θ = 60.64
    3.    
      KapPreM Mathp2qa19c
      Tan θ = 6 ÷8
      θ = 36. 87

    4. Total surface area
      2 (8 x 6) + 2 (4.5 x 8) + 2(4.5 x 6)
      = 222cm2
  20.      
    1.    
      60º 120º 180º 240º 300º 360º 420º 480º 540º
      Cos x 1.00 0.50 -0.50 -1.00 -0.5 0.50 1.00 0.50 -0.50 -1.00
      2Cos(x+30) 1.73 0.00 -1.73 -1.73 0.00 1.73 1.73 0.00 -1.73 -1.73
    2.        
      1. Period = 360º
      2. phase angle = 30º
    3.     
      KapPreM Mathp2qa20

    4. 37.5º ≤ x ≤ 217.5º
      397.5º ≤ x ≤ 540º
  21.          
    1.         
      1.  5/20 = 1/4
      2.   5/20 + 9/20
        =14/20 = 7/10
    2.               
      1.   ( 5/20 X 4/19)+(6/20 X 5/19) + (9/20 X 8/19)
        = 20/380 + 30/380 + 72/380
        = 122/380 = 61/190
        =
        1- 61/90
        =
        29/90
  22.                    
    1. <QTS = <SQR = 35° B1
      Angles in alternate segments 

    2. <QST = 55° B1  
      <QRS = 55 - 35° B1  
        = 20°

    3. <QVT = <SQV = 30°
    4. 90 – (30° + 35)  = 25°
    5. 180 – (60° + 65°) = 55°

  23. ½ x 6x h = 9✓
    h= 3✓
    P1p2 = 5.6 cm
    KapPreM Mathp2qa23
  24.          
    1. T3 = a + 2d
      T9 = a + 8d
      T25 = a + 24d

    2.      
      1. a+8d/a+2d = a+24d/a+8d
        a2 + 2bad + 48d2 = a2 + 16ad + 64d2
        10ad = 16d2
        a = 16d/10 or 8/5d
        a + 6d + 2(a + 5d) = 78
        3a + 16d = 78
        24d + 80d = 390
        104d = 390
        d = 3.75
        a = 8/5 x 3.75
        a = 6

      2. S9 = 4.5 (12 + (8 X 3.75)
        4.5 (12 + 30)
        = 4.5 (42)
        = 189
        (10 Marks)

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