Mathematics Paper 2 Pre Mock Questions and Answers - Mokasa I Joint Examination July 2021

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INSTRUCTIONS TO THE CANDIDATES

  • Write your name and school and index number in the spaces provided above
  • This paper contains two sections; Section 1 and Section 11.
  • Answer all the questions in section 1 and only five questions from Section 11
  • Show all the steps in your calculations, giving your answers at each stage in the spaces below each question.
  • Marks may be given for correct working even if the answer is wrong.
  • Non-Programmable silent calculators and KNEC Mathematical tables may be used EXCEPT where stated otherwise.

FOR EXAMINERS’S USE ONLY
Section 1

Question

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

Total

Marks

                                 

Section 1I

Question

17

18

19

20

21

22

13

24

Total

Marks

                 

QUESTIONS

SECTION I (50 marks):
Answer all questions in this section

  1. Use logarithm tables to evaluate Q1 JYGAUYTGDA( 4 marks)
  2. Solve the equation 62x+1 = 23x+1 ( 3 marks)
  3. Kevin truncated 0.00627 to 3 decimals and 487.74 to 3 significant figures. Calculate his percentage error in calculating product of numbers in truncated values to 1 decimal places. ( 3marks)
  4. A new laptop depreciates at 8% per annum in the first year and 12% per year in the second year. If its value at the end of the second year was sh121,440. Calculate the original value of the laptop. (3marks)
  5. Rationalize the denominator and simplify Q2 GHHAD( 3 marks)
  6. In the figure below angle A=68º, B= 390, BC= 8.4cm and CN is the bisector of angle ACB. Calculate the length CN to 1decimal place. ( 3 marks)
    Q6 UJYG A UYGD
  7. In the figure below R, T and S are points on a circle centre O. PQ is a tangent to the circle at T, POR is a straight line and angle QPR =20º. Find the size of angle RST (3marks)
    q7 jygauyd
  8. Use binomial expansion to find the value of (1.02)5 correct to 3 decimal place. ( 4 marks)
  9. Make x the subject of the equation (3 marks)
    q9 iiugtauygd
  10. The equation of the circle is given by x2 + y2 + 8x -2y -1 = 0 . Determine the radius and the centre of the circle. (3marks)
  11. Given that the minor arc of a circle subtends an angle of 140º at the centre of a circle of radii 3.5cm. Calculate the area of the major segment correct to 4 significant figures ( 3 marks)
  12. Given that the matrix q12 juaguydais a singular matrix, find the values of x. (3marks)
  13. The mass of a mixture A of peas and millet is 72 kg. The ratio of peas to millet is 3:5 respectively;
    1. Find the mass of millet in the mixture. (1mark)
    2. A second mixture of B of peas and millet of mass 98 kg is mixed with A. The final ratio of peas to millet is 8:9 respectively. Find the ratio of peas to millet in B (2marks)
  14. Draw a line AB= 8cm long. Divide the line proportionally into5 equal parts. Locate a point Y on the line AB such that AY: YB = 3:2. ( 3 marks)
  15. A solid prism is made of a pentagonal cross section of sides 10cm. If the prism is 30cm long calculate area of the cross section hence the volume of the prism (3 marks)
  16. Given that X = 2i + j -2k, y = -3i + 4j – k and z= 5i + 3j + 2k and that p= 3x – y + 2z, find the magnitude of vector p to 3 significant figures. (3marks)


SECTION II (50 Marks)
Answer any five questions in this section

  1. The masses in kilograms of patients who attended a clinic on a certain day were recorded as:
    38 52 46 48 60 59 62 73 49 54 49 41 57 58 69 72 60 58 42 41
    79 62 58 67 54 60 65 61 48 47 69 59 70 52 63 58 59 49 51 44
    67 49 51 58 54 59 39 59 54 52
    1. starting with class 35-39, make a frequency distribution table for the data indicating the class and frequency. ( 3 marks)
    2. state the modal class ( 1 mark)
    3. Calculate the mean mass ( 3 marks)
    4. Calculate the median mass ( 3 marks)
  2. The income tax rates of a certain year were as shown below:

    Monthly taxable income in Ksh

    Tax rate in %

    0-9680

    10

    9681-18800

    15

    8801-27920

    20

    27921- 37040

    25

    37041 and above

    30


    In that year, Sayao monthly earnings were as follows; basic salary Ksh. 30 000, house allowance Ksh.15 000, and medical allowance of Ksh 3,500. He is entitled to a monthly tax relief of Ksh. 1056.
    1. Calculate Sayao’s taxable income ( 2 marks)
    2. Calculate his P.A.Y.E ( 5 marks)
    3. A part from P.A.Y.E, other deductions is education insurance policy Ksh. 1500 and Ksh 2500 as cooperative shares. Find his net income at end of the month. ( 3 marks)
  3. A Quantity P varies partly as the square of m and partly as n. When p= 3.8, m = 2 and n = -3, When p = - 0.2, m = 3 and n= 2.
    1. Find
      1. The equation that connects p, m and n (4marks)
      2. The value of p when m = 10 and n = 4 (1mark)
    2. Express m in terms of p and n (2marks)
    3. If P and n are each increased by 10%, find the percentage increase in m correct to 2 decimal place. (3marks)
  4.                    
    1. Complete the table below by filling in the blank spaces ( 2 marks)

         x

      00

      300

      600

      900

      1200

      1500

      1800

      2100

      2400

      2700

      3000

      3300

      3600

      y=Cos x

      1.00

      0.87

       

      0.00

       

      1.00

       

      -0.87

           

      0.87

       

      y=3sinx

      0.00

         

      3.00

                   

      -1.50

      0.00

    2. using the scale 1cm to represent 300 on the x-axis and 2 cm to represent 1 unit on the vertical axis, draw on the graphs of cosx and 3sinx ( 5 marks)
      q15 jygyuasd
    3. use your graph to solve the equation cosx =3sinx ( 2 marks)
    4. What is the difference in the values of y=cosx and y=3sinx at x=1200 ( 1 mark)
  5. The 5th term of an AP is 16 and the 12th term is 37.
    1. Find;
      1. The first term and the common difference ( 3 marks)
      2. The sum of the first 21 terms (2 marks)
    2. The second, fourth and the seventh term of an AP are the first 3 consecutive terms of a GP. If the common difference of the AP is 2.
      Find:
      1. The common ratio of the GP ( 3 marks)
      2. The sum of the first 8 terms of the GP (2 marks)
  6. In driving to work, John has to pass through three sets of traffic lights. The probability that he will have to stop at any of the lights is ¾
    1. Draw a tree diagram to represent the above information. (2 marks)
    2. Using the diagram, determine the probability that on any one journey, he will have to stop at:
      1. All the three sets. (2 marks)
      2. Only one of the sets (2 marks)
      3. Only two of the sets (2 marks)
      4. None of the sets. (2 marks)
  7. The figure below shows a lampshade in the form of a conical frustum
    q23 jygauyda
    The top and bottom radii are 7cm and 14cm respectively. The slant height AB is 20cm. Calculate:
    1. The slant height of the original cone correct to two decimal places ( 2 marks)
    2. The height h, of the lampshade ( 2 marks)
    3. The curved surface area of the lampshade ( 3 marks)
    4. The volume of the lampshade correct to 4 significant figures ( 3 marks)
  8. Gary bought 5 tins of plums and 3 tins of peaches from a supermarket for Ksh.75, while Mike bought 3 tins of plums and 5 tins of peaches for Ksh.77
    1. Set up the simultaneous equations which represent the given information (2 marks)
    2. Write down the matrix equation ( 2 marks)
    3. Using the matrix method, find the cost of
      1. 4 tins of plums ( 5 marks)
      2. 2 tins of peaches ( 2 marks)

MARKING SCHEME

SECTION I (50 marks):
Answer all questions in this section

  1. Use logarithm tables to evaluate Q1 JYGAUYTGDA( 4 marks)

    No  Standard form  Log  
    0.4239
    149.6
    log6 = 0.7782
    4.332
    4.239 × 10-1
    1.496 × 102
    7.782 × 10-1
    4.335 × 10º
    1.6272 +
    2.1750
    1.9022 -
    1.8911
    1.9111
       3
    0.6370
    M1
    M1
    M1 

    =4.335M
  2. Solve the equation 62x+1 = 23x+1 ( 3 marks)
    (2x + 10)log6 = (3x + 1)LOG2
    2x + 1 = log2 M1
    3x + 1    log6
    2x + 1 = 0.3869
    3x + 1
    2x + 1 = (3x + 1)(0.3869) M1
    2x + 1 = 1.1607x + 0.3869
    2x - 1.1607x = 0.3869 - 1
    0.8393x =  -0.6131
    x = -0.7305 A1
  3. Kevin truncated 0.00627 to 3 decimals and 487.74 to 3 significant figures. Calculate his percentage error in calculating product of numbers in truncated values to 1 decimal places. ( 3marks)
    Actual = 0.00627 × 487.74
    =3.0581298
    Truncated = 0.006 × 488 M1
    =2.928
    % = Actual - Truncated
                    Actual
    3.0581298 - 2.928
         3.0581298
    = 0.1301298 × 100
        3.0581298
    =4.255
    =4.3% 1dp  A1
  4. A new laptop depreciates at 8% per annum in the first year and 12% per year in the second year. If its value at the end of the second year was sh121,440. Calculate the original value of the laptop. (3marks)
    Let the original value be P
    A = P(1 - r/100)n
    P(1 - 8/100)1
    A = P(92/100)
    = 0.92P
    2nd year
    A = P(1 - r/100)n
    0.92P(1 - 12/100)1
    0.92P(0.88)
    A = 0.8096P
    0.8096P = 121,440 M1
    P = 121,440
           0.8096
    P = shs.150,000 A1
  5. Rationalize the denominator and simplify Q2 GHHAD( 3 marks)
    (√3 + 2√5)(√5 + √3)
      (√5 - √3)(√5 + √3)
    Numerator
    √3(√5 + √3) +2√5(√5 + √3)
    √15 + 3 + 10 + 2√15
    13 + 3√15  M1
    Denominator
    (√5 - √3)(√5 + √3)
    5 - 3 = 2 M1
    13 + 3√15
          2
    6.5 + 1.5√15    A1
  6. In the figure below angle A=68º, B= 390, BC= 8.4cm and CN is the bisector of angle ACB. Calculate the length CN to 1decimal place. ( 3 marks)
    q6 ujygujgde
    ACN = 180º - (68 + 39) = 365º
                       2
      8.4   =    x    
    sin68º   sin39º
    x =    8.4    × sin39º = 5.701
           sin68º
       CN    5.701   
      sin68     sin75.5
    cn = 5.701 × sin68
           sin75.5
    CN = 5.459
    CN = 5.5(1dp) cm
  7. In the figure below R, T and S are points on a circle centre O. PQ is a tangent to the circle at T, POR is a straight line and angle QPR =20º. Find the size of angle RST (3marks)
    q7 uyguaygda
    RST = 55º
    Angle substended by a chord at the centre is thrice what is substends at the circumference
  8. Use binomial expansion to find the value of (1.02)5 correct to 3 decimal place. ( 4 marks)
    (1 + x)5 = 15(x)0 + 14(x)1 + 13(x)2 + 12(x)3 + 11(x)4 + 10(x)5 
    x = 0.02
    1 + x + x2 + x3 + x4 + x5
    1    5    10   10    5        1
    1 + 5x + 10x2 + 10x3 + 5x+ x5 M1
    1 + 5(0.02) + 10(0.02)2 + 10(0.02)3 + 5(0.02)+ (0.02) M1
    1 + 0.1 + 0.004 + 0.00008 + 0.0000008 + 0.0000000032 M1
    = 1.1040808
    = 1.104(3 dp) A1
  9. Make x the subject of the equation (3 marks)
    q9 iiugtauygd
    (t/s)2 = (b/√x - 4)2
    t2 =   b2
    s2   (x - 4)
    t2(x - 4) = s2b2
    t2x - 4t2 = s2b2
    t2x = s2b2 + 4t2 M1
    x = s2b2 + 4t2   A1
               t  
  10. The equation of the circle is given by x2 + y2 + 8x -2y -1 = 0 . Determine the radius and the centre of the circle. (3marks)
    x2 + 8x + 16 + y2 - 2y + 1 = 1 + 16 + 1
    (x + 4)2 + (y - 1)2 = 18
    Centre (-4,1)  radius = √18 = 4.243 units
  11. Given that the minor arc of a circle subtends an angle of 140º at the centre of a circle of radii 3.5cm. Calculate the area of the major segment correct to 4 significant figures ( 3 marks)
    360 - 140 = 220
    220 x 22 x 3.52 = 23.52778
    360    7
    = 23.53 cm2
  12. Given that the matrix q12 juaguyda is a singular matrix, find the values of x. (3marks)
    x( x- 1) - 0 = 0
    x2 - 1 = 0
    x2 = 1 
    x = √1
    x = 1  or x = -1
    x = ±1
  13. The mass of a mixture A of peas and millet is 72 kg. The ratio of peas to millet is 3:5 respectively;
    1. Find the mass of millet in the mixture. (1mark)
      Total mass = 72 kg
      peas: millet
      3:5
      a) millet = 5/8 x 72 = 45kg
    2. A second mixture of B of peas and millet of mass 98 kg is mixed with A. The final ratio of peas to millet is 8:9 respectively. Find the ratio of peas to millet in B (2marks)
      A + B mixture = 72 + 98 = 170 kg
      mass of peas in A  = 3/8 x 72 = 27kg
      A and B 8:9 of 170 kg
      peas = 8/17 x 170 = 80kg
      millet 9/17 x 170 = 90kg
      In mixture 
      B:millet = 90 - 45
      =45kg
      In mixture
      B :peas 
      =80 - 27
      =53kg
      Ratio = 53:45
  14. Draw a line AB= 8cm long. Divide the line proportionally into5 equal parts. Locate a point Y on the line AB such that AY: YB = 3:2. ( 3 marks)
    q14 ujgada
  15. A solid prism is made of a pentagonal cross section of sides 10cm. If the prism is 30cm long calculate area of the cross section hence the volume of the prism (3 marks)
    q15 utfgauygdyta
    tan36º = 5
                  h
    h =     5    
          tan36º
    h = 6.882
    5(1/2 x 6.882 x 10) 
    cross sectionesd are = 172.05cm2
  16. Given that X = 2i + j -2k, y = -3i + 4j – k and z= 5i + 3j + 2k and that p= 3x – y + 2z, find the magnitude of vector p to 3 significant figures. (3marks)
    q16 ujhfgjhgadj
    = 19.7

SECTION II (50 Marks)
Answer any five questions in this section

  1. The masses in kilograms of patients who attended a clinic on a certain day were recorded as:
    38 52 46 48 60 59 62 73 49 54 49 41 57 58 69 72 60 58 42 41
    79 62 58 67 54 60 65 61 48 47 69 59 70 52 63 58 59 49 51 44
    67 49 51 58 54 59 39 59 54 52
    1. starting with class 35-39, make a frequency distribution table for the data indicating the class and frequency. ( 3 marks)
      class tally  frequency x fx c.f 
      35-39 ll 2 37  74  2
      40-44 llll 4 42 168  6
      45-49 llll lll 8 47 376 14
      50-54 llll llll  9 52 468 23
      55-59 llll llll 11 57  627 34
      60-64 llll ll 7 62 434 41
      65-69 llll 5 67 335  46
      70-74 lll 3 72 216 49
      75-79 l 1 77 77 50
              Σfx = 2775  
    2. state the modal class ( 1 mark)
      55 - 59
    3. Calculate the mean mass ( 3 marks)
      mean = Σfx = 2775
                             50
      = 55.5
    4. Calculate the median mass ( 3 marks)
      54.5 + 2 x 5 = 55.41
               11
  2. The income tax rates of a certain year were as shown below:

    Monthly taxable income in Ksh

    Tax rate in %

    0-9680

    10

    9681-18800

    15

    8801-27920

    20

    27921- 37040

    25

    37041 and above

    30

    In that year, Sayao monthly earnings were as follows; basic salary Ksh. 30 000, house allowance Ksh.15 000, and medical allowance of Ksh 3,500. He is entitled to a monthly tax relief of Ksh. 1056.
    1. Calculate Sayao’s taxable income ( 2 marks)
      30,000 + 15,000 + 3,500 = 48,500
    2. Calculate his P.A.Y.E ( 5 marks)
      9680 × 10 = 968
                 100
      9120 × 15= 1368
                 100
      9120 × 20 = 1824
                 100
      9120 × 25= 2280
                 100
      11460 × 30 = 3438
                   100
      Gross tax = 9878
      P.A.Y.E = 9878 - 1056
      = ksh. 8822 p.m
    3. A part from P.A.Y.E, other deductions is education insurance policy Ksh. 1500 and Ksh 2500 as cooperative shares. Find his net income at end of the month. ( 3 marks)
      Total deductions = 8822 + 2500 + 1500
      = 12, 822
      Net income = 48,500 - 12,822 = ksh. 35,678
  3. A Quantity P varies partly as the square of m and partly as n. When p= 3.8, m = 2 and n = -3, When p = - 0.2, m = 3 and n= 2.
    1. Find
      1. The equation that connects p, m and n (4marks)
        p = xm2 + yn
        3.8 = 4x - 3y
        -0.2 = 9x + 2y
          7.6 = 8x - 6y
        -0.6 = 27 + 6y  +
           7 = 35x
        x = 7/35 = 1/5 = 0.2
        3.8 = 0.8 - 3y
        3 = =-3y
        y = -1
        P = 0.2m2 - n
      2. The value of p when m = 10 and n = 4 (1mark)
        P= 0.2 x 100 - 4
        =20 - 4
        =16
    2. Express m in terms of p and n (2marks)
      P = 0.2m2 - n
      0.2m2 = P + n
      m2 = p + n
                0.2
      q19 iyha uyhda
    3. If P and n are each increased by 10%, find the percentage increase in m correct to 2 decimal place. (3marks)
      q20 iygauygda
  4.                    
    1. Complete the table below by filling in the blank spaces ( 2 marks)

         x

      00

      300

      600

      900

      1200

      1500

      1800

      2100

      2400

      2700

      3000

      3300

      3600

      y=Cos x

      1.00

      0.87

      0.50

      0.00

      -0.50

      -0.89

      -1.00

      -0.87

      0.50 0.00 0.50

      0.87

      1.00

      y=3sinx

      0.00

      1.50 2.60

      3.00

      2.60 1.50 0.00 -1.50 -2.60 3.00 -2.60

      -1.50

      0.00

    2. using the scale 1cm to represent 300 on the x-axis and 2 cm to represent 1 unit on the vertical axis, draw on the graphs of cosx and 3sinx ( 5 marks)
      m20 iygahuydga
    3. use your graph to solve the equation cosx =3sinx ( 2 marks)
      x = 18º, 198º
    4. What is the difference in the values of y=cosx and y=3sinx at x=1200 ( 1 mark)
      2.60 - - 0.5 = 3.1 
  5. The 5th term of an AP is 16 and the 12th term is 37.
    1. Find;
      1. The first term and the common difference ( 3 marks)
        Tn = a + (n - 1)d
        T5 = a + 4d = 16
        T12 = a + 11d = 37 
          7d = 21
        d = 3
        a + 4 (3) = 16
        a + 12 = 16
        a = 4
      2. The sum of the first 21 terms (2 marks)
        S21 = 21 {2(4) + 20(3)}
                 2
        = 714
    2. The second, fourth and the seventh term of an AP are the first 3 consecutive terms of a GP. If the common difference of the AP is 2.
      Find:
      1. The common ratio of the GP ( 3 marks)
        a + d, a + 3d, a + 6d
        a + 2  a + 6  a + 12
        a + 6 = a + 12
        a + 2    a + 6
        (a + b)2 = (a + 2)(a + 12)
        a + 12a + 36 = a2 + 14a + 24
        12= 2a
        a = 6
        comm ratio
        r = 12 = 11
              8       2
      2. The sum of the first 8 terms of the GP (2 marks)
        S8 = 8((3/2)8 - 1 ) = 197.03 = 394.063
                     11/2 -1           1/2
  6. In driving to work, John has to pass through three sets of traffic lights. The probability that he will have to stop at any of the lights is ¾
    1. Draw a tree diagram to represent the above information. (2 marks)
      q22 jgvaytfdtfa
    2. Using the diagram, determine the probability that on any one journey, he will have to stop at:
      1. All the three sets. (2 marks)
        P(sss) = 3/× 3/4 × 3/4 = 27/64
      2. Only one of the sets (2 marks)
        P(ss1s1) or P(s1ss1) or P(s1s1s)
        (3/× 1/4 × 1/4) + (1/× 3/4 × 1/4) + (1/× 1/4 × 3/4) = 3/64  × 3 = 9/64
      3. Only two of the sets (2 marks)
        P(sss1) or P(ss1s) or P(s1ss) 
        (3/× 3/4 × 1/4) + (3/× 1/4 × 3/4) + (1/× 3/4 × 3/4) = (9/64)3 = 27/64
      4. None of the sets. (2 marks)
        P(s1s1s1) = 1/× 1/4 × 1/4 = 1/64
  7. The figure below shows a lampshade in the form of a conical frustum
    q23 n agtydga
    The top and bottom radii are 7cm and 14cm respectively. The slant height AB is 20cm. Calculate:
    1. The slant height of the original cone correct to two decimal places ( 2 marks)
      L + 20 = 14
          L         7 
      14L = 7L + 140
      7L = 140
      L = 20
      Slant height 
      = 20 + 20 = 40cm
    2. The height h, of the lampshade ( 2 marks)
      q23b jyguagd
      402 = H2 + 142
      1600 = H2 + 196
      1600 - 196 = H2
      H2 = 1404
      H = 37.47
      202 = X2 + 49
      400 - 49 = X2   
      X = 18.73
      37.44 - 18.73
      h = 18.74cm
    3. The curved surface area of the lampshade ( 3 marks)
      T1R1 - T1rL
      22/7 × 14 × 40  - 22/7 × 7 × 20
      1760 - 440 = 1320cm2
    4. The volume of the lampshade correct to 4 significant figures ( 3 marks)
      1/3πR2H - 1/3πr2h
      1/× 22/× 142 × 37.47  - 1/× 22/× 72 × 18.73
      = 7693.84 - 961.47
      = 6732.37cm3
  8. Gary bought 5 tins of plums and 3 tins of peaches from a supermarket for Ksh.75, while Mike bought 3 tins of plums and 5 tins of peaches for Ksh.77
    1. Set up the simultaneous equations which represent the given information (2 marks)
      let plums be x and peaches be y
      5x + 3y = 75
      3x + 5y = ??
    2. Write down the matrix equation ( 2 marks)
      q24 jygydf
    3. Using the matrix method, find the cost of
      1. 4 tins of plums ( 5 marks)
        q24 b jmhvgad
      2. 2 tins of peaches ( 2 marks)
        2 tins of peaches
        10 × 2 = 20
        IA = A
        x = 9 
        y   10
        x = 9
        y = 10
        4 tins of peaches  = 9 × 4 = ksh. 36

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