Mathematics Paper 2 Questions and Answers - Momaliche Joint Pre Mock Exams 2022

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QUESTIONS
SECTION I

  1. Solve the quadratic equation by completing the square method.
    x2 – 7x + 10 = 0 (3mks)
  2. Simplify √5- √3 by rationalizing the denominator. (2mks)
                 √5+ √3
  3. Grade x rice costs shs 75 per kg and grade y rice costs shs 50 per kg. The two grades are mixed in the ratio such that the blend costs sh 70 per kg. Find the ratio in which grade x rice was mixed with grade y rice. (3mks)
  4. Given that A = ∜d-c2g make c the subject of the formula. (3mks) 
                             b+c2f
  5. Solve the equation 2 sin (3x + 60) = 1 for 0º ≤ x ≤ 180º (3mks)
  6.      
    1. Expand and simplify (2 – x)8 up to 4th term. (2mks)
    2. Use the simplified expression in (a) above to estimate the value of (1.99)8 giving your answer correct to 4 significant figures. (2mks)
  7. Solve the equation 2 + log3 (2x -7) = log3(5x - 7) . (3mks) 
  8.      
    1. Construct a triangle ABC in which AB is 2.2cm, BC is 3.5cm and angle ABC is 60º(2mks)
    2. A point D moves such that it is on the same side of BC as A. Construct the locus of D such that the area of ΔABC = area of ΔBCD. (2mks)
  9. The equation of a circle is x2 + y2 – 6x + 8y – 11 = 0. Find the coordinates of the centre of the circle and its radius. (3mks)
  10. The length of a rectangle is 8.3 cm and its width is 5.45cm. Calculate
    1. The relative error in area of the rectangle (3mks)
    2. The percentage error in area (1mk)
  11. A triangle xyz whose area is 15.3cm2 mapped onto Δx1y1z1 by a transformation whose matrix is m =1. Find the area of Δx1y1z1 .(3mks)
  12. Solve the triangle (3mks)
    2
  13. Fifteen men working at the rate of 9 hours per day take 20 days to complete a job. Find the number of days 27 men would take to complete the same job working at the rate of 10 hours a day. (3mks)
  14. The 5th term of an arithmetic sequences is 71 and the 7th term is 59. Find
    1. The first term and the common difference. (2mks)
    2. The 10th term (2mks)
  15. Use matrix methods to solve the simultaneous equations. (3mks)
    3x – 4y = 2
    6x + y = 13
  16. Factorise: 2x2 – x – 10 (2mks)

SECTION (II) (50MARKS)

  1. The following table shows the distribution of marks obtained by 50 students of a certain school.
    Marks 45-49 50-54 55 - 59 60- 64 65 - 69 70-74 75- 79
    No. of students 3 9 13 15 5 4 1
    1. State the modal class (1mk)
    2. By using an assumed mean of 62, calculate
      1. The mean (5mks)
      2. The standard deviation (4mks)
  2. The figure below is a triangle OAB, where OA = a and OB = b. A point R divides AB in the ratio 2:5 and a point T divides OB in the ratio 1:3. OR and AT intersect at D.
    3
    1. Find in terms of a and b
      1. BT (1mk)
      2. OR (2mks)
      3. AT (1mk)
    2. Given that AD = KAT and RD = hRO where k and h are scalars. Find the values of k and h, hence express AD in terms of a and b (5mks)
  3. John travels to work by either boda-boda or by tuk-tuk. If he travel by tuk-tuk on every one day, there is a probability of 0.75 that he travels by a boda boda the following day. If he travels by a boda boda on any one day, then he travels by tuk-tuk the following day with a likelihood of 5/6 . There is a chance of 2/3 that he travels by tuk-tuk on Tuesday.
    1. Draw a tree diaagram to illustrate the possible outcomes in 3 days. (2mks)
    2. Find the probability that he travel by;
      1. Boda-boda on Wednesday (2mks)
      2. Tuk-tuk on Wednesday (2mks)
      3. Boda – boda on Thursday (3mks)
      4. Tuk-tuk on Thursday (1mk)
  4.      
    1. Fill the table below for the curves given by y = 3sin(2x + 30º) and y = cos2x for x value in the range 0º ≤ x ≤ 180º (2mks)
      15º 30º 45º 60º 75º 90º 105º 120º 135º 150º 165º 180º
      Y = 3 sin (2x + 30) 1.50     2.60           -2.60     1.50
       Y =cos 2x 1.00         0-0.87     -0.50       1.00
    2. Draw the graph of y = 3 sin (2x + 30º) Y = 3sin (2x + 30º) and y = cos 2x on the same axes. (4mks)
      x-axis 1cm rep 15º
      y-axis 1cm rep 0.5 units
    3. Use your graph to solve the equation 3sin (2x +30º) = cos 2x (2mks)
    4. Determine the following from your graph
      1. Amplitude of y=3sin(2x+30º) (1mk)
      2. The period of y=3sin(2x+30º) (1mk)
  5. An arithmetic progression has the first term as a and the common difference as d.
    1. Write down in terms of a and d the 3rd, 9th and 25th term of the progression (1mk)
    2. The progression is increasing, and the 3rd, 9th and 25th terms form the first three consecutive terms of a geometric series. If the sum of the 7th and twice the 6th term of arithmetic progression is 78, calculate
      1. The first term and the common difference of the AP. (6mks)
      2. The sum of the first nine terms of the AP. (3mks)
  6. The figure below is a right rectangular based pyramid VABCD where AB =5cm,
    BC =7cm and VC =VB=VA=VD=13cm and O is a point on the base of the pyramid vertically below V.
    Calculate
    1. The length of AC (2mks)
    2. VO the height of the pyramid. (2mks)
    3. The angle between the edge VB and the plane ABCD (3mks)
    4. The angle between the planes VBC and ABCD (3mks)
  7. Three quantities L, M and N are such that L varies directly as M and inversely as the square root of N.
    1. Given that L = 2250 when M = 450 and N = 64, write down an equation connecting L, M and N. (4mks)
    2. If M decreased by 16% and N increased by 44%, calculate the percentage change in L. (3mks)
    3. In soccer competition, the number of goals (G) scored in a penalty shoot-out is partly constant and partly varies as the skill (S) of the player. Given that G = 8 when S = 2 and G = 12 when S =4, find the value of G when S = 6. (3mks)
  8. The table below shows income tax rates
    Monthly taxable pay (k£) Rate of tax ksh per £
    1- 435
    436 – 870
    871 – 1305
    1306 – 1740
    Excess over 1740
    2
    3
    4
    5
    6
    A company employee earns a monthly basic salary of Ksh 28,000. He is also entitled to the following monthly allowances: house allowance of Ksh 9000, a medical allowance of sh 2000 and a commuter allowance of shs 1480.
    1. Calculate his total income tax. (5mks)
    2. He is entitled to a personal tax relief of Ksh 1056 per month. Determine the net tax. (1mk)
    3. If he received a 50% increase in his total income, calculate the corresponding percentage increase on the income tax. (4mks)


MARKING SCHEME

  1. x2 - 7x + 10 = 0
    x2-7x = -10 
    x2 - 7x + 49/4 = -10 + 49/4
    (x - 7/2)2 = 9/4
    x - 7/= ± 3/2
    X = 7/2 ± 3/2
    Either x = 7/2 + 3/2 = 10/2 = 5
    or
    x = 7/3/24/= 2
  2. 75x + 50y = 70
        x + y
    75x + 50y = 70x +70y
    5x = 20y
    x/y = 20/5 = 4
    x:y = 4:1
  3. √5 - √3√5 - √3 
    √5 + √3   √5 - √3 
    (√5 - √3)( √5 - √3)
    5 - √15 + √15 - 3
    5 - √15 - √15 + 3 = 8 - 2√15
               5 - 3                2
    = 4 - √15
  4. A = 4d - c2g
              b + c2f
    A4 = d - c2g
            b + c2f
    A4b + A4c2f = d - c2g
    A4c2f + c2g = d - Ab
    c2(A+ g) = d - Ab
    cd - A
             A4+ g
    c = ± √d - A
                A4+ g
  5. 2sm(3x+60 = 1
    (3x+60) = 0.5
    3x+60 = 30
    3x = -30
    x = -10
    3x + 60 = 180 -30 = 150
    3x = 150 - 60
    X = 30
    3x+60 = 360 +30 = 390
    3x = 390 - 60 = 330
    x = 110º
    3x + 60 =510
    3x = 450
    x = 150º
  6.        
    1. (2 - x)8 = 2- 27x+ 26x2 - 25x3
      = 256 - 1024x + 1792x- 1792x2
    2. (1.99)8 = (2-0.01)8
      = 256 - 1024 (0.01)+ 1792(0.01)2 - 1792(0.01)3
      = 256- 10.24 +0:1792- 0.001792
      = 245.937408
  7. 2 + log3(2x-7) = log3(5x - 7)
    2log3+ log3(2x - 7) =log3(5x - 7)
    log39 + log3(2x - 7) = log3(5x - 7)
    log3(9(2x - 7)) = log3(5x - 7)
    9(2x - 7) = 5x - 7
    18x - 63 = 5x - 7
    13x = 56
    x = 44/3

  8. 4
    Area of ΔABC = ½ x 2.2 x 3.5sin60
    3.334
    ΔBCD = ½ x 3.5 x h = 3.334
    h = 1.9cm
    locus of Δ
  9. x2 + y2 -6x + 84 - 11 = 0
    x- 6x + y+ 8y = 11
    x4 - 6x + 9 + y2 + 8y + 16 = 11 +9+16
    (x - 3)2 + (y + 4)2 = 36 = 62
    (x - a)+ (y - b)2 = r2
    Centre (3, -4)
    Radius - 6 units.
  10. 8.3 ± 0.05cm by 5:45 ± 0.005 cm
    1. Max. Area = 8.35 x 5.455
      = 45.549250 cm2 
      Min Area = 8.25 x 5.445
      = 44.92125 cm2
      Actual area = 8.3 x 5.45 = 45.235 cm2
      AE = 45.54925 - 44.92125
                            2
      = 0.314
      R.E = 0.314
               45.235
      =0.006941527578
    2. Percentage error = RE x 100
      = 0.006941527578 x 100
      = 0.6941527578%
      ≈ 0.6942%(4sf) 
  11. Area of Δxyz = 15.3 cm2
    Area scale factor = Determinant of matrix
    5ASF = 3 x 2 - (-1x - 2)
    = 6 - 2 = 4
    Area of ΔX'Y'Z = 4 x 15.3
    = 61.2cm2

  12. 6 9.4 
    =  16:6  
    SinB  Sin110
    SinB = 9.4Sin110 = 0.5321
                    16.6
    β = 32.1°
    θ = 180-32.1 =147.9°
    x= 9.42+10.8- 2 x 9.4 x 10.8 cos110°
    = 88.36 + 116.64 - 203.04 Cos110
    = 88:36+ 116.64 + 203.04 x 0.3420
    = 274.43968
    x = 16.566cm(3d.p)
  13. M   H   D
    15  9   20
    27  10  ?
    No. of days = 15 x 20 x 9
                             27 x 10º
    = 10 days
  14. Tn = a + (n - 1)d
    1. T5 = a +(5-1) d = 71 (i)
      T7 = a +(7-1) d = 59 (ii)
      a + 4d = 71
      a + 6d = 59
      -2d = 12
      d = -6
      a + 4x - 6 = 71
      a = 95 .
    2. T10 = a5 + (10-1) x -6
      = 95 - 54
      = 41
  15. 3x - 4y = 2
    6x + y = 13
    7
  16. 2x2 - x - 10
    2x2 - 5x + 4x - 10
    = x(2x - 5) + 2(2x - 5)
    =(2x - 5)(x + 2)
  17.      
    1. Modal class 60-64
    2.        
      Class x f d = x - 62 d2 fd fd2
      40-49 47 3 -45 225 -45 675
      50-54 52 9 -10 100 -90 900
      55-59 57 13 -5 25 -65 325
      60-64 62 15 0 0 0 0
      65-69 67 5 5 25 25 125
      70-74 72 4 10 100 40 400
      75-79 77 1 15 225 15 225
          48≈50     ∑fd = -120 ∑fd2= 2650
      1. Mean
        x = A + ∑fd
                    ∑f
        = 62 + -120
                    50
        = 62 - 2.4 = 59.6
      2. Standard deviation 
        2550 - (-120)2
            50       50 
        = √47.24
        = 6.873
  18.      
    1.          
      1. BT = ¾BO = -¾b
      2. OR =  5  OA +  2  OB
                2+5        2+5
        = 5/7a + 2/7b
      3. AT = AO + OT
        = -a + ¼b/¼b - a
    2. AB = K(¼b-a) = ¼kb - ka
      AD = AR + RD
      = 2/7AB + h(-5/7a - 2/7b)
      = 2/7(b - a) + (-5/7ha - 2/7hb)
      = -2/7a - 5/7ha + 2/7b -2/7hb
      = (-2/7 - 5/7h)a + (2/7 - 2/7h)b
      k = -2/7 - 5/7h → k =2/7 + 5/7h
      ¼k = 2/7 - 2/7h
      ¼(2/7 + 5/7h) = 2/7 - 2/7h
      h = 6/13
      k = 8/13
      AD = 8/13 (b-a) = 2/13b - 8/13a
  19.      
    1.     
      8
    2.      
      1. P(TB or BB)
        =(2/3 x ¾) + (1/3 x 1/6)
        =½ + 1/18 = 9 +1 = 10/18
                            18
      2. P(TT or BT)
        =(2/3 x ¼) + (1/3 x 5/6)
        = 2/12 + 5/18 = 6 +10 = 16/36
                                36
      3. P(TTB or TBB or BTB or BBB)
        =(2/x ¼ x ¾) + (2/3 + ¼ x 1/6) + (1/3 x 5/x ¾) + (1/3 x 1/x 1/)
        = 6/48 + 6/72 + 15/72 + 1/100
        = 184/432
      4. 1 - 23/54 = 31/54
  20.      

    1. 15º 30º 45º 60º 75º 90º 105º 120º 135º 150º 165º 180º
      Y = 3 sin (2x + 30º)   2.60 3.00   1.50 0 -1.50 -2.60 -3.00   -1.50 0  
      Y =cos 2x   0.87 0.50 0 -0.50   -1.00 -0.87   0 0.50 0.87  
    2.     
      9
    3. x = 84º or 174º 
    4.    
      1. Amplitude 3
      2. Period 180º
  21.        
    1. T3 = a + 2d
      T25 = a + 24d
    2.      
      1. a +2d, a +8d, a + 24d
        a + 8d = a + 24d
        a + 2d    a + 8d
        (a + 8d)(a + 8d) = (a + 24d)(a + 2d)
        a2 + 16ad + 64d2 = a2 + 26ad + 48d2
        64d2 = 48d2 = 26ad - 46ad
        16d2 = 10ad
        10a = 16d
        a = 1.6d
        a + 6d + 2(a + 5d) = 78
        3a + 16d = 78
        4.8d + 16d = 78
        20.8d = 78
        d = 3.75 
        a = 6
      2. S9 = 9/2 [2 x 6 + (8 - 1)15/4]
        = 9/2[12 + 30]
        = 189
  22.            
    1. AC = √52 + 72
      = √74
      = 8.602cm
    2. VO = √132 - 4.3012
      = √150.501
      = 12.27
    3. BD = AC = 8.602
      BO = ½BD = 4.301
      ∠VBO = θ
      10
      Cosθ = 4.301
                    13
      θ = Cos-10.3308
      = 70.68º
    4. Midpoint of BC
      VM = √MO2 + OV2
      = √2.52 + 12.27
      = √156.8029
      = 12.52
      11
      Cos x =  2.5  
                  12.52
      x = 78.48
  23.      
    1. L = aM
            √N
      2250 = 450a
                  √64
      a = 2250 x 8 = 40
               450
      L = 40M
            √N
    2. M1 = 0.84M , N1 = 1.44N
      L1 = 0.84aM = 0.84am
              √1.44N     1.2√N
      = 0.7L
      %change = (0.7 - 1)L x 100%
                              L
      = -0.3 x 100% = -30%
    3. G = a+bs
      8 = a + 2b (i)
      12 = a + 4b (ii)
      -4 = 2b 
      b = 2
      a = 8 - 4 = 4
      G = 4 + 2 x 6
      = 4 + 12
      = 16
  24.           
    1. Total income tax 
      Taxable income = 28000 + 900 + 2000 + 1480
                                                      20
      = 2024
      Tax 
      435 x 2 = 870
      435 x 3 = 1305
      435 x 4 = 1740
      435 x 5 = 2175
      Rem  284 x 6 = 1704 
      Total income tax = 7794
    2. Net tax = 7794 - 1056
      = 6738
    3. Increase in income 
      = 150 x 2024 = 3036
         100
      Total income tax = 870 + 1305 + 1740 + 2175 + 1296 x 6
      = 6090 + 7776
       = 13866
      % increase in income tax = 13866 - 7794 x 100
                                                         7794
      = 77.906%

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