Mathematics Paper 2 Questions and Answers - KCSE 2022 Prediction

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INSTRUCTIONS TO CANDIDATES

  • This paper consists of two sections; Section I and Section II.
  • Answer all questions in section I and only five questions from section II.
  • Show all the steps in your calculations, giving the answers at each stage in the spaces provided below each question.
  • Marks may be given for correct working even if the answer is wrong.
  • Non-programmable silent electronic calculators and KNEC mathematical tables may be used, except where stated otherwise.
  • Candidates should answer the questions in English.


Questions

SECTION I (50 marks)
Answer all the questions in this section in the spaces provided.

  1. The roots of a quadratic equation are x = 11/3 and x = −½. Form the quadratic equation in the form where ax2 + bx + c = 0 where a,b and c are integers. (2 marks)
  2. Find the percentage error in estimating the volume of a cuboid whose dimensions are 8 cm by 4.20 cm by 5.6 cm. (3 marks)
  3. The cash price of a television set is sh. 25 000. A customer paid a deposit of sh. 3 750 . He paid the amount owing in 24 equal monthly installments. If he was charged simple interest at the rate of 40% p.a How much was ;
    1. The carrying charge? (2 marks)
    2. Each installment? (1 mark)
  4. Without using mathematical tables or a calculator, evaluate 3 log4(√3) − ½ log43 + 3log42 − log46   (3 marks)
  5. Given that 2−3n²(a + x)−n = 20−3n² , make n the subject of the formula. (3 marks)
  6. In the figure below, A, B, C, D and E are points on a circle. Point C is due north of point D and point E is due west of point D. Angle CAB = 27º. The angle of elevation of point B from point E is 87º.
    Maths PVM Q6 2122 PP2
    Find the size of angle of elevation of point B from the point D. (3 marks)
  7. Given that x = 2 − √3, find the value of (x − 1/x)3        (3 marks)
  8. Expand (x − y)4. Hence use the expression to find the value of 1.584 , giving your answer correct to 4 significant figures. (3 marks)
  9. The diagram below shows a wedge ABCDEF.
    Maths PVM Q9 2122 PP2
    Calculate the angle between the lines BD and BF, correct to 1 decimal place. (4 marks)
  10. State the amplitude and the phase shift of the curve y = 2 Cos(2x+ 90°)     (2 marks)
  11. The tax authority collected the following amount of tax from different firms in a particular market.
     Amount of Taxes (Ksh)   10 000   11 000   12 000   13 000   14 000 
     No. of firms     3     12     18      12       3
    Determine the quartile deviation of the data. (3 marks)
  12. The position of towns P(0º,51ºW) and Q(0º,37ºE).Find the distance between the two towns in kilometres, correct to one decimal place. (3 marks)
  13. A trader bought three kinds of nuts at Ksh. 10 per kg, Ksh. 8.40 per kg and Sh. 6 per kg. He mixed them in the ratio 3: 5: 4 respectively and sold the mixed nuts making a profit of 25% in the process. At what price per kg did he sell them? (3 marks)
  14. In the figure below, AB and CB are tangents to the circle, centre O and radius 6.5 cm. The arc AC subtends an angle of 140º at the centre of the circle.
    Maths PVM Q14 2122 PP2
    Calculate the area of the shaded region, correct to 2 decimal places (Use π = 3.142)     (4 marks)
  15. A square QRST with vertices Q(1,1),R(3,1),S(3,3) and T(1,3) is transformed by the matrix to Maths PVM Q15 2122 PP2 to Q'R'S'T' Find:
    1. the are of square QRST; (2 marks)
    2. the area of image Q'R'S'T' (2 marks)
  16. Find the radius and the centre of a circle whose equation of a circle is x2 + y2 − 2x + 8y − 8 = 0, hence draw the circle on the grid provided below. (4 marks)

SECTION II (50 marks)
Answer any five questions from this section in the spaces provided.

  1.  
    1. Complete the table below for values of y = x3 − 7x2 + 8x + 16 for − 2 ≤ x ≤ 6      (2 marks)
       x   − 2   − 1   0   1   2   3   4   5   6 
       y = x3 − 7x2 + 8x + 16   − 36       18         6  
    2. On the grid provided, draw the graph of the function y = x3 + 7x2 − 7x − 15 for  − 2 ≤ x ≤ 6
      Use a scale of 2 cm to represent 1 unit on the x – axis and 1 cm represent 5 units on the y – axis.   (3 marks)
    3. Using the graph, determine:
      1. the average rate of change between x = 1.2 and x = 3.1 (2 marks)
      2. The instantaneous rate of change of the curve at x = − 0.5 , correct to one decimal place.      (3 marks)
  2.  
    1. An increasing arithmetic progression (AP) has first term as a and a common difference d. The ratio of the fifth term to the twelfth term of the A.P is 6:13.Given that product of first term a and the third term is 32. Find the;
      1. Value of a and the value of d. (4 marks)
      2. Sum of the first five terms of the A.P. (2 marks)
    2. Determine the least number of terms that must be added together so that the sum of the progression is more than 700. (4 marks)
  3. The table below shows some values of the curves y = 2 cos(½x) and y = sin(½x)
    1. Complete the table for values of y = 2 cos(½x) and y = sin(½x), correct to 1 decimal place.     (2 marks)
       x°    −360    −270    −180    −90   0   90   180   270   360 
       y = sin(½x)      0        0           0
       y = 2cos(½x)    −2.0       0          0     −2.0 
    2. On the grid provided, draw the graph of y = 2 cos(½x) and y = sin(½x) for −360° ≤ x ≤ 360° on the same axes. (5 marks)
    3. Use your graph in (b) above to find the values of; 
      1. x in 2cos(½x) − sin(½x) = 0     (2 marks)
      2. y when  2cos(½x) = sin(½x)       (1 mark)
  4. The runway in Wilson airport is 3.22 km long and it is wide enough to accommodate the landing of two aircrafts simultaneously, An aircraft x landed at the runway with a speed of 144 km/hr and maintained this speed till the end of the runway when it took off again, Another aircraft y landed at runway half a minute later with a speed of 252 km/hr and followed x on the parallel runway throughout to the end of the runway when it also took off again.
    1. How far from starting point was x at the instant y overtook x. (4 marks)
    2. Find the distance between the two planes at the instant when y is beginning to take off.(3 marks)
    3. Selina took a train from her home to the Nairobi. At a railway station she met another train travelling in the opposite direction at 25 km/hr. If the train she boarded was travelling at a speed of 15 km/hr and it took 10.8 seconds to completely pass the other train. Calculate the length of each train, if the train Selina boarded was twice as long as the train moving in the opposite direction. (3 marks)
  5. The charge C in shillings per person for a certain seminar is partly fixed and partly inversely proportional to the total number of N of people.
    1. Write down an expression for C in terms of N. (2 marks)
    2. When 100 people attended the charge was Ksh 8,700 per person. While for 35 people charge is Ksh 10,000 per person. Calculate the fixed charges and hence state the equation connecting charge C. (4 marks)
    3. If a person had paid the full amount and doesn’t attend, the fixed charge is refunded. A group of people paid but 10% of them did not attend. After the refund the organizers remained with Ksh574,000.Find the number of people initially in the group. (4 marks)
  6. In certain game, 2/3 of the players are boys while the rest are girls. 80% of the boys and 90% of the girls are right handed. The probability that a right handed player will break the rules of the game is 0.1.Otherwise it is 0.3, these probabilities being independent of the sex of the player.
    1. Draw a tree diagram to represent this information. (2 marks)
    2. Use your tree diagram in (a) above determine;
      1. The probability that a player chosen at random from the game is left handed. (2 marks)
      2. The probability that a game rule is broken is a session. (2 marks)
      3. The probability of a girl breaking a rule of the game. (2 marks)
      4. The probability that a boy who is right handed doesn’t break the rules and a girl who is left handed break rules of the game. (2 marks)
  7. In the diagram drawn below, point D and E are such that AD = 1/3OB and OE = 1/3OA
    Maths PVM Q23 2122 PP2
    1. Given that vectors OA = a and OB = b, express in terms of a and b:
      1. OD (1 mark)
      2. BE (1 mark)
      3. BD (2 marks)
    2. Line OD and BE intersect at X such that OX = kOD and BX = hBE and where k and h , express OX in two different ways (2 marks)
    3. Using the expressions in (b) above, determine.
      1. the values of k and h. (3 marks)
      2. Hence state the ratio in which k divides OD. (1 mark)
  8. A plan for a national stadium is in the form a pentagon ABCDE inscribed in a circle, a hockey pitch is to be constructed outside the stadium. The centre of the pitch is fixed at a point X. Given that the angle DAC = 36º, angle  ACB = 42° AE = ED, BC = AB and that the tangent at C meets ED produced at X.
    1. Calculate the angles of the pentagon.    (4 marks)
    2. Giving reasons, calculate ∠DXC.    (3 marks)
    3. Given that DX = 20m, calculate the area of triangle CDX.     (3 marks)


 



 



 



Marking Scheme

  1. (x − 4/3)(x + ½) = 0
    x2 − 5/6x − 2/3 = 0 
    6x2 − 5x − 4 = 0 
  2. Actual volume = (8 × 4.20 × 5.6) = 188.16cm3 
    Maximum volume = (8.5 × 4.205 × 5.65) = 201.945125cm3
    Minimu volume = (7.5 × 4.195 × 5.55) = 174.616875cm3 
    Absolute error = 201.945125 − 1740616875
                                                  2
                           = 13.664125
    Percentage error = (13.664125) × 100%
                                        188.16
                               = 7.262%
    Alternative method
    (0.5 + 0.005 + 0.005) × 100%
       8      4.20       5.6
    = 7.261906%
    1. Sh.(25 000 – 3730) = Sh. 21 250
      Carrying charge = (21250 × 40/100 × 2)
                                 = Sh. 17 000
    2. 21250 + 17000 = Sh. 1593.75
                 24
  3. 3 log4(√3) − ½ log43 + 3log42 − log4
    Maths PVM Ans4 2122 PP2
    = log423 
    Let m = log423 ⇒ 22m = 23 
    2m = 3
    m = 1½
    ∴ log423 = 1½
  4.   2−3n²(a + x)−n = 20−3n²
    (a + x)−n = (103n)−n 
    a + x = 103n 
    log(a + x) = 3n log 10
    n = 1/3 log(a + x)
  5. The required angle is BDE
    Now ∠EBC = 90° angles in a semicircle in triangle EBC
    ∠BEC = 27°, hence;
    ∠ECB = 90 − 27 = 63°(angles in a traingle)
    But ∠BDE = ∠BCE, angles in the same segment.
    Hence ∠BDE = 63°
  6. 1   1      × 2 + √3 = 2 + √3 = 2 + √3
    x    2 − √3    2 + √3        1
    x − 1/x = (2 − √3) − (2 + √3) = −2√3
    (x − 1/x)3 = (−2√3)3 
                   =(−2)3 (√3)2
                   = −24√3
  7. (x  − y)4 = x4 + 4x3 ( −y) + 6x2  − y2 + 4x( −y)3 + ( −y)4 
    (1.58)4 = (2  − 0.42)2 
    24  − 4(2)2 (0.42) + 6(2)2  − 4(2)(0.42)3 + (0.42)4 
    = 16  − 13.44 + 4.2336  − 0.592704 + 0.013069123
    = 6.213965123
    ≅ 6.214
  8. BF =  √(42 + 22)
         = 2√5
    DE = √(52 + 22)
          = √29
    Cos β =  41 + 20 − 29  
                  2 × √41 × √20
             = 56.0°
  9. y = 2Cos(2x + 90°)
    Amplitude = 2
    Phase shift = − 180° = 45°(left of 45°)
                               4
  10. Q148/4 = 12th item = 11,000
    Q3 = ¾ × 48 = 36th item
         = 13,000
    Q.D = 13,000 − 11,000
                          2
           = 1,000
  11. Latitude difference = 51° + 37° = 88°
    Distance in kilometres; 88/360 × 22/7  × 2  × 6370
    = 9787.6km
    1. Side of QRST = √((1 − 1)2 + (3 − 1)2)
                             = 2
                      ∴ Area = 2 × 2 = 4
    2. The area of image Q'R'S'T'
      = det Maths PVM Q15 2122 PP2 × 4
      = 6 × 4 
      =24sq
  12. AB = 6.5 tan 70°
          = 17.86cm
    Area of quadrilateral OABC = 2(½ × 6.52 tan 70°)
                                                = 116.08cm2 
    Area of sector OAC = 140/360 × 3.142 × 6.52 
                                     = 51.62cm2 
           Shaded area = 116.08 − 51.62
                                = 64.47cm2 
  13. C fraction in 1 minute = 1/30 − 1/401/120 
                C takes 120 minutes
    B fraction in 1 minute = 1/30 − 1/451/90 
               B takes 90 minutes
    A fraction in 1 minute = 1/30 − 1/905/360 
          A takes = 360/5 = 72 minutes
  14.  x2 − 2x + ( −2/2) + y2 + 8y + (8/2)2 = 8 + 1 + 16
      (x − 1)2 + (y + 4)2 = 52 
    Centre (1, −4)
    Radius 5 units
    PVM 2122 maths PP2 ans 16
    1.  
       x   − 2   − 1   0   1   2   3   4   5   6 
       y = x3 − 7x2 + 8x + 16   − 36     0  16   18   12        6  28 
    2.  
      PVM 2122 maths PP2 Ans 17b
    3. Average rate of change;
      3 − 17 
         3.1−1.2
      = − 7.37
    4. Tangent at x = −0.5
      Gradient =  −6 − 30  
                       −1.5 − 0.8
                    = 15.7
  15.  
    1.  
      1. a + 4d  ⇒ a = 2d .........(i)
        a + 11d   13
        a(a + 2d) = 32 ...................(ii)
        2d(2d + 2d) = 32
                   8d2 = 32
                     d2 = 4
                     d = ±2
                     d = 2
                    a = 2d
                       = 2 × 2 
                      = 4
      2. S55/2{2 × 4 + (5 − 1)2}
             = 5/2(8 + 8)
            =40
    2. n/2 { 2 × 4 + (n − 1)2} > 700
                        n/2 {6 + 2n} > 700
               2n2 + 6n − 1400 > 0
                   n2 + 3n − 700 > 0
      n2 + 28n − 25n − 700 > 0
      n(n + 28) − 25(n + 28) > 0
                n − 25) (n + 28) > 0     
      ∴ n − 25 > 0
        n > 25
         n = 25 + 1
            = 26
  16.   
    1.   
       x°    −360    −270    −180    −90   0   90   180   270   360 
       y = sin(½x)      0  −0.7   − 1.0  − 0.7   0  −0.7   1.0  0.7     0
       y = 2cos(½x)    −2.0  −1.4     0  1.4  2.0    1.4    0  − 1.4    −2.0 
    2.  
      PVM 2122 Maths PP2 Q19b
    3.  
      1. −234° ±1° and 126°±1°
      2. values of y 
        y1 = − 0.9
        y2 = 0.9
  17.  
    1. Distance in 1 minute 144 × 30 = 1.2km
                                           3600
      D252km/h = 1.2 + x   D144km/h = x
      S = 252 km/h    S = 144km/h
      T = 1.2 + x        T =   x  
               252                144
      1.2 + x   =    x  
         252         144
      1.75x − x = 1.2
      0.75x = 1.2
             x = 1.6 km
      D252km/h = 1.2 + x 
                    = 1.2 + 1.6
                    = 2.8 km
      Alternatively;
      Relative speed = 252 − 144
                              = 108 km/h
      T =  1.2  hrs
            10.8    90
      D = 1/90 × 144
          = 2.8km
    2. Remaining distance = 3.22 – 2.8 = 0.42 km
      Distance = 144 × 0.42 = 0.24km
                                   252
      (0.42 − 0.24) = 0.18km
    3. Distance = 2x + x = 3x
      RS = (25 + 15) = 40km/h
      3x = 10.83
      40    3600
      x = 40 × 10.8
             3 × 3600
      = 0.04km  = 40m
      Train 1     = 40m
      Train 2; 2 × 40 = 80m
  18.  
    1. C = k + m/N 
    2. 8,700 = k + m/100 
      10,000 = k + m/35
      8700 − m/100 = 10000 − m/35 
      100m − 35m = 4,550,000
           m = 4550000
                       65
              = 70,000
      k = 85,000 − m/100 
         = 85,000 − 70000
                             100
        = 8000
      C = 8000 + 70000
                             N
    3. Let the number of people be x; P = 90/100x = 0.9x
      Amount per person = 57400
                                           0.9x
      57400 = 800 + 70000
      0.9x                   0.9x
      1/0.9x(574000 − 70000) = 8000
      x =   504000   
            0.9 × 8000
         = 70
  19.  
    1.  
      Maths PVM Ans22a 2122 PP2
    2.  
      1. P(BL) or P(GL)
        P(BL) or P(GL) = (2/3 × 2/10) + (1/3 × 1/10)
                                = 1/6  
      2. P(BRB) or P(BLB) or P(GRB) or P(GLB)
        = (2/3 × 8/10 × 1/10) + (2/3 × 2/10 × 3/10) + (1/3 × 9/10 × 1/9) + (1/3 × 1/10 × 3/10)
        4/751/251/301/100 
        41/300   
      3. P(GRB) or P(GLB)
        P(GRB) or P(GLB) = (1/3 × 9/10 × 1/10) + (1/3 × 1/10 × 3/10)
                                      = 9/3001/100 
                                      = 1/25  
      4. P(BRB') or P(GLB')
        P(BRB') or P(GLB') = (2/3 × 8/10 × 9/10) + (1/3 × 1/10 × 3/10)
                                         = 144/300 × 3/300 
                                         = 3/625 
  20.  
    1.  
      1. OD = a1/3b
      2. BE = −b1/3a1/3a − b 
      3. BD = BO + OA + AD
        = −b + a1/3b
        = a − 2/3b
    2. OX = k(a1/3b) = ka1/3kb
      OX = OB + hBE 
      = b + h(1/3a − b) = b(1 − h)+1/3ha
    3.  
      1. ka1/3kb = b(1 − h) + 1/3ha
        k1/3h
        1/3k = 1 − k = 3(1 − h)
        1/3h = 3 − 3
        31/3h = 3
        h9/10 
        k1/3 × 9/10 
            = 3/10 
      2. OX:XD = 3:7
  21.   
    1.  
      Maths PVM Ans24a 2122 PP2
      ∠ABC = 180° − 84° = 96°
      ∠ACD = 180° − (84 + 36)° = 60°
      ∠DEA = 180° − 60° = 120°
      ∠EDA = ∠EAD = 180 − 120 = 30°
                                         2
      ∠BCD = ∠ACD + ∠ABC 
                = 60+ 42 = 102°
      ∠CDE = ∠CDA + ∠EDA 
                 = 84 + 30 
                =114°
      ∠EDA = 120°
      ∠EAB = ∠EAD + ∠DAC ∠ CAB 
                = 30 + 36 + 42
               = 108°
    2. ∠CAB = ∠BCY = 42° Alternative angles are equal
      ∠XDC = 180 − (30 + 84) = 66°, Angles on straight line are supplementary/ add up to 180°
      ∠DXC = 180 − (66 + 36) = 78°, sum of angles in a triangle add up to 180°
    3. DC = sin 78 (  28   )  = 33.28m
                           sin 36
      A = ½ × 33.28 × 20 sin 66°
         = 305.05m2

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