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

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Instructions to candidates

  • The paper contains TWO sections: Section I and Section II.
  • Answer ALL the questions in Section I and any five questions from Section II
  • Show all the steps in your calculations, giving your answers 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 1 (50 MARKS)
Answer all the questions in this section

  1. Use a calculator to find V if 1  1     −     1          (2mks)
                                                V   23.9      38.45
  2. Solve for X in Log (7X – 3) + 2 Log 5 = 2 + Log (X+3) (3mks)
  3. A quantity P is partly constant and partly varies as the square of Q. when Q = 2, P = 40 and when Q = 3, P = 65. Determine the equation connecting P and Q (3mks)
  4. Expand (1 − 1/2X)6 up to the fourth term; hence use your expansion to evaluate 0.9966 correct to 4 decimal places. (4mks)
  5. Simplify √5 + 3 . Give the answer in the form of a + b√c where a, b and c are integers (3mks)
                  √5 − 2
  6. Given that X−5, X−3 and 2X−3 are three consecutive terms of a geometric progression, find the possible values of X and the ratio (2X+1):(X+2) (4mks)
  7. The figure below is a segment of a circle cut off by a chord AB. Line CD is a perpendicular bisector of chord AB.
    Maths BGMDM PP2 Q7 2122
    If AB is 24cm and CD is 8cm, calculate the radius of the circle. (3mks)
  8. By completing the square, solve for x in the equation 2x2 − 6 = x. (3mks)
  9. Given that y = b  − bx² make x the subject (3mks)
                           cx2 − a
  10. The base and height of a right-angled triangle are 4cm and 5cm respectively. Calculate the percentage error in its area. (3mks)
  11. Given that P = Maths BGMDM PP2 Q11a 2122, find ;
    1. Its inverse (1mk)
    2. The value of x and y if Maths BGMDM PP2 Q11b 2122(3mks)
  12. The equation of a circle is given by x2 + y2 + 6x – 10y − 30 = 0. Determine the radius and center of the circle (3mks)
  13. Find the value of X which satisfies the equation 52x − 6 × 5x + 5 = 0 (3mks)
  14. A scooter mixes oil and petrol in the ratio 5:19. If petrol costs Ksh. 130 per liter and oil costs Ksh. 250 per liter, find the cost of a liter of the mixture. (2mks)
  15. Solve the pair of equations simultaneously (4mks)
    2x – y = 3
    x2 – xy = −4
  16. The cash price of a water pump is Ksh. 38,000. Mr. Ahero opts to buy the pump on hire purchase terms by paying a deposit of Ksh. 6,500 and 24 equal monthly installments. Calculate the amount of each installment, if simple interest of 20% p.a is charged. (3mks)

SECTION II (50 MARKS):
Attempt any five questions in this section

  1. The first term of an arithmetic sequence is equal to the first term of the geometric sequence. The second term of the arithmetic sequence is equal to the fourth term of the geometric sequence, while the tenth term of the arithmetic sequence is equal to the seventh term of the geometric sequence.
    1. Given that a is the first term and d is the common difference of the arithmetic sequence while r is the common ratio of the geometric sequence, write down two equations connecting the arithmetic and geometric sequences. (2mks)
    2. Find the value of r that satisfies the geometric sequence (4mks)
    3. Given that the tenth term of the geometric sequence is 5120, find the values of a and d (2mks)
    4. Calculate the sum of the first 20 terms of the arithmetic sequence (2mks)
  2. Three quantities R, S and T are such that R varies directly as S and inversely as the square of T.
    1. Given that R = 480 when S = 150 and T = 5, write an equation connecting R, S and T (3mks)
    2. Find,
      1. the value of R when S = 160 and T = 1.6 (3mks)
      2. the percentage change in R if S increases by 5% and T decreases by 20% (4mks)
  3. The table below shows income tax rates
     Monthly income in Kenya shillings (Ksh)   Tax rate % in each shilling 
     Up to 9680  10%
     From 9681to18800  15%
     From 18801 to 27920  20%
     From 27921 to 37040  25%
     From 37041 and above  30%
    In that year Okumu’s salary amounted to K£ 45,000 p.a and he received allowances totaling Ksh. 300,000 p.a. He was entitled to:-
    1. Monthly personal relief of Ksh. 1,056
    2. Monthly insurance relief at the rate of 15% of the premium paid
      Okumu paid a monthly premium of Ksh. 2,500 towards his life insurance policy
      Calculate
      1. His gross monthly income in Ksh (2mks)
      2. The monthly income tax he pays (5mks)
      3. His net monthly income, if his other monthly deductions were: - Ksh. 4,800 to HELB, Ksh. 5,000 to his co-operative and Ksh. 2,800 towards a bank loan repayment. (3mks)
  4. Square OABC with vertices O(0,0),A(2,0), B(2,2) and C(0,2) is mapped onto O’(0,0), A’(2,0), B’(5,2) and C’(3,2) by the matrix T = Maths BGMDM PP2 Q20 2122
    1. Find T (3mks)
    2. Draw O’A’B’C’ and reflect it on the line x + y = 0 to obtain O”A”B”C” (4mks)  (attach graph paper)
    3. What single matrix P maps OABC to O”A”B”C” (3mks)
  5. In the triangle PQR below L and M are points on PQ and QR respectively such that PL: LQ = 1:3 and QM: MR = 1:2, PM and RL intersect at X. Given that PQ = b and PR=c,
    Maths BGMDM PP2 Q21 2122
    1. Express the following vectors in terms of b and c
      1. QR (1mk)
      2. PM (1mk)
      3. RL (1mk)
    2. By taking PX = hPM and RX = kRL where h and k are constants find two expressions of PX in terms of h, k, b and c. Hence determine the values of the constants h and k. (6mks)
    3. Determine the ratio LX : XR (1mk)
  6. During a traffic crackdown, 1,000 motor cycles were sampled. 250 of these were found to lack necessary driving gear, 200 had no valid insurance and 300 lacked the driving license. Taking the sample to represent all motorcycles in the country;
    1. Represent the information in a tree diagram (3mks)
    2. Find the probability that, a motorcyclist at any given time
      1. Has no driving license (3mks)
      2. Lacks a valid insurance but is in proper driving gear and has a valid driving license (2mks)
      3. Has none of the offence (2mks)
  7. In the figure below, K L M and N are points on the circumference of a circle centre O. The points K, O, M and P are on a straight line. PQ is a tangent to the circle at N. Angle KOL = 130° and angle MKN = 40°
    Maths BGMDM PP2 Q23 2122
    Find the values of the following angles, stating the reasons in each case:
    1. <MLN (2mks)
    2. <OLN (2mks)
    3. < LNP (2mks)
    4. <MPQ (2mks)
    5. <KNQ (2mks)
  8. Complete the table below for y = Sin 2x and y = Sin(2x+30)° giving values to 2 d.p (1mk)
     x°   0°   15°   30°   45°   60°   75°   90°   105°   120°   135°   150°   165°   180° 
     Sin 2x   0.00         0.87          −0.87        0.00
     Sin (2x+30)°   0.50        0.50        −1.00         0.50
    1. Draw the graph of y = Sin 2x and y = Sin(2x+30)° on the same axis (4mks)
    2. Use your graph to solve Sin(2x+30)°− Sin 2x = 0 (1mk)
    3. Describe the transformation which maps the wave Sin 2x onto the wave Sin (2x +30) (2mks)
    4. State the amplitude and period of y = a cos (bx +c) (2mks)


MARKING SCHEME

 NO   WORKING   MARKS   COMMENT 
 1  1/V= 0.01583
V =      1     
      0.01583
 V = 63.17
    M1
    A1
 
        2  
 2

Log (7X – 3) + 2 Log 5 – Log (X+3) = 2
Log 25(7X – 3)= 2
         (X+3) 
102 = 25(7X – 3)
              (X+3)
100x +300 = 175x – 75
X =15

   M1


   M1
 
    A1

 
         3  
 3 P = k + cQ2
40 = k + 4c
65 = k + 9c
c = 5, k = 20
p = 20 + 5 Q2

   M1
   A1

   B1

 Both eqns
 Both values
       3  
 4  1(1/2X)+ 6(1/2X)+ 15(1/2X)2 + 20(1/2X)3 
1 − 3 + 15   −  20 
      x     4x2     8x3
1 −   3   +  3   −      1     
      125   125    781250
= 0.9760

   M1
  M1

  M1
  A1

 
       4  
 5

 (√5+ 3) × (√5+ 2)
 (√5- 2)     (√5+ 2)
  5+2√5+3√5+6
         5 - 4
11 + 5√ 5

  M1

  M1
  A1
 
        3  
  6  = X−3 = 2X−3
    X−5     X−3
x2 − 6x +9 = 2x2 −13x +15
(x − 6)(x −1) = 0
x = 6 or x = 1
13:87 or 1:1

   M1


   M1
   A1
   B1

 
        4  
 7  Maths BGMDM PP2 ans 7 2122
122 = (r − 8)2 = r2
 144 + r2 − 16r + 64 = r2 
  R= 13
   M1
  
   M1
   A1
 
         3  
 8  x2 – ½ x + (¼ )2 =3 + (¼ )2
(x – ¼ )2 = ± 49/16
x = 3/2 0r -4
   M1
   M1
   A1
 
         3  
 9  (cx2 – a) y = b – bx
cx2 + bx = b + ay
x2 = b+ ay
        cy+b
x = Maths BGMDM PP2 ans9 2122
   M1
   M1
   
    A1
 ✓Formation of the equations 
 ✓attempt to solve
        3  
 10 Min area = ½ × 4.5 × 3.5 = 7.875

Max area = ½ × 5.5 × 4.5 = 12.375
Absolute Error = ½ (7.875 + 12.375)
Percentage Error = 8.4375 × 100% = 84.38%
                                   10

   M1


   M1A1
 
        3  
 11
  1.  ½ Maths BGMDM PP2 ans 11a 2122
  2. Maths BGMDM PP2 ans 11b 2122(x/y) = (3/2)
½ Maths BGMDM PP2 ans 11a 2122 Maths BGMDM PP2 ans 11b 2122 (x/y) = ½ Maths BGMDM PP2 ans 11a 2122(3/2)
½ Maths BGMDM PP2 ans 11c 2122 (x/y) = ½ (6/-8)
Maths BGMDM PP2 ans 11d 2122 (x/y) = (3/−4)
X = 3 , y = −4

   M1


  M1

  M1

  A1

 
      4  
 12  x2 +6x + 9 + y2 – 10y +25 = 30 + 9 + 25
(x + 3)2 + (y – 5)2 = 82
Centre (-3, 5) Radius = 8 units
 M1
 M1
 A1
 
       
 13  Let 5x = y
y2 – 6y + 5 = 0
(y – 5) (y – 1) = 0
y = 5 or 1
5x = 5 or 5x = 1
x = 1 or x = 0

   M1

   A1
 
   B1

 
       3  
 14   (250 × 5) + (130 × 19) = 155
              5+19
 M1A1  
       2  
 15  y = 2x – 3
x2 – x(2x – 3) = – 4
(x – 4) (x – 1) = 0
x =4 or x = 1
y = 5 or y = –1
  M1
  M1
  A1
  B1
 
      4  
 16  31500 + (31500 × 20/100 × 24/12 ) = 44,100
44100  = 1,837.50
   
24
   M1
 M1A1
 
       3  
 17
  1. a + d = ar3
    a + 9d = ar6
  2. r3 = a+d
             a
    a + 9d = a(a+d)2 
                        a
    a2 + 9ad = a + 2ad + d2
    7a = d
    r3 = a+7a  = 8
             a
    r = 2
  3. ar9 = 5120
    a(29) = 5120
    a = 10
    d = 70
  4. S20 = 20/2{2(10)  + 19×70}
          = 10(20 + 1330)
          = 13,500

   B1
   B1


   M1
   
   M1
   M1

   A1

   M1

   A1
   M1

   A1

  a and d
      10  
 18
  1. 480 = 150 K
                25
    K = 802
    R = 80S
            T2
  2.  
    1. R = 80 ×160
                1.62
         =5,000
    2. R1 = 80 × 1.05S
                   (0.8T)2
                           S 
      R1 = 131.25 T2
      Change in R = 51.25S
                                  T2
      % Change in R = 51.25 ×100%
                                     80
                               = 64.0625%

   M1
   M1
   A1

 

   M1M1

   
    A1

   M1

 
 
   M1

   M1

 
       10  
 19
  1. (45,000×20)+(300,000)
                  12
    = 100,000
  2. 9,680 ×10% = 968
    9,120 × 15% = 1,368
    9,120 ×20% = 1,824
    9,120 × 25% = 2,280
    62,960 × 30% = 18,888
    Total = 25,328
    Net tax = 25, 328 – (1,056 + 15% of 2,500)
                 = 23,897
  3. 100, 000 – (23,897 + 2,500 + 4,800 + 5,000 + 2,800)
       = 61, 003 

   M1


   A1

   M1

   M1

   M1
   M1
    A1
  MN1M1
   A1

 
       10  
 20
  1.  Maths BGMDM PP1 Q20ai 2122
    a = 1      b = 1.5
    c = 0       d = 1
    Maths BGMDM PP1 Q20aii 2122
  2.  
    Maths PP2 20b BDM 2122
  3.  
    Maths BGMDM PP2 ans20ci 2122
    w = 0      x = −1
    y = −1     z = −1.5
    Maths BGMDM PP2 ans20cii 2122
   
 21
  1.  
    1. QR = c – b
    2. PM = 1/3 (2b − c)
    3. RL = ¼ b − c
  2. PX =2/3 h b+ 1/3 h c
    PX= ¼ k b + (1 – k) c
    2/3 h b+ 1/3 h c = ¼ k b + (1 – k) c
                     h= 3/8 k
    1/3 h = 1 – k
    1/8k = 1 – k
    k = 8/9
    h = 1/3

   B1
   B1
   B1

   B1
   B1
   B1

   B1
   B1

   B1

  One unknown for both values
       10  
 22
  1.  
    Maths Ans 22 PP2 BDM 2122
  2.  
    1. (25/100 × 2/10 × 3/10)+(25/100 × 8/10 × 3/10)+(75/100 × 2/10 × 3/10) = (75/100 × 8/10 × 3/10)
      3/200 + 3/50 + 9/200 + 9/50 
      = 3/10 
    2. (75/100 × 2/10 × 7/10) + (75/100 × 8/10 × 7/10)
      = 21/100 + 21/5063/100 
    3. (75/100 × 8/10 × 7/10)
      = 21/50
 







 
  




  M1
 
  M1
  A1
  M1
   A1

  M1
   A1
 
 23
  1.  <MLN = 40° (angles subtended by the same arc NL are equal
  2. <OLN = 25° (65° – 40°)
  3. <LNP = 65° (40 + 25 angles subtended in alternate segment)
  4. <MPQ = 10°(180°− (130+40))
  5. <KNQ = 50° (Supplementary angles)
   
 24
 x°   0°   15°   30°   45°   60°   75°   90°   105°   120°   135°   150°   165°   180° 
 Sin 2x   0.00   0.05   0.87   1.00   0.87   0.50   0.00    −0.50   −0.87   −1.00    −0.87    −0.50   0.00
 Sin (2x+30)°   0.50  0.87  1.00  0.87  0.50  0.00  −0.50   −0.87 −1.00    −0.87   −0.50   0.00  0.50
  1.  
    Maths Pp2 ans 24 BDM 2122
  2. x = 37.5°, x = 127.5°
  3. a translation of Maths BGMDM PP1 Ans 24c 2122
  4. amplitude a
    Period 360/b
   

 


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