Mathematics Paper 2 Questions and Answers - Maranda Pre-Mock Examinations 2022

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

1. An empty tank of capacity 99 792 litres is to be filled with water using a cylindrical pipe of diameter 0.042m. The rate of flow of water from the pipe is 6 m/s. Find the time in hours it would take to fill up the tank. (take π = ²²/₇)(3 marks)
2. The first term of a Geometric Progression (GP) is 5. The common ratio of the G.P. is 4. The product of the last two terms of the G.P. is 25 600. Determine the number of terms in the G.P. (4 marks)
3. The expression dx² -56x+16 is a perfect square, where d is a constant. Find the value of d.(2 marks)
4. Make v the subject of the formula in s =    cv     (3 marks)
                                                                  √dv² - f 
5. The figure below shows a circle and a point Q outside the circle.
   
   Using a ruler and pair of compasses, construct a tangent to the circle from Q. (4 marks)
6. Four quantities W, X, Y and Z are such that W varies directly as the square root of X and inversely as the square of the sum of Y and Z. Quantity X is decreased by 19% while quantities Y and Z are each increased by 16%. Find the corresponding percentage change in W correct to 2 decimal places. (4 marks)
7. The figure below represents a prism PQRSTUVW of length 7 cm. The cross section QRUV of the prism is a trapezium in which VU= 27 cm, QR = 24 cm, QV = 5 cm and ∠VUR - ∠ORU = 90°.
   
   Calculate correct to 1 decimal place the angle between the line UP and the plane VUTW. (3 marks)
8. The cash price of a refrigerator is Ksh 45 000. A customer bought the refrigerator on hire purchase terms by paying a deposit of Ksh 18 000 followed by 15 equal monthly instalments of Ksh 2 300 each. Annual interest, compounded quarterly, was charged on the balance for the period of 15 months. Determine, correct to 1 decimal place, the rate of interest per annum. (4 marks)
9. The table below shows the values of t and the corresponding values of h for a given relation.
   1. On the grid provided, draw a graph to represent the information on the table given (2 marks)
      

      t	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0	4.5
      h	8.0	4.2	2.9	2.2	1.7	1.4	1.2	1.1	1.0

   2. Use the graph to determine, correct to 1 decimal place, the rate of change of h at t = 2 (2 marks)
      
10. The equation of a trigonometric wave is y=2 cos(bx -60)°. The wave has a period of 120°.
    1. Determine the value of b. (1 mark)
    2. Deduce the phase angle of the wave. (1 mark)
11. A point Ris 1800 nm to the East of a point T(30°S,170°E). Find the longitude of R to the nearest degree (3 marks)
12. A box contains 4 blue beads and 7 yellow beads. The beads are identical except for the colours. Two balls are picked at random without replacement.
    1. Draw a tree diagram to show all the possible outcomes. (1 mark)
    2. Determine the probability that the balls picked are of different colours. (2 marks)
13. The figure below shows triangle ABC.
    
    Using a ruler and a pair of compasses, locate a point X on the triangle such that X is 3 cm from line BC and is equidistant from lines AB and AC. Measure length XC. (3 marks)
14. The position vectors of points RS and T are OP=3i- +1.5k, OS = 6i-2.5j +3k and OT -41 -1.5j +2k. Show that R, S and T are collinear points.(3 marks)
15. In a transformation an object of area y cm² is mapped on to an image whose area is 9y cm². Given that the matrix of the transformation is, find the possible values of y. (3 marks)
16. Given that sin(y +20)º =-0.7660, find y to the nearest degree, for 0° ≤ y ≤ 360° (3 marks)

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

17. Pump R can fill an empty water tank in 6 hours while pump S can fill the same tank in 7 hours. On a certain day, when the tank was empty, both pumps were opened for 17 hours.
    1. Determine the fraction of the tank that was still empty at the end of the 17 hours. (4 marks)
    2. Pump R was later opened alone to completely fill the tank. Determine the time it took pump R to fill the remaining fraction of the tank. (2 marks)
    3. The two pumps R and S are operated by different proprietors. Water from the full tank was sold for Ksh 31 500. The money was shared between the two proprietors in the ratio of the quantity of water supplied by each.
       Determine the amount of money received by the proprietor of pump R (4 marks)
18. A rectangular plot measures 75 m by 36 m A lawn, rectangular in shape, is situated inside the plot with a path surrounding it as shown in the figure below.
    
    The width of the path is ym between the lengths of the lawn and those of the plot and 2y m between the widths of the lawn and those of the plot.
    1. Form and simplify an expression in y for the area of the
       1. Lawn; (2 marks)
       2. Path. (1 mark)
    2. The ratio of the area of the path to the area of the lawn 3:2
       1. Form an equation in y and hence solve for y. (4 marks)
       2. Determine the perimeter of the lawn. (3 marks)
19. In the figure below, points P, Q, R, S and T lie on the circumference of a circle centre 0. Line UPV is a tangent to the circle at A. Chord ST of the circle is produced to intersect with the tangent at U. Angle UPT = 28º, ∠RST = 100° and ∠ORQ = 50°
    
    
    1. Determine the size of:
       1. ∠PTR (3 marks)
       2. ∠PTO (3 marks)
    2. Given that PQ = 6 cm, ST=5.4 cm and TU=3.5 cm Calculate correct to 1 decimal place:
       1. The radius of the circle (2 marks)
       2. The length of line PU (2 marks)
20. The table below shows income tax rates in a certain year.
    

    Monthly taxable income in Kenya Shillings	Tax rates
    0 - 13458	10%
    13459 - 26351	15%
    26352 - 39244	20%
    39245 - 52137	25%
    52138 and above	30%

    In the year, the monthly earnings of Kaliech were as follows:
    Basic salary Ksh 75 500
    House allowance Ksh 13 600
    Kaliech contributes 12.5% of his basic salary to a pension scheme. This contribution is exempted from taxation. He is entitled to a personal tax relief of Ksh 2 400 per month.
    Calculate:
    1. Kaliech's monthly taxable income (2 marks)
    2. The tax payable by Kaliech that month. (6 marks)
    3. Kaliech's net pay that month. (2 marks)
21. The vertices of the triangle shown on the grid are P'(-1,1), Q'(-6,5) and R'(0,2). Triangle P'QR' is the image of triangle PQR under a transformation whose matrix is (^-2₁ ¹₀)
    
    
    1. Find the coordinates of triangle POR (4 marks)
    2. Triangle P'O'R" is the image of triangle P'O'R' under a transformation matrix (^-1₀ ⁰₂).
       Determine the coordinates of triangle P"Q"R". (2 marks)
    3. On the same grid provided draw triangles PQR and P'Q"R" (2 marks)
    4. Determine a single matrix that maps PQR onto P'O'R" (2 marks)
22. Workers in an institution commute from their homes to the institution. The table below shows the distances in kilometres, covered by the workers.
    

    Distance (km)	4-6	7-9	10-12	13-15	16-18	19-21	22-24
    Number of workers	4	15	21	k	13	9	5

    The mean distance covered was 13.4 km.
    1. Determine the value of k and hence the standard deviation of the distances correct to 2 decimal places. (6 marks)
    2. Calculate, correct to 2 decimal places, the interquartile range of the distances. (4 marks)
23.    
    1. Complete the table below giving the values correct to 1 decimal place.(2 marks)
       

       xº	0	30	60	90	120	150	180	210	240	270	300	330	360
       y= 3 sin(2/3x) –3 cos(2/3x)	-3	-1.8		1.1	2.4	3.5		4.2	3.8	3	1.8		-1.1
       y=1–2 sinx	1	0		-1		0	1	2	2.7			2	1

    2. On the grid provided and using the same axis, draw the graphs of y = 3 sin(²/₃x) - 3cos(²/₃x) and y=1-2cosx for the range 0° ≤ x ≤ 360° (4 marks)
       
    3. Using the graphs in part (b):
       1. Find the values of x for which sin(²/₃x) = ½ + cos(²/₃x) (2 marks)
       2. Determine the range of x which 3 sin(²/₃x) -3 cos(²/₃x) > 1 - 2cosx (2 marks)
24. A particle moves along a straight line such that its displacement s metres after t seconds is given by s = t³ - pt² + qt + 4. Given that its velocity, v after 5 seconds was 28 m/s and its acceleration, a after 5 seconds was 20m/s².
    1. Determine the value of p and q (5 marks)
    2. Find the values of t when the particle is momentarily at rest. (3 marks)
    3. Calculate the displacement of the particle at t = 5 seconds (2 marks)

MARKING SCHEME

1. 99792l = 99.792m³
   v = r²h
   ²²/₇ x 0.021² x 6t = 99.792
   t = 12000s
         3600
   3¹/₃hrs
2. last term = arn-1 = 5 x 4^n-1
   2nd last term = ar^n-2 = 5 x 4^n-2
   5 x 4^n-1  x 5 x4^n-2 = 25600
   4^2n-3 = 4⁵ /2^4n-6 = 2¹⁰
   n = 4terms
3. 4ac = b²
   4 x d x 16 = -562
   d = ³¹³⁶/₆₄
   d = 49
4. 52 =  c²v² 
           dv²-f
   s2²dv² - s²f = c²v² 
   s2²dv² - c²v²  = s²f 
   v²(s²d - c²) = s²f
   v² =   s²f  
         s²d-c² 
   v = ±√  s²f  
            s²d-c²
5.        
   
6. w =  k√x  
        (y + z)²
   w_new =       k√0.81x    
             (1.16y + 1.16z)²
   =     0.9k√x     
      1.3456(y+z)²
   %change in W = (1125 - 1 ) x 100%
                               1682
   = 33.12%decrease
7. 
   
   sin = 4
   25.32cm
   = 9.1
8. H.P.P = 18000 + (15+2300)
   = 52500
   52500 - 18000 = 45000 - 18000(1 +   r   )5
                                                            400
   34500 = 27000 (1 +    r   )5
                                   400
   1.278 = 1 +   r  )5
                     400
   1.0503 = 1 +   r   
                       400
   r = 20.1%
9.        
   1.      
      
   2. (0.7,4)
      (3,1.8)
      rate of change at t = 2
      = 1.8 - 4 
         3 - 0.7
      = ^-2.2/_2.3
      = -1.0
10.      
    1. ³⁶⁰/_b = 120
       b = 3
    2. phase angle = 60º
11. 60 x θcos30 = 1800
    θ = 34.64º
    360 -(170 + 34.64)
    155ºW
12.    
    1.        
       
    2. P(BY) orP(YB)
       (⁴/₁₁ x ⁷/₁₀) + (⁷/₁₁ x ⁴/₁₀)
       ²⁸/₅₅
13. 
    
    xc = 6.2 1cm     
14.     
    
15. 2y² - 1(y + 1) = ^9y/_y
    2y² - y - 1 = 9
    2y² - y - 10 = 0
    2y² - 5y + 4y - 10 = 0
    y(2y - 5) + 2(2y - 5) = 0
    (y + 2)(2y - 5) = 0
    y = -2 or y = 2
16. sin(y + 20) = sin 230, sin310
    y = 210º, 290º
17.      
    1. In 1h R fills ⁴/₅, S fills ²/₁₅
       In 1h both fill ⁴/₂₅ + ²/₁₅ = ²²/₇₅
       In 1⁷/₈h both fill ¹⁵/₈ x ²²/₇₅ = ¹¹/₂₀
       fraction empty = 1 - ¹¹/₂₀
       = ⁹/₂₀
    2. ⁴/₂₅ is filled in 1hr
       ⁹/₂₀ is filled in ⁹/₂₀ x ²⁵/₄ x 1
       2hrs 48mins 45secs/ 213/16hrs
    3. fraction of tank filled S
       1h = ²/₁₅
       17/8 = ¹⁵/₈ x ²/₁₅ = ¼
       fraction filled by R = 1 - ¼ = ¾
       amount for R = ¾ x 31500º)
18.      
    1.    
       1. area = (75-4y)(36-2y)
          2700-144y-150y+8y²
          2700-294y+8y²
       2. (75 x 36) - (2700 - 294y + 8y²)
          2700 - 2700 + 294y - 8y²
          294y - 8y²
    2.      
       1. (294y - 8y²) = ³/₂(2700 - 294y + 8y²)
          588y - 16y² = 8100 - 882y + 24y²
          40y² - 1470y + 8100 = 0
          4y² - 147y + 810 = 0
          y = 147±√(-147)² - 4 x 2 x 810
                               2 x 4
           y =147 ±123
                      8
          y = 33.75m or y =3m
       2. length = 75 - 4 x 3 = 63m
          width = 36 - 2 x 3 = 30m
          P = 2(63 + 30)
          = 186m
19.    
      
    1.          
       1. ∠PQT = 28º(alternate segment theorem)
          ∠RQT = 180 - 100 = 80º (opp ∠s of cyclic quad add to 180)
          ∠PTR = 180-(80+28)
       2. ROQ = 180 -(50x2)=80º 
          QTR = ⁸⁰/₂ = 40º
          PTQ = 72 - 40 = 32º
    2.    
       1. ⁶/_sin32º = 2R
          R = 5.7cm
       2. (PU)² = TU.SU
          PU = √3.5(3.5 + 5.4)
          = 5.6cm
20.      
    1. ⁸⁷⁵/₁₀₀ x 75500 + 136000
       Ksh 79662.50
       OR
       75500 + 13600 - 12.5/100 x 75500
       Ksh 79662.50
    2. 1st slab 13458 x 10/100 = 1345.8
       2nd slab 12893 x 15/100 = 1933.95
       3rd slab 12893 x 20/100 = 2578.6
       4th slab 12893 x 25/100 = 3223.25
       5th slab 27525.5 x 30/100 =  8257.65  
                                                    17339.25
       17339.25 - 2400 
       Ksh 14939.25
    3. 75500 + 13600 - 12.5/100 x 75500 + 14939.25
       Ksh 64,723.25
       or 79662.50 - 14939.25
21.  
         
    1. 
       
    2.       
       
    3.     
    4. 
       
       a+b = 1
       5a+4b=6
       5a+5b=5
       5a+4b=6
       b = -1
       a = 1+1=2
       c+d = -2
       5c+4d = -10
       5c+5d=-10
       5c+4d= -10
       d = 0
       c= -2
       Matrix = (2-2 -10)
22. 
      

    Distance (km)	4-6	7-9	10-12	13-15	16-18	19-21	22-24
    Number of workers	4	15	21	k	13	9	5
    c.f	4	19	40	58	71	80	85

    1. 
       

       class	f	x	fx	fx2
       4-6	4	5	20	100
       7-9	15	8	120	960
       10-12	21	11	231	2541
       13-15	k	14	14k	3528
       16-18	13	17	221	3757
       19-21	9	20	180	3600
       22-24	5	23	115	2645
       	85			1731

       s.d = √17131 - 13.4²
                     85
       = 4.69
       x =∑fx = 887 + 14k = 13.4
             ∑f        67+k 
       k = ^10.8/_0.6 = 18
    2. Q =¼ x 85 = 21.25
       9.5 + (21.25 - 19) x 3
                       21
       9.821
       Q3 =¾ x 85 = 63.75
       15.5 + (63.75 - 58) x 3
                         13
       = 16.827
       Q - Q1 = 16.827 - 9.821 = 7.01    
23.      
    1.      
       

       xº	0	30	60	90	120	150	180	210	240	270	300	330	360
       y= 3 sin(2/3x) –3 cos(2/3x)	-3	-1.8	-0.4	1.1	2.4	3.5	4.1	4.2	3.8	3	1.8	0.4	-1.1
       y=1–2 sinx	1	0	-0.7	-1	-0.7	0	1	2	2.7	3.0	2.7	2	1

    2.        
       
    3.      
       1. sin(²/₃x) - cos(²/₃x) = ½
          3sin(²/₃x) - 3cos(²/₃x) = ³/₂
          3sin(²/₃x) - 3cos(²/₃x) = y
          y = ³/₂
          x = 97º to 101º or 307º to 311º
       2. 54º < x < 270º
24.    
    1. v = ^ds/_dt = 3t² - 2pt + q
       3(5)² - 2p(5) + q = 28
       75 - 10p + q = 28
       -10p + q = -47
       a = dv/dt = 6t - 2p
       6(5) - 2p = 20
       -2p = -10
       p = 5
       -10(5) + q = -47
       q = 3
    2. s = t³ - 5t² + 3t + 4
       v = ^ds/_dt = 3t² - 10t + 3 = 0
       3t² - 9t - t + 3 = 0
       3t(t-3) -1(t-3)=0
       (3t-1)(t-3) = 0
       t = ¹/₃ or t = 3
    3. s = 5³ - 5(5)² + 3(5) + 4
       s = 125 - 125 + 15 + 4
       = 19m