Mathematics Paper 2 Questions and Answers - Sunrise 2 Evaluation Mock Exams 2022

Instructions to candidates

1. Write your name and index number in the spaces provided above.
2. Sign and write the date of examination in the spaces provided above.
3. This paper consists of TWO sections: Section I and Section II.
4. Answer ALL the questions in Section I and only five questions from Section II.
5. Show all the steps in your calculations, giving your answers at each stage in the spaces provided below each question.
6. Marks may be given for correct working even if the answer is wrong.
7. Non-programmable silent electronic calculators and KNEC Mathematical tables may be used, except where stated otherwise.
8. Candidates should check the question paper to ascertain that all the pages are printed as indicated and that no questions are missing.
9. Candidates should answer the questions in English.

For examiner’s use only
Section I

Section II

QUESTIONS

1. A rectangular block has a square base whose side is exactly 8cm. Its height measured to the nearest millimetre is 3.1cm. Find in cubic centimetres the greatest possible error in calculating its volume (3mks)
2. Simplify (3mks)
   
3. Find the area bounded by the curve y= x3 + 5, the x axis and lines x=1 and x = 3 (3mks)
4. A bag contains 2 green balls, 3 red balls and one blue ball. Another bag contains 4 green balls, 5 red balls and 3 blue balls. A ball is chosen at random from a bag. Find the probability that the chosen ball is blue. (3mks)
5. P varies partly as the square of V and partly as the cube of V. When V= 2, P = -20 and when V = -3, P = 135. Find the relationship between P and V. (3mks)
6. The second term of a G.P is 6, and the fifth term is 48, find the common ratio and the 3rd term of the G.P. (3mks)
7. Nancy pays for a car on hire purchase in 15 monthly instalments. The cash price of the car is Ksh. 300,000 and the interest rate is 15% p.a. A deposit of Ksh 75,000 is made. Calculate her monthly repayments. (4mks)
8. Solve for θ in the equation 2 sin (2θ + 10) = -1 for 0º ≤ θ ≤ 360º. (3mks)
9. The diagram below shows a garden XYZ
   
   
   1. Draw the locus of points equidistant from sides XY and XZ. (1mk)
   2. Draw the locus of points equidistant from points X and Z. (1mk)
   3. A bead is lost in the garden within a region which is nearer to point X than to point Z and closer to side XZ than to side XY. Show by shading the region where the bead can be located. (1mk)
10. Expand the given binomial up to the term with x4: (1+3x)⁶. Use your expansion to evaluate (1.3)⁶ correct to 4 decimal places. (4mks)
    (1 + 3x)⁶
    = 1(1)⁶ + 6(1)⁵(3x)¹ + 15(1)⁴(3x)² + 20(1)³(3x)³ + 15(1)²(3x)⁴ +  ........................
     = 1  + 18x + 135x2 + 540x2 + 1215x4
    (1.3)⁶ = (1 + 0.3)⁶ = (1 + 3x)⁶
    3x = 0.3 : x = 0.1
    => (1 + 0.3)⁶ = 1 + 18(0.1) + 135(0.1)² + 540(0.1)³ + 1215(0.1)⁴
    = 4.8115
11. Line PQ is the diameter of a circle. Find the equation of the circle in the form ax²+bx+cy²+dy-e =0, given the coordinates P(0,2) and Q(6,2). (3mks)
12. Determine the amplitude, the period and phase angle for the graph y=3sin(1/4x-90º). (3mks)
13. Make x the subject of the formular. (3mks)
    
14. Under a transformation represented by a matrix  , a triangle of area 10cm² is mapped onto atriangle of area 110cm². Find x. (3mks)
15. Evaluate without using Mathematical tables or a calculator. (3mks)
    
16. A body starts from rest and after t seconds its velocity in ms-1 was recorded as shown below;
    

    T in (sec)	0	1	2	3	4	5	6
    Velocity	0	0.29	5.4	7.7	9.7	11.4	12.7

    Use the trapezoidal rule to estimate the distance covered by the body between 1 and 6 seconds (3mks)
17.                          
    1. Complete the table below to 2 decimal places. (2mks)
       

       X	00	300	600	900	1200	1500	1800	2100	2400	2700	3000	3300	3600
       –  Cos x	– 1		– 0.5		0.5	0.87		0.87			-0.5	0.87
       Sin ( x – 300)		0.0	0.5			0.87	0.5		– 0.5			– 0.87	– 0.5

    2. Draw the graphs of y = sin (x – 30º) and y = – Cos x on the same axes, for 0º ≤ x ≤ 360º. (5mks)
       
    3. Use your graph to solve the equation sin (x - 300) + Cos x = 0. (3mks)
18. Figure below is a pyramid on a rectangular base. PQ=16cm, QR = 12cm and VP = 13cm.
    
    Find
    1. The length of QS. (2mks)
    2. The height of the pyramid to 1 decimal place. (2mks)
    3. The angle between VQ and the base. (2mks)
    4. The angle between plane VQR and the base. (2mks)
    5. The angle between planes VQR and VPS (2mks)
19. The following table shows the distribution of marks obtained by 50 students in a test.
    

    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

    By using an assumed mean of 62, calculate
    1. The mean (5mks)
    2. The variance (3mks)
    3. The standard deviation (2mks)
20.                  
    1. Hellen’s earnings are as follows:
       Basic salary sh. 38000 per month
       House allowance sh. 14000 per months
       Travelling allowance sh.8500 per month and
       Medical allowance Ksh.3300 per month.
       She is given a personal relief of Ksh. 12672 per annum
       The table for payable tax is shown below
       Income in K£ p.a Payable tax rate in Kshs per K£
       0-6000 2
       6001-12000 3
       12001-18000 4
       18001 -24000 5
       24001-30000 6
       30001-36000 7
       36001-42000 8
       42001-48000 9
       Over 48000 10
       Calculate
       
       1. Hellen’s taxable income in K£ p.a (2mks)
       2. Her P.A.Y.E (5mks)
       3. Hellen is deducted the following items per month
          NHIF Ksh.320
          Cooperative shares Ksh.2000
          Loan repayment Ksh5000
          Determine her net salary per month (3mks)
21. In the figure below, O is the centre of the circle. A, B, C and D are points on the circumference of the circle. A, O, X and C are points on a straight line. DE is a tangent to the circle at D. Angle BOC= 48º and angle CAD = 36º.
    
    
    1. Giving reasons or otherwise, find the value of the following angles:-
       1. Angle CBA (1mk)
       2. Angle BDE (2mks)
       3. Angle CED (3mks)
    2. It is also given that AX = 12 cm, XC = 4 cm, DB = 14 cm and DE = 20 cm. Calculate:
       1. DX (2mks)
       2. AE (2mks)
22. The position of two towns A and B on the earths surface are (36ºN, 49ºE) and (36ºN, 131ºW) respectively.
    1. Find the difference in longitude between town A and town B. (2mks)
    2. Given that the radius of the earth is 6370, calculate the distance between town A and B.
       1. In nm. (2mks)
       2. In kilometers. (2mks)
    3. Another town C is 840km East of town B on the same latitude as town A and B. Find the longitude of town C. (4mks)
23.                        
    1. Using a ruler and compasses only, construct triangle ABC such that AB=4cm, BC=5cm and ∠ABC = 120º. Measure AC. (4mks)
    2. On the diagram, construct a circle which passes through the vertices of the triangle ABC. Measure the radius of the table. (4mks)
    3. Construct a perpendicular from the centre of the circle to the line BC. Measure the length of the perpendicular. (2mks)
24. A metal sphere has a radius 5cm and a density of 2.4g/cm³.
    1. Calculate the mass of the ball in kg (4mks)
    2. The ball is dropped into a cylindrical container which is partially filled with water. The ball is fully sub-merged. If the cylinder has a base radius of 8cm, calculate the change in the water level. (3mks)
    3. The sphere is melted down to form a metal cylinder of same radius. Calculate the height of the cylinder formed. (3mks)

MARKING SCHEME

1. A rectangular block has a square base whose side is exactly 8cm. Its height measured to the nearest millimetre is 3.1cm. Find in cubic centimetres the greatest possible error in calculating its volume (3mks)
   Maximum volume = 64 x 3.15 = 201.6cm³
   Actual volume = 64 x 3.1 = 198.4cm³
   Error = 201.6 – 198.4
   = 3.2cm³
2. Simplify (3mks)
   
   
3. Find the area bounded by the curve y= x3 + 5, the x axis and lines x=1 and x = 3 (3mks)
   
   35.25 – 5.25
   = 30 sq. units
4. A bag contains 2 green balls, 3 red balls and one blue ball. Another bag contains 4 green balls, 5 red balls and 3 blue balls. A ball is chosen at random from a bag. Find the probability that the chosen ball is blue. (3mks)
   
5. P varies partly as the square of V and partly as the cube of V. When V= 2, P = -20 and when V = -3, P = 135. Find the relationship between P and V. (3mks)
   

   P = a V²   + b  V³   :   4a  + 8b   = ‑20
                                      9a  - 27b  = 135
      36a  + 72b  = -180
   -  36a – 108b  = 540
                180b = -720
                 b = - 4
   -20 = 4a  - 32 
         4a  = 12
             a = 3
     P = 3V²  +   4V³  

6. The second term of a G.P is 6, and the fifth term is 48, find the common ratio and the 3rd term of the G.P. (3mks)
   

   ar  =  6  a = 6/r
   ar⁴   = 48
   6/r  x r⁴  =  48
      6 r³  = 48
         r³  = 8 
           r = 2.
        2a  = 6
          a = 3
   3^rd term = ar²   
    = 3 x 2²  = 12

7. Nancy pays for a car on hire purchase in 15 monthly instalments. The cash price of the car is Ksh. 300,000 and the interest rate is 15% p.a. A deposit of Ksh 75,000 is made. Calculate her monthly repayments. (4mks)
   

   P = 300,000 – 75,000
                       = 225,000 
                    A=225,000 x 1.15^1.25
   Monthly Instalment = 225,000 x 1.15 ^1.25 
                                            15
   = 225,000 x 1.191  =    267975
                15                       15
   = Kshs17,865  

8. Solve for θ in the equation 2 sin (2θ + 10) = -1 for 0º ≤ θ ≤ 360º. (3mks)
   

   2sin(2θ + 10) =  -1
   sin(2θ + 10) = -0.5
   2θ+10 = 210,330,570, 690 
   2θ=200, 320,560,680
   θ= 100^o,160^o,280^o,340^o

9. The diagram below shows a garden XYZ
   
   
   1. Draw the locus of points equidistant from sides XY and XZ. (1mk)
      
   2. Draw the locus of points equidistant from points X and Z. (1mk)
   3. A bead is lost in the garden within a region which is nearer to point X than to point Z and closer to side XZ than to side XY. Show by shading the region where the bead can be located. (1mk)
10. Expand the given binomial up to the term with x4: (1+3x)⁶. Use your expansion to evaluate (1.3)⁶ correct to 4 decimal places. (4mks)
    Ans = 4.8115
11. Line PQ is the diameter of a circle. Find the equation of the circle in the form ax²+bx+cy²+dy-e =0, given the coordinates P(0,2) and Q(6,2). (3mks)
    
12. Determine the amplitude, the period and phase angle for the graph y=3sin(1/4x-90º). (3mks)
    amplitude 3 units
    period = 360/1/4 = 1440º
    phase angle  = 90º
13. Make x the subject of the formular. (3mks)
    
    p² = bx² - ax
                x²
    p² = b
    -a
     x
    a = b - p²
    x
    = x = a
        b - p²
14. Under a transformation represented by a matrix  , a triangle of area 10cm² is mapped onto atriangle of area 110cm². Find x. (3mks)
    A.S.F = 110/10 = 11
    => 5x² - (-6x) = 11
    5x² + 6x - 11 = 0
    (5x + 11) (x - 1) = 0
    x = 1 or -11/5
15. Evaluate without using Mathematical tables or a calculator. (3mks)
    
    log5² - log16^1/2 + log 40²
    log 25 x 1600
               4
    log 10 000
    log 10⁴
    4 log 10
    = 4
16. A body starts from rest and after t seconds its velocity in ms-1 was recorded as shown below;
    

    T in (sec)	0	1	2	3	4	5	6
    Velocity	0	0.29	5.4	7.7	9.7	11.4	12.7

    Use the trapezoidal rule to estimate the distance covered by the body between 1 and 6 seconds (3mks)
    A = 1/2h (sum of ends + 2(sum of middle)
    = 1/2 x 1 (0.1 + (0.29 + 12.7) + 2(5.4 + 7.7 + 9.7 + 11.4)
    = 1/2 (12.99 + 2(34.21)
    = 40.695m
17.                          
    1. Complete the table below to 2 decimal places. (2mks)
       

       X	00	300	600	900	1200	1500	1800	2100	2400	2700	3000	3300	3600
       –  Cos x	– 1	0.87	– 0.5	0	0.5	0.87	1	0.87	0.5	0	-0.5	0.87	-1
       Sin ( x – 300)	0	0.0	0.5	0.87	1	0.87	0.5	0	– 0.5	-0.87	-1	– 0.87	– 0.5

    2. Draw the graphs of y = sin (x – 30º) and y = – Cos x on the same axes, for 0º ≤ x ≤ 360º. (5mks)
       
    3. Use your graph to solve the equation sin (x - 300) + Cos x = 0. (3mks)
18. Figure below is a pyramid on a rectangular base. PQ=16cm, QR = 12cm and VP = 13cm.
    
    Length of line Qs = √162 + 122
    = √256 + 144 = √400
    = 20 cm
    Height of the pyramid
    

    = √13²  - 10² = √169 – 100 = √69
    =  8.3cm ( 1d.p)
    Cos q = 10/13  = 0.7692
    q  = 39.72°

    Find
    1. The length of QS. (2mks)
    2. The height of the pyramid to 1 decimal place. (2mks)
    3. The angle between VQ and the base. (2mks)
    4. The angle between plane VQR and the base. (2mks)
       

        h²  = 13² -6² = 169 - 36
       h = √133 = 11.53 cm( 2d.p)
       Cos ∝ =      8        = 0.6938
                       11.53
         ∝   = 46.06°

    5. The angle between planes VQR and VPS (2mks)
       

       θ = 2 x sin^-1 8/11.53

              = 2 x 43.93

              = 87.87⁰

19. The following table shows the distribution of marks obtained by 50 students in a test.
    

    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

    By using an assumed mean of 62, calculate
    1. The mean (5mks)
       Let A  = 62      
       

       Marks	f	x	d = x -A	fd	d2	fd2
       45 - 49 50 -54 55 - 59 60 – 64 65 – 69 70 – 74 75 - 79	3 9 13 15 5 4 1	47 52 57 62 67 72 77	-15 -10 -5 0 5 10 15	-45 -90 -65 0 25 40 15	225 100 25 0 25 100 225	675 900 325 0 125 400 225
       	f = 50			∑fd = -120		∑fd2 =  2650

       = A + ∑fd
                 ∑f
       = 62 + -120
                     50
       = 62 - 2.4
       = 59.6
       
    2. The variance (3mks)
    3. The standard deviation (2mks)
20.                  
    1. Hellen’s earnings are as follows:
       Basic salary sh. 38000 per month
       House allowance sh. 14000 per months
       Travelling allowance sh.8500 per month and
       Medical allowance Ksh.3300 per month.
       She is given a personal relief of Ksh. 12672 per annum
       The table for payable tax is shown below
       Income in K£ p.a Payable tax rate in Kshs per K£
       0-6000 2
       6001-12000 3
       12001-18000 4
       18001 -24000 5
       24001-30000 6
       30001-36000 7
       36001-42000 8
       42001-48000 9
       Over 48000 10
       Calculate
       
       1. Hellen’s taxable income in K£ p.a (2mks)
          (38000 + 14000 + 8500 + 3300) x 12 = 38,280 ksh p.a
                               20
       2. Her P.A.Y.E (5mks)
          6000 x 2 =12000
          6000 x 3 = 18000
          6000 x 4 = 24000
          6000 x 5 = 30000
          6000 x 6 = 36000
          6000 x 7 = 42000
          2280 x 8 = 18240
          Gross tax p.a = 180,240
          less relief = 12672
          P.A.Y.E  = 167,568 p.a.
          = 13,964 p.m
       3. Hellen is deducted the following items per month
          NHIF Ksh.320
          Cooperative shares Ksh.2000
          Loan repayment Ksh5000
          Determine her net salary per month (3mks)
          total ded = (13969 + 320 + 2000 + 1000)
          = ksh 21 284
          Net salary = 63800 - 21284
          = ksh 42, 516
21. In the figure below, O is the centre of the circle. A, B, C and D are points on the circumference of the circle. A, O, X and C are points on a straight line. DE is a tangent to the circle at D. Angle BOC= 48º and angle CAD = 36º.
    
    
    1. Giving reasons or otherwise, find the value of the following angles:-
       1. Angle CBA (1mk)
          90º -> angle subtended by the diameter of the circumference
       2. Angle BDE (2mks)
          36º - alternate segment theorem
       3. Angle CED (3mks)
          180 - (36 + 36 + 61)
          180 - 133º
          = 47º angles in a traingle add upto 180º
    2. It is also given that AX = 12 cm, XC = 4 cm, DB = 14 cm and DE = 20 cm. Calculate:
       1. DX (2mks)
          DX = X
          X6 = (14 - X)
          AX.XC = BX.XB
          12 x 4 = x(N - x)
          48 = 14x - x²
          x² - 14x + 48 = 0
          (x - 8)(x - 6) = 0
          x = 8 or x = 6 cm
       2. AE (2mks)
          AE.CE = DE²
          20² = 16(16 + a)
          400 = 256 + 169
          CE = a
          144 = 169
          a = 9 cm
22. The position of two towns A and B on the earths surface are (36ºN, 49ºE) and (36ºN, 131ºW) respectively.
    1. Find the difference in longitude between town A and town B. (2mks)
       131 + 49
       = 180
    2. Given that the radius of the earth is 6370, calculate the distance between town A and B.
       1. In nm. (2mks)
          60 x ∝ Cos θ
          = 60 x 180 Cos 36º
          = 60 x 180 x 0.8090
          = 8737.38 pm
       2. In kilometers. (2mks)
          ∝ x 2π R Cos θ
          360
          180 x 2 x 3.142 x 6370 Cos 36º
          360
          = 16192.103 km
    3. Another town C is 840km East of town B on the same latitude as town A and B. Find the longitude of town C. (4mks)
       

       ∝ x 2 x 3.142 x 6370 Cos 36º
       360
       = 840 km
       ∝ = 840 x 360º
       628 x 6370 Cos 36º
       = 9.34º

       Longitude of town c
       13 1º – 9.34º
       = 121.66º

23.                        
    1. Using a ruler and compasses only, construct triangle ABC such that AB=4cm, BC=5cm and ∠ABC = 120º. Measure AC. (4mks)
       
       AC = 7.8 ± 0.1
    2. On the diagram, construct a circle which passes through the vertices of the triangle ABC. Measure the radius of the table. (4mks)
       Radius = 4.5 ± 0.1 cm
    3. Construct a perpendicular from the centre of the circle to the line BC. Measure the length of the perpendicular. (2mks)
       length of perpendicular = 4.5 ± 0.1
24. A metal sphere has a radius 5cm and a density of 2.4g/cm³.
    1. Calculate the mass of the ball in kg (4mks)
       f = m/v
       v = 4/3πr³
       = 4/3 x 22/7 x 5³
       = 523.5987756cm³
       = fv
       = 523.5987756 x 2.4
       = 1256.637061g
       = 1.256637061 kg
    2. The ball is dropped into a cylindrical container which is partially filled with water. The ball is fully sub-merged. If the cylinder has a base radius of 8cm, calculate the change in the water level. (3mks)
       v = πr²h = 523.5987756 cm³
       h = 523.5987756
                π x 8 x 8
       h = 2.604166667
    3. The sphere is melted down to form a metal cylinder of same radius. Calculate the height of the cylinder formed. (3mks)
       v = πr²h = 523.5987756 cm³
       h = 523.5987756
              π x 25
       h = 6.667 cm