Mathematics Paper 2 Questions and Answers - Catholic Diocese of Kakamega Mock Exams 2023

INSTRUCTIONS TO CANDIDATES :

1. This paper consists of two sections: Section I and Section II.
2. Answer all questions in Section I and only five questions in section II.
3. Show all steps in your calculations, giving your answer at each stage in the spaces provided below each question.
4. Marks may be given for correct working even if the answer is wrong.
5. Non-programmable silent electronic calculator preferably Casio fx82MS and KNEC mathematical tables (7th Edition) may be used except where stated otherwise.
6. Candidates should answer questions in English.
   
   

SECTION A: (50 MARKS)

1. The radius and height of a cone are stated as 21.36 cm and 35.7 cm respectively. Find the percentage error in calculating its volume.
   (2 marks)
2. Solve for x without using mathematical tables: ¹/₂ log₂ 81+ log₂ (x² - ^x/₃) = 1
   (3 marks)
3. Under a transformation whose matrix  a triangle whose area is 12.5cm² is mapped onto a triangle whose area is 50cm². Calculate the absolute value of a, hence state the matrix A.
   (3 marks)
4. Simplify          3      +    1    leaving your answer in the form a + b√c where a, b and c are rational numbers.
                     √5 - 2        √5
   (3 marks)
5. The radius of a cone increases in the ratio 4:3 while its height decreases in the ratio 3:4. Determine the percentage change in the volume of the cone
   (3 marks)
6. The mid-point of AB is (1,^-3/₂,2). If the position vector of point A is – i + j + k, find the magnitude of AB correct to 1 decimal place.
   (3 marks)
7. A squared hollow of side 2 cm was bored at the middle of the top surface of a cube up to the bottom surface. If such through hollow was made on each of the sides of the cube, find the volume of the remaining solid.
   
   (3 marks)
8.  
   1. Write down the simplified expansion form of (2 - ¹⁄₂ x)⁵
      (2 marks)
   2. Use the expansion up to the fourth term to estimate the value of (1.98)⁵ (2 marks)
9. The figure below shows a circle centre O. Line PQM is a tangent to the circle at Q. Given that angle PQR is 105°. Find the values of angles a and b.
   
   (2 marks)
10. Given that a is an acute angle and cos a = √5 find without using mathematical tables or a calculator,tan (90 - a)°.
                                                                          5
    (2 marks)
11.  
    The fourth and the tenth terms of an arithmetic progression are 30 and -30 respectively.
    
    1. Determine the first term and the common difference.
       (2 marks)
    2. Find the maximum possible number of terms such that the sum is a positive integer.
       (2 marks)
12. A quantity y is partly constant and partly varies as the square root of x.
    
    1. Using constants M and N, write down an equation connecting y and x.
       (1 mark)
    2. If x = 9 when y = 24 and x = 36 when y = 32, find the values of the constants M and N.
       (3 marks)
13. Whitney made a regular octagon from a squared grid board of size x cm by cutting four isosceles triangles whose equal sides are y cm each as shown in the figure below.
    
    
    1. Write down an expression for the area of the octagon in terms of x and y.
       (1 mark)
    2. Find value of x in surd form.
       (1 mark)
    3. Find the area of the octagon to the nearest whole number.
       (2 marks)
14. A circle passing through point (-3, -1) has its centre at (4, 3). Determine the equation of the circle in the form x²+ y²+ ax + by + c = 0, where a, b and c are constants.
    (3 marks)
15. Murebe takes 8 hours to water nursery beds for the school Agriculture project. However, with the help of his friend Esito, it takes them 6 hours only. How long will it take Esito to water the nursery beds alone?
    (3 marks)
16. The probabilities that Omboki, Owuor and Mbati will miss morning preps are 1  , 1 ,  1 respectively.
                                                                                                                              8   10  12   
    Calculate the probability that on any morning;
    
    1. One of them will miss morning preps.
       (2 marks)
    2. At most one of them will miss morning preps.
       (2 marks)
       
       

SECTION II: (50 MARKS)
Answer ONLY five questions from this section

17. The table below shows income tax rates for a certain month this year.
    

    Monthly income in Kshs	Tax rate in each shillings
    1- 12000	15%
    12,001 - 20,600	20%
    20,601 - 29,400	25%
    29,401 - 38,201	30%
    38,200 - and above	35%

    
    A monthly tax relief of Khs 2,400 was allowed. Dreck's taxable income in the last band was Ksh 34,750 in month. Dreck's salary included Rental house allowance of 13,000 and commuter allowance of 6,000. He contributed 7.5% of his basic salary to a pension scheme and 2.5% of his basic salary to a Housing Fund levy which are exempted from taxation.
    
    1. Calculate
       
       1. His taxable income per month.
          (2 marks)
       2. The amount of tax he paid in a month.
          (5 marks)
    2. Dreck was also deducted 1,400 for NHIF, 5,000 for Savings and 20,000 for Development loan that
       month. Calculate his net pay.
       (3 marks)
18.  
    1. Completet the table below for y=x3+ 4x2 -5x-5
       

       X	-5	-4	-3	-2	-1	0	I	2
       Y

       
       (2 marks)
    2. On the grid provided draw the graph y=x³ + 4x² -5x-5
       
       (3 marks)
    3. Use the graph to solve the equation x³ + 4x² -5x-5=0
       (2 marks)
    4. By drawing a suitable straight line on the graph, solve the equation x² + 4x² - 5x -5 = -4x-1
       (3 marks)
19.  
    1. Use the trapezium rule with 7 strips to estimate the area enclosed by the curve
       y=¹/₂x²+3 and the lines x = 1, x = 8 and the x-axis.
       (3 marks)
    2. Use the mid-ordinate rule with 7 ordinates to estimate the area enclosed by the curve y = x2 + 3 and the lines x = 1, x = 8 and the x-axis.
       (3 marks)
    3. Determine the exact area bounded the curve and the lines in section a) above
       (2 marks)
    4. Calculate the percentage error made in the estimated area from the trapezoidal rule
       (2 marks)
20.  
    1. Complete the table below for y = sin 2x and y = sin (2x +30)° giving values to 2 decimal places.
       (2 marks)
       

       x	0	15	30	45	60	75	90	105	120	135	150	165	180
       sin 2x
       sin (2x+30)

    2. Draw the graph of y = sin 2x and y = sin (2x +30)° on the grid provided
       (4 marks)
       
    3. Use the graph to solve sin (2x+30)°- sin 2x = 0
       (1 mark)
    4. Determine the transformation that maps sin 2x onto sin (2x+30)
       (1 mark)
    5. State the period and amplitude of y = sin (2x+30)°
       (2 marks)
21.  
    Construct a parallelogram ABCD in which AB = 9cm, AD = 5cm and ≤ BAD = 60°.
    (3 marks)
    
    1. Measure the length of AC
       (1 mark)
    2. Draw the locus of point P which moves such that it is equidistant from A and C
       (1 mark)
    3. Draw the locus of points Q on the same side as D which moves such that angle AQB = 90°
       (3 marks)
    4. Given that AP PC shade the region bounded by the locus of P and Q
       (2 marks)
22. The relationship between two variables P and Q is given by the equation P = aQ" where a and n are constants. The results on the variation of P and Q are tabulated below.
    

    Q	5	10	15	20	25
    P	17.7	31.9	44.9	57.5	69.4

    1. Write down a linear equation connecting P and Q.
       (1 mark)
    2. Complete the table above to 2 decimal places
       (2 marks)
    3. Draw a suitable linear graph to verify that the assumed relation is approximately true
       (3 marks)
    4. Hence determine the values of a and n correct to 2d.p
       (3 marks)
    5. State the formula connecting P and Q.
       (1 mark)
23. Figure below is a pyramid on a rectangular base. PQ-16cm, QR = 12cm and VP 13cm.
    
    Find
    1. The length of QS.
       (2 marks)
    2. The height of the pyramid to 1 decimal place.
       (2 marks)
    3. The angle between VQ and the base.
       (2 marks)
    4. The angle between plane VQR and the base.
       (2 marks)
    5.  The angle between planes VQR and VPS
       (2 marks)
24.  
    1.  
       1. Taking the radius of the earth, R = 6371km and π =  ²²/₇, calculate the shortest distance between two cities P(60°N, 60°W) and Q(60°N, 60°E) along a great circle.
          (3 marks)
       2. If it is 1.20pm at P, what is the local time at Q
          1. (3 marks)
    2. An aeroplane flew due south from a point A(50°N, 55°E) to a point B, the distance covered by the aeroplane was 8000km, determine the position of B.
       (4 marks)
       
       

MARKING SCHEME

NO	WORKINGS	AWARD	COMMENTS
1.	P.E = (0.005/21.36 + 0.005/21.36 + 0.05/35.7) x 100              = 0.1868725018% = 0.1869% (accept 4 sig fig)	B1 B1 B1	for reciprocal correct attempt to add correct logs for 1.094
2 .	log29+ log2(x2 - x/3) = log2 2 9(x2 – x/3)=2 9x2 - 3x - 2 = 0 x = 3 ± 9/18 = 2/3 or -1/3	M1 Α1	expression for P.E Allow 0.163464262
3.	a(a-2) - - 2a = 50/12.5 a2 - 2a + 2a = 4 a2 =4 /a/ = 2 A = ( 02 -22)	M1 M1 Α1	Single logs on both sides Correct attempt to solve
4 .	4√5 - 2 5 - 2√5 4√5 - 2  +  5 + 2√5 5 - 2√5      5 + 2√5 20√5 + 40 - 10 - 4√5          25 - 20 30 + 16√5       5 6 + 16√5         5	B1 M1 Α1	Single fraction Expanded numerator over a rationalized denominator
5 .	1/3 x 16/9 x 3/4 = 4/9 4/9 - 1/3  x  100      1/3 33 1/3	B1 M1 Α1	new volume expression
6.		B1 B1 A1	for B for AB expression for magnitude
7 .	(4 x 4 x 4) - (2 x 2 x 4 + 2 x 2 x 4) 64-32 32	M1 M1 A1
8 .	a) 25 - (5)24(1/2x) + (10)23(x)2 - (10)22(1/2x)3+(5)21(1/2x)4 - (1) (1/2x)5 32 - 80x + 20x5 - 5x3 + 0.625 x4-- 0.03125 x5 b) x = 0.04 32 − 80(0.04) + 20(0.04)2 - 5(0.04)3 28.83168	M1 A1 M1 A1	Expression Simplified form Substitution
9 .	<QSR = 75° < b = 90º - 75º = 15º < a = 180º – (105 +15º) = 60°	B1 B1	For < b For < a
10 .	opp = √(52 - √5) = 2√5 tan(90 - a) = √5 / 2√5 = 1/2	B1 B1
11 .	a) 30 = a + 3d   - 30 = a + 9d     60 = - 6d d = -10 a = 60 b) 0 = n/2 {2(60) + (n - 1) (-10)} 5n=65 n=13	M1 A1 M1 A1	Correct attempt to solve For both a and d Equating to zero
12 .	a) y = M + N√x b) 24 = M + 3N     32 = M + 6N     - 8 = - 3N N = 8/3 M = 16	B1 A1 M1 B1	Equation Correct attempt to solve
13 .	a) x2-2y2 b) 2y2= 441 y = √220.5 x = 21 + 2√220.5 c) (21 +2√220.5)2 – (√220.5)2 = 2350	B1 B1 M1 A1
14 .	r = √(72 + 42) = √65 (x −4)2 + (y − 3)2 = 65 x2 - 8x + y2 - 6y - 40 = 0	M1 M1 A1
15 .	1/x = 1/6 - 1/8 1/x = 1/24  x = 24 hrs	M1 M1 A1
16.	a) = (1/8 x 9/10 x 11/12) + (7/8 x 1/10 x 11/12) + (7/8 x 9/10  x 1/12)  = 218/960 b) (1/8 + 9/10 + 11/12) + (7/8 + 1/10 + 11/12) + (7/8 + 9/10  + 1/12) + (7/8 + 9/10  + 11/12)  = 911/960	M1 A1 M1 A1	correct probabilities allow equivalents correct probabilities allow equivalents
17 .	a) i .Taxable income = 38,200 + 34,750                = 72,950          ii. 1st band = 12,000 x 15/100 = 1800              2nd band = 8,600 x 20/100 = 1720              3rd band = 8,800 x 25/100 = 2200              4th band = 8,800 x 30/100 = 2640              5th band = 34,750 x 35/100 = 12,162.5 Net tax = 18,922.5 -2,400 b) basic salary = 72,950 - 19,000 = 53,950 Tax exempt = 10/100 x 53,950 = 5,395 net pay = 72,950 – (12,162.5 + 5,000+ 20,000+ 1,400 + 5,395) = 38,992.5	M1 A1 M1 M1 M1 A1 B1 M1 A1	Taxable income 1st band 2nd and 3rd bands 4th and 5th bands Less tax relief Tax paid Income exempted Net pay
18 .		B2 B1 S1 P1 C1 L1 B1 B1 L1 B1	All values correct At least 5 correct linear/uniform/sufficient All point plotted accurately Smooth curve Allow ± 0.1 y = - 4x - 1 drawn Allow ± 0.1
19.	a) x 1 2 3 4 5 6 7 8 y 3.5 5 7.5 11 15.5 21 27.5 35 A = 1⁄2 {(3.5 +35) + 2(5+7.5+ 11+ 15.5 + 21+ 27.5+ 35)}   = 106.75 b) x 1.5 2.5 3.5 4.5 5.5 6.5 7.5 y 4.125 6.125 9.125 13.125 18.125 24.125 31.125 A = 1 {4.125 +6.125 + 9.125 + 13.125 + 18.125 + 24.125 + 31.125} = 105.875 c)     ∫81 (1/2 x  +  3)dx = [ x3 + 3x]                                     6  = 106 1/6 d)       106.75 - 106 1/6  x  100               106 1/6 = 0.5494450549	M1 M1 A1 B1 M1 A1 M1 A1 M1 A1
20 .		B1 B1 P1 C1 P1 C1 B1 B1 B1 B1	Sinx Sin(2x+30) Sin2x Sin(2x+30) Allow ± 1º
21.
22.			Linear equation Log P Log Q Scale Plotting Line of best fit expression formula
23 .	a) QS = √(162 + 122) = 20cm b) h = √(132 + 102) = 8.3cm c) Ø =cos-1(10/13) = 39.71° d) α =Tan-1(8.3/8) = 46.05º e) β =2Tan-1(8/8.3) = 2 x 43.95 = 87.89°	M1 A1 M1 A1 M1 A1 M1 A1 M1 A1
24.	a) i) Central angle, β = 2sin-1(cos60sin120/2) = 51.96 PQ = 51.96 x 2 x 22 x 6371           360             7 5780.18km ii) longitude difference = 60 + 60 = 120 time difference = 120 x 4 = 480 min time at Q 1.20 + 8.00 = 9.20pm b) 50 + α x 2 x 22 x 6371 = 8000        360            7 50 + α = 71.92 α = 71.92 - 50 α = 21.92 B(21.92°S, 55°E)	B1 M1 A1 M1 M1 A1 M1 M1 A1 B1	Central angle Expression Longitude difference Time difference Time at Q Equation Correct attempt to solve Latitude at B Position of B