Mathematics Paper 2 Questions and Answers - Kapsabet Mocks Exams 2023

• The paper contains two sections. Section I and Section II.
• Answer ALL the questions in section I and any five questions in section II.
• Answers and working must be written on the question paper in the spaces provided below each question.
• Show all steps in your calculations below each question.
• Marks may be given for correct working even if the answer is wrong.
• Non programmable silent electronic calculators and KNEC mathematical table may be used, except where stated otherwise.

                                                                                        SECTION I (50 MARKS)

                                                                          Answer all the questions from this section

1. Use Logarithms correct to four significant figures to evaluate. (4marks)
   
2. Find the percentage error in the total length of four rods measuring 12.5cm, 24.5cm, 12.9cm and 10.1cm all the nearest 0.1cm. (3 marks)
3. In the figure below QT is a tangent to the circle at Q. PXRT and QXS are straight lines. PX = 6cm, RT = 8cm, QX = 4.8cm and XS = 5cm.
                                                          
   Find the length of QT    (3 marks)
4. Use the trapezium rule with seven ordinates to find the area bounded by the curve y = x² + 1 lines x = −2, x = 4 and x – axis (3 marks)
5. Given that make p the subject of the formula (3 marks)
6. Solve for x in the equation below:
   Log 3(x + 3) = 3 log 3 + 2 (3 marks)
7. The points (5, 5) and (−3, −1) are ends of a diameter of a circle centre A. Determine:
   1. The coordinates of A. (1 mark)
   2. The equation of a circle expressing it in form x² + y² + ax + by + c = 0 (2 marks)
8. A transformation is represented by the matrix  . This transformation maps a triangle ABC of the area 12.5cm² onto another triangle A′B′C′. Find the area of triangle A′B′C′.    (3marks)
9. Two taps A and B can fill a water bath in 8 minutes and 10 minutes respectively. Tap A is opened for 2 minutes then closed. Tap B is later opened for one minute then closed. How long will the two taps take running together to fill the remaining part of the water bath?  
     (3 marks)
10.  
    1. Expand and simplify (1-3x)⁵ up to the term in x³ (2 marks)
    2. Hence use your expansion to estimate (0.97)⁵ correct to 4d.p. (2 marks)
11. Solve for x in the equation:
    2cos4x = − 1 for 0° ≤ x ≤ 180° (3 marks)
12. Wanjiku 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. (3 marks)
13. The gradient function of a curve is given dy = 3x² – 8x + 2. If the curve passes through the point, (2, –2), find its equation. (3 marks)
                                                                     dx
14. Rationalize the denominator and simplify  (3 marks)
      2 √2  
    √5 + 2
15. The sum of two numbers is 24. The difference of their squares is 144. What are the two numbers?    (3marks)
16. The data below represents the marks scored by 15 form 4 students in an exam:
    58, 61, 40, 37, 39, 40, 41, 43, 44, 37, 70, 44, 47, 36 and 52
    Calculate the interquartile range of the above data                                      (3 marks)
    
                                                                                   SECTION II (50 MARKS)
                                                                    Answer five questions only from this section
17. The following table shows the rate at which income tax was charged during a certain year.
    

    Monthly taxable income in Ksh.	Tax rate %
    0 - 9860                                 9861 - 19720                               19721 - 29580                               29581 - 39440                               39441 - 49300                               49301 - 59160                                    over 59160	10         15         20         25         30         35         40

    A civil servant earns a basic salary of Ksh.35750 and a monthly house allowance of sh.12500. The civil servant is entitled to a personal relief of sh.1062 per month. Calculate:
    1. Taxable income (2 marks)
    2. Calculate his net monthly tax (5 marks)
    3. Apart from the salary the following deduction are also made from his monthly income.
       WCPS at 2% of the basic salary
       Loan repayment Ksh.1325
       NHIF sh.480
       Calculate his net monthly earning. (3 marks)
18. The diagram below represents a cuboid ABCDEFGH in which FG = 4.5 cm, GH = 8 cm and HC = 6 cm
                                      
    Calculate:
    1. The length of FC                                                            (2 marks)
    2.  
       1. The size of the angle between the lines FC and FH       (2 marks)
       2. The size of the angle between the lines AB and FH      (3 marks)
    3. The size of the angle between the planes ABHE and the plane FGHE       (3 marks)
19. A plane S flies from a point P (40°N, 45°W) to a point Q (35°N, 45°W) and then to another point T (35°N, 135°E).
    1. Given that the radius of the earth is 6370km find the distance from P to Q in Km.
       (Take π = ²²/₇) (2 marks)
    2. Find in nm
       1. The shortest distance between Q and T. (2 marks)
       2. The longest distance between Q and T (to the nearest tens) (2 marks)
    3. Find the difference in time taken when S flies along the shortest and longest routes if its speed is 420 knots (4 marks)
20. The probability that a pupil goes to school by a boda-boda is ²/₃ and by a matatu is ¹/₄. If he uses a boda-boda the probability that he is late is ²/₅ and if he uses matatu the probability of being late is ³/₁₀. If he uses other means of transport the probability of being late is ³/₂₀.
    1. Draw a tree diagram to represent this information. (3marks)
    2. Find the probability that he will be late for school. (3marks)
    3. Find the probability that he will be late for school if he does not use a matatu. (2marks)
    4. What is the probability that he will not be late to school? (2marks)
21. A farmer has 50 acres of land.  He has a capital Shs. 2,400 to grow carrots and potatoes as cash crops.  The cost of growing carrots is Shs.40 per acre and that of growing potatoes is Shs.60 per acre.  He estimates that the respective profits per acre are Shs.30 (on carrots) and Shs. 40 (on potatoes). By letting x and y to represent carrots and potatoes respectively:-
    1. Form suitable inequalities to represent this information. (4marks)
    2. By representing this information on a graph, determine on how many acres he should grow each crop for maximum profit.     (4marks)
       
    3. Find the maximum profit.             (2 marks)
22. An arithmetic progession is such that the first term is –5, the last term is 135 and the sum of the progression is 975.
    1. Calculate
       1. The number of terms in the series (4 marks)
       2. The common difference of the progression (2 marks)
    2. The sum of the first three terms of a geometric progression is 27 and first term is 36.       
       Determine the common ration and the value of the fourth term (4 marks)
23. In the figure below E is the midpoint of BC. AD: DC 3:2 and F is the meeting point of BD and AE.
                                                                   
    1. If AB = b and AC = c, find:
       1. BD (2marks)
       2. AE (2marks)
    2. If BF = t BD and AF = n AE. Find the value of t and n. (5marks)
    3. State the ratio of BD to BF. (1mark)
24. Given that y = 2sin 2x and y = 3cos (x + 45°)
    1. Complete the table below.  (2mks)
       

       x	0°	20°	40°	60°	80°	100°	120°	140°	160°	180°
       2sin x	0		1.97		0.68	−0.68	−1.73		−1.28	0.00
       3cos (x+ 45°)	2.12	1.27		−0.78		−2.46			−2.72	−2.12

    2. Use the data to draw the graphs of y = 2 sin 2x and y = 3 cos (x + 45°) for 0°≤ x ≤ 180° on the same axes. (4marks)
    3. State the amplitude and period of each curve. (2marks)
    4. Use the graph to solve the equation 2 sin 2x – 3cos (x + 45°) = 0 for 0°≤ x ≤ 180°    (2marks)

                                                                              MARKING SCHEME 

SECTION I

	WORKING	MARKS	GUIDELINES
1.	No.  Log   24.36   0.066547  1.48   10-1x9.045   0.9045   1.3867  −2.8231     0.2098     0.1703     x       2    0.3406    0.2098     0.3406   -1.8692 x 1/3    3̅  + −2.8692     3           3   = −1.9564      = 0.9045	M1      M1        M1      A1	logs All           Addn & Subtr            Attempt to   divided by 3
		4
2.	Error = ½ x 0.1 = + 0.05cm  Actual length = 12.5 + 24.5 + 12.9 + 10.1 = 60.0  Max length = 12.55 + 24.55 + 12.95 + 10.15   = 60.20  Min length = 12.45 + 24.45 + 12.85 + 10.05   = 59.80  A.E. = Max – Min     = 60.20 – 59.80                     2                        2   P.E. = 0.2 x 100                   60          = 0.3%	M1  M1  A1
		3
3.	X R = 4.8 x 5                 6          = 4  QT2 = PT x RT  QT2 = 18 x 8  QT = √144  QT = 12cm	M1     M1   A1
		3
4.	X   −2    −1    0    1    2    3     4   y     5      2     1    2    5   10    17   Area =½ ( 5 + 17 + 2(2 + 1 + 2 + 5 + 10) = 31 sq. Units	M1   M1   A1
		3
5.	x2  =     t p              2μ + p  2μx2 + px2 = tp  2μx2 = tp − px2  2μx2 = p(t − x2)  p = 2μx3       t − x3	M1       M1          A1	(squaring on    both sides)
6.	log (3x + 9) = log 33 + log 100        log (3x + 9) = log 2700            3x + 9 = 2700            3x = 2691            3x = 2691             3       3              x = 897	M1   M1        A1
		3
7.	(a)             A (1, 2)   (b) (x - a)2 + (y - b)2 = r2         (5 - 1)2 + (5 - 2)2 = r2         42 + 32 = 52          radius 5 units         (x - 1)2 + (y - 2)2 = 52         x2 - 2x + 1 + y2 - 4y + 4 = 25         x2 - 2x + y2 - 4y - 20 = 0	A1           M1                    A1
		3
8.	Determinmant = 2 – 12 = –10   A.S.F = Ι −10Ι             = 10  10 x 12.5 = 125 cm2	M1  M1  A1
		10
9.	Tap A                                Tap B     1/8 x 2 = 1/4                    1/10 x 1 = 1/10      1  + 1  = 10 + 4 = 7      4   10        40       20     Remaining part 13                                20     in a minute 1 + 1 = 9                        8   10  40     13 x 40 = 26 = 28/9 min      20     9      9             or    time = 2 min 53 sec.	M1         M1       A1
		3
10.	i) 15 - 5(3x) + 13 x 10 (3x)2 - 12 x 10 (3x)3 + 1 − 15x + 90x2 - 270x3 +    ii) (0.97)5 = (1 - 0.03)5              3x = 0.03                x = 0.01      (0.97)5 = 1 - 15(0.01) + 90(0.01)2 - 270(0.01)3  = 0.8587	M1  A1  M1  A1
		3
11.	cos 4x = −½    cos-1 − ½= 60o     x = 30o, 127.5o, 150o	M1  M1  A1
		3
12.	P = 300,000 - 75000       = 225,000     A = 225,000 x 1.151.25          = 225,000 x 1.151.25                         15           225000 x 1.190 = 267950                   15                     15                                                           = Ksh.17863	M1     M1 . A1
		3
13.	dy/dx = 3x2 − 8x + 2        y = x3−4x2 + 2x + c    At x = 2   y= −2     −2 = 8-16+4+c       C = 2        y = x3- 4x2 + 2x + 2	M1   M1   A1
		3
14.	2√5(√5 − 2)              (√5 + 2)(√5 − 2)        10 − 4√10             5 − 4        10 − 2√10                1        10 − 2√10	M1   M1       A1	For conjugate
		3
15.	x + y = 24    x2 + y2 = 144    x2 – (24 –x )2 = 144    x2 – [576 -48x + x2] = 144    x2 -576 + 48x – x2 = 144    48x = 720    x =15    y = 24 -15        = 9    The two numbers are 9 and 15	M1     M1       A1
		3
16.	36,37,37,39,40,40,41,43,44,44,47,52,58,61,70   Q1 = 39    Q3 = 52    Interquartile range = (52 − 39)                                  = 13	M1  M1  A1	Arranging in   ascending or   descending   orde
	SECTION II
17.	a)  taxable income       35750 + 12500 = 48250 = sh.48250    b) 9860 x 10/100 = 986      9860 x 75/100 = 1479      9860 x 20/100 = 2976      9860 x 25/100 = 2465      8810 x 30/100 = 2643                                 9545        Total less relief 1062                             sh.8483pm c) WCPS = 2/100 x 35750 = 715      Total deduction      (8483 + 715 + 1325 + 480) = 11000      Net salary = 48250                        − 11000                      sh.37250 p.m	M1       A1    M1    M1   M1    M1   A1   M1  M1  A1
		10
18.	(a)       (b).  i).tanθ = 6/9.2             tan θ = 0.6522                   θ = 330                             ii). Tan θ = 8/4.5               Al             Tan θ 1.7750                    θ = 60.600            (c). Cosine rule          62 = 102 + 82 − 2 x 8 x 10 cos θ          36 = 100 +64 -160 Cosθ          36 = 164 -160 cos θ          Cos θ = 128/16          Cos θ =0.8                    = 36.91	M1                  A1               M1   A1    M1   M1   A1    M1       M1            A1
		10
19.	(a) 10/360 x 2 x 22/7 x 6370                            = 1112km   (b)  (i) 110 x 60                            = 6600nm         (ii) 180 x 60 x cos 350                            = 8850nm.   (c) 420 = 6600                     T1        T1 = 6600                 420                           = 15hr 43min         420 = 8850                       T2        T2 = 21h 4min        T2 – T1 = 21hr 4min – 15hr 43min        = 5hrs 21min	M1    A1   M1  A1  M1     A1    M1   M1    M1    A1
		10
20.	(a)                                          (b)  P(BL) or P(ML) or P(OL)         = 2/3 x 2/5 + 1/4 x 3/10 + 1/12 + 3/20        = 4/15 + 3/40 + 1/80           = 17/48  (c) P(BL) or P(OL)         = 4/15 + 3/40 + 3/20          = 4/15 + 1/80          = 67/240  (d) P(Not late to school) = 1 – P(Late to school)                                         = 1 − 17/48 = 31/48	B1    B1 M1     A1    M1   A1   M1      A1   M1   A1
		10
21.	(a) Let the carrots be x, potatoes y and the total profit be p. The inequalities that represents         this information are: x + y = ≤ 50                                      40x + 60y ≤ 2400                                      x ≥ 0. and                                        y ≥ 0   (b)                            maximum profit✔                       P= 30x + 40y = 30( 30) + 40 ( 20)                         = sh 1700✔	B1  B1  B1  B1                                                           M1 A1
		10
22.	(a) (i) nth term = a + (n – 1)d         Last term = − 5 + (n – 1)d = 135                              (n – 1)d = 140      Sum of nth term = n/2(2a + (n – 1)d           n/2(− 10 + 140) = 975           n/2 x 130 = 975           n = 975 x 2                 130           n = 15     Alternatively.     Sum = n/2(a + L)             = n/2(−5 + 135) = 975             = n/2(130) = 975             = (130n) = 975                   2           n = 15       (ii) nth term = a + (n – 1)d            − 5− 14d = 135                    14d = 140                        d = 10   (b) s = 27, a = 36        s = a + ar + ar2      27 = 36 + 36r + 3r2        3 = 4 + 4r + 4r2       4r2 + 4r + 1 = 0       (2r + 1)2 = 0                   r = −½	M1         M1     M1     A1      M1  M1   M1   A1   M1      A1    M1    M1    M1      A1
		10
23.	(c)  BF= tBD               BF = 5               BD    8             ∴BD:DF = 8 : − 3
		10
24.	(a) x  0°  20°  40°   60°   80°   100°   120°   140°  160°  180°  2sin x  0  1.28  1.97  1.73  0.68  −0.68  −1.73  −1.97  −1.28  0.00  3cos (x+ 45°)  2.12  1.27   0.17  −0.78   −1.72  −2.46   −2.90  −2.99  −2.72  −2.12 (b)          c) y = 2 sin 2x                 Amplitude = 2                       Period = 180°             y = 3 cos(x + 45°)                    Amplitude = 3                          Period = 360°             (d) 2 sin 2x – 3 cos ( x + 450) = 0                 X = 20°
		10