Mathematics Paper 2 Questions and Answers - Lainaku 1 Joint PreMock Exams 2023

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

1. Use logarithms, correct to 4 decimal places, to evaluate    (4mks)
                        3√83.46x0.0054
                                 1.52²
2. Given that the ratio of x:y = 4:5, find to the simplest form the ratio of (7x −2y):(x + 2y)  (2mks)
3. Ruto bought a plot of land for Ksh.280,000. After 4 years, the value of the plot was Ksh.495,000. Determine the rate of appreciation, per annum, correct to one decimal place. (3mks)
4. The height in centimeters, of 100 tree seedlings in a tree nursery are shown in the table below.
   

   Height(cm)	10-19	20-29	30-39	40-49	50-59	60-69
   Number of trees	9	16	19	26	20	10

   Find the quartile deviations of the heights. (3mks)
5. The equation of a circle is given by 4x² − 12x + 4y² − 8y − 3 = 0. Determine the coordinates of the centre of the circle and the radius of the circle. (4mks)
6. Simplify the expression √48  , leaving the answer in the form a√b + c where a, b and c are integers (3mks)
                                      √5 + √3             
7. In the figure below R, T and S are points on a circle centre O. PQ is a tangent to the circle at T, POR is a straight line and <QPR = 20°.
                                                              
         Giving reasons find the size of <RST (2mks)
8.  
   1. Expand the expression (1 + ½x)⁵ in ascending powers of x, leaving the coefficients as fraction in their simplest form. (2mks)
   2. Use the first three terms of the expansion in (a) above to estimate the value of (1.05)⁵ correct to 4 significant figures (2mks)
9. Make t the subject of the formula in s = √3d(t − d) (3mks)
                                                                           8
10. A trader bought maize for Ksh 20 per kilogram and beans for Ksh 60 per kilogram. She mixed the maize and beans and sold the mixture at Ksh 48 per kilogram. If she made a 60% profit, determine the ratio of maize:beans per kilogram in the mixture. (3mks)
11. The cash price of a digital television is Ksh. 27,500. A customer decided to buy it on hire purchase terms by paying a deposit of Ksh 17,250. Determine the monthly rate of compound interest charged on the balance if the customer is required to repay by six equal monthly instalments of Ksh. 2,100 each. (3mks)
12. The first, the third and the seventh terms of an increasing arithmetic progression are three consecutive terms of a geometric progression. If the first term of the arithmetic progression is 10, find the common difference of the arithmetic progression. (4mks)
13. The lengths of two similar pieces of wood were given as 12.5 m and 9.23 m. Calculate the absolute error in calculating the difference in length between the two bars. (3mks)
14. Solve for x given that ¹/₃log₂8 + log₂(2x − 4) = 5 (3mks)
15. In nomination for a committee, two people were to be selected at random from a group of 3 men and 5 women. Find the probability that a man and a woman were selected. (2mks)
16. Pipe A can fill an empty tank in 3 hours while, pipe B can fill the same tank in 6 hours. When the tank is full it can be emptied by pipe C in 8 hours. Pipe A and B are opened at the same time when the tank is empty. If one hour later, pipe C is opened, find the total time taken to fill the tank. (4mks)

                                                                                       SECTION II (50 marks)

                                                          Answer only five questions in this section in spaces provided

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

    Monthly taxable income in Ksh	Tax rate (%) in each shilling
    1-11,180	10
    11,181-21,714	15
    21,715-32,248	20
    32,249-42,782	25
    Over 42,782	30

    
    
    1. During the year, Njuguna’s monthly income was as follows: Basic salary Ksh 40,000, House allowance Ksh 11,090 and Commuter allowance Ksh 7,000.
       Calculate
       1. Njuguna’s total monthly taxable income. (1mk)
       2. Total income tax charged on Njuguna’s monthly income (4mks)
    2. Njuguna’s net monthly tax was Ksh. 10,750.80. Determine the monthly tax relief allowed. (1mk)
    3. A proposal to expand the size of the first income tax band by 50% while retaining the size of the next three bands was made. The tax rates would remain as before in each band. Using the proposal, calculate:
       1. The tax Njuguna would pay in the first band. (2mk)
       2. The tax Njuguna would pay in the last tax band. (2mks)
18.  
    1. Given that A = (3 x x + 1 2) and B = (1 2 3 0), find the values of x for which AB is a singular matrix. (3mks)
    2. Otieno bought 3 exercise books and 5 pens for a total of Ksh 165. If Otieno had bought 2 exercise books and 4 pens, he would have spent Ksh 45 less. Taking letter e to represent the price of an exercise book and letter p to represent the price of a pen
       1. Form two equations to represent the above information. (2mks)
       2. Use matrix method to find the price of an exercise book and that of a pen. (3mks)
       3. The principal of Njabini boys decided to reward 36 students each with 2 exercise books and one pen. Calculate the total amount of money he paid. (2mks)
19. Three quantities X, Y and Z are such that X varies directly as the square root of Y and inversely as the fourth root of Z. When X=64, Y=16 and Z= 625.
    1. Determine the equation connecting X, Y and Z. (4mks)
    2. Find the value of Z when Y=36 and X=160 (2mks)
    3. Find the percentage change in X when Y is increased by 44% and Z decreased by 19% correct to one decimal place. (4mks)
20. The figure below shows a parallelogram OPQR with O as the origin, OP = p and OR= r. Point T divides RQ in the ratio 1:4. PT meets OQ at S.
                                        
    1. Express in terms of p and r the vectors
       1. OQ (1mk)
       2. OT (1mk)
    2. Vector OS can be expressed in two ways; i) OS = mOQ ii) OS = OT + nTP, where m and n are constants. Express OS in terms of
       1. m, p and r (1mk)
       2. n, p and r (1mk)
          Hence find the
       3. value of n and m (5mks)
       4. ratio OS:SQ (1mk)
21. A quadrilateral ABCD has vertices at A(1,1), B(4,2), C(1,3), D(2,2).
    1. Draw ABCD on the grid provided (1mk)
    2. Give that X = [1 0 0−1] Y= [−1 0 0 1] V = [0 −1 −1 0] . Find the coordinates of A¹B¹C¹D¹, A¹¹B¹¹C¹¹D¹¹ and A¹¹¹B¹¹¹C¹¹¹D¹¹¹ the images of ABCD under combined transformation VXY. Show all your working of coordinates below;
       1. Coordinates of A¹B¹C¹D¹ and draw it on the grid. (2mks)
       2. Coordinates of A¹¹B¹¹C¹¹D¹¹ and draw it on the grid. (2mks)
       3. Coordinates of A¹¹¹B¹¹¹C¹¹¹D¹¹¹ and draw it on the grid. (2mks)
    3. Showing your working find a single matrix that will map ABCD onto A¹¹¹B¹¹¹C¹¹¹D¹¹¹. (3mks)
        
22.  
    1. Complete the table below for y = x³ + 4x² − 5x −5 (2mks)
       

       X	−5	−4	−3	−2	−1		1	2
       y = x3 + 4x2 − 5x −5						−5

    2. On the grid provided, draw the graph of y = x³ + 4x² − 5x − 5 for −5 ≤ x ≤ 2 (3mks)
    3.  
       1. Use the graph to solve the equation x³ + 4x² − 5x − 5 = 0 (2mks)
       2. By drawing a suitable straight line graph, solve the equation x³ + 4x² −x − 4 = 0 (3mks)
          
23. A polytechnic planned to buy x lockers for a total cost of Ksh 16,200. The supplier agreed to offer a discount of Ksh 60 per locker. The polytechnic was then able to get three extra lockers for the same amount of money.
    1. Write an expression in terms of x, for the:
       1. Original price of each locker; (1mk)
       2. Price of each locker after the discount. (1mk)
    2. Form an equation in x and hence determine the number of lockers the polytechnic bought. (5mks)
    3. Calculate the discount offered to the polytechnic as a percentage (3mks)
24.  
    1. Using ruler and compasses only construct triangle ABC such that AB=4 cm, BC= 5cm and <ABC=120°. (3mks)
    2. Measure AC (1mk)
    3. On the same diagram draw a locus of points equidistant from point A and point C and label the locus as L₁. (1mk)
    4. Draw on the same diagram a locus of points L₂ equidistant from point C and point B and lable the locus as L₂ (1mk)
    5. Label the point where L₁ and L₂ meet as O. Using O as a centre draw a locus of points L₃ touching points A, B and C. Measure the length from point O to L₃. (2mks)
    6. Draw the locus of points L₄ equidistant from line AC and Line AB. Extend L₄ to meet line BC and lable where they meet point D. Measure the length AD. (2mks)

                                                                                          MARKING SCHEME

NO	WORKING	MARKS	REMARKS
1		M1    M1    M1    A1	For all logs correct    For addition and subtraction    For dividing by 3    For 0.5799
	Total	4
2	7(4) − 2(5): 4 + 2(5)  28-10 : 4+10        18:14          9:7	M1    A1	Substituting for x and y.   For Answer
	Total	2
3	495,000 = 280,000(1+ R/100)4   495,000 − 280,000 (1+R/100)4   280,000    280,000   4√1.767857 = 1+ R/100   1.1531−1= R/100   0.1531 = R/100   R = 15.3%	M1      M1    A1	For correct substitution    For 4√1.767857    For answer.
	Total	3
4	Lower quartile = 19.5 + (25 − 9)10                                              16                          = 19.5 + 10 = 29.5   Upper quartile = 49.5 + (75-70)10                                              20                          = 49.5+2.5                          = 52           Quartile Deviation = 52 − 29.5                                                  2                                        = 11.25	M1    M1    A1	For lower quartile    For upper quartile      For 11.25
	Total	3
5	4/4x2 −12/4x + 4/4y2 − 8/4y −3/4y = 0  x2− 3x + y2 − 2y = 0.75  x2− 3x +1.52 + y2 − 2y + 1 = 0.75 + 2.25 + 1  (x −1.5)2 + (y−1)2 = 4  Center = (1.5, 1)  Radius = 2	M1      1MK    A1  A1	For dividing by 4     For completing square.   For center For radius
	Total	4
6	√48 = √16 x 3 = 4√3    4√3(√5 − √3)    (√5 + 3)(√5 − √3)    (4√15 −12)            2     2(2√15 − 6)             2                = (2√15 − 6)	M1    M1        A1	For 4√3    For multiplying b conjugate      For (2√15 − 6)
	Total	3
7	<RST=55°    Reason; Angles in alternate segments are equal.	B1  B1	For 55°  For reason
	Total	2
8	(a)  BC for power 5 = 1, 5, 10, 10, 5, 1       1+5(½x)1+10(½x)2 +10(½x)3 + 5(½x)4 + (½x)5       1+ 5/2x + 5/2x2 + 5/4x3 + 5/16 x4 +1/32x5 (b) 1 + 5/2x + 5/2x2 are the1st 3 terms      1+ ½x = 1.05      ½x = 0.05          X= 0.1      1+ 5/2(0.1) + 5/2(0.1)2      1 + 0.25 + 0.025 = 1.275	M1    A1        M1    A1	Inserting binomial coefficients.  In simplest form.        For X= 0.1    For the answer
	Total	4
9	s2 = 3d(t−d)8   8s2 = 3dt−3d2   3dt = 8s2+3d2	M1    M1    A1	For squaring both sides    For making 3dt subject    For any of the two expressions of t.
	Total	3
10	Cost price before profit
	Total	3
11	Amount = MI = 2100 x 6 = 12,600 Ksh   Principal = 27,500 −17250 = 10, 250 Ksh   12,600 = 10,250(1+R/100)6      12,600 = 10,250(1+R/100)6      10,250    10,250      6√1.2293 = 1+ R/100            1.035 = 1 + R/100          1.035 − 1 = R/100                       R = 3.5 %	M1   M1     A1	Substituting amt and principal in formula.  Finding 6th root on both sides.    For 3.5%
	Total	3
12	a, a + 2d, a +6d   a +6d  = a + 2d   a +2d         a   Replacing a with 10  10 + 6d  = 10 + 2d  10 + 2d         10  10(10 + 6d) = (10 + 2d)(10 + 2d)  100 + 60d = 100 + 40d + 4d2   60d = 40d + 4d2    20d = 4d2    4d      4d                     d = 5	M1    M1    M1    A1	For the three terms    For 10 +6d  = 10+2d         10 +2d        10    For working out for d
	Total	4
13	Actual difference = 12.5 – 9.23 = 3.27   Max difference = 12.55 − 9.225 = 3.325   Minimum difference = 12.45 − 9.235 = 3.215   Absolute error = 3.325 − 3.215 = 0.055                                        2	M1  M1  A1	For max difference  For minimum difference
	Total	3
14	Log281/3 + Log2(2x − 4) = Log232   Log22 + Log2(2x − 4)=Log232    2(2x−4) = 32     4x − 8 = 32             X = 10	M1    M1      A1	For Log232    For the equation 2(2x − 4) = 32
	Total	3
15	P(Woman or Man) = P(MW) or P(WM)                                  = ( 3/8 x 5/7 ) + (5/8 x 3/7)                                  = 15/56 + 15/56 = 15/28	M1     A1	For working
	Total	2
16	Work done by A per hour = 13 Work done by B per hour = 16 Work done by A per hour = 18 In one hour of A and B =13 + 16 = 12 Remaining work= 12 Work done by all pipes in 1 hour=13 + 16-18= 38 1hr 38 12 Time by all pipes =12 ÷ 38 = 1 2 x 83 = 43 = 113 hours Total time taken = 1 hr + 113 hrs = 213 hours	M1    M1    M1    A1	Work done by A and B    Work done by all pipes.    Time by all pipes.    Total time taken
	Total	4
17	(i) (a)  40,000 + 11, 090 + 7,000  = 58,090 Ksh             1st band 11,180 x 10%    = 1,118 Ksh             2nd band 10,534 x 15%   = 1,580.10 Ksh             3rd band 10,534 x 20%    = 2,106.80 Ksh             4th band 10,534 x 25%    = 2,633.50 Ksh             Remaining 10,534 x 30% = 4,592.40 Ksh                                          Total tax  12,030.80 Ksh (ii)   Tax relief = 12,030.80 – 10,750.80 = 1,280 Ksh (iii) (a) 11,180 x 150%     = 16,770 Ksh             Tax for the band = 16,770 x 10% = 1, 677 Ksh        (b) 58,090 – 48372 = 9,718 Ksh                   9718 x 30% = 2,915.40 Ksh	M1  M1  M1  M1  A1  A1  M1  A1  M1  A1	For taxable income  For 1st and 2nd bands  For 3rd and 4th bands  For remaining amount.  For total tax  For tax relief
	Total	10
18	(a)  3 x x+1 2 1 2 3 0 =3+3x 6 x+7 2x+2        (3+3x)(2x+2)-(6x+42) = 0         6x2 + 6x − 36 = 0         Dividing by 6 we get         x2 + x − 6 = 0         Solving         x2 + 3x − 2x − 6 = 0         x = 2 or x = −3 (b) (i)       3e + 5p = 165               2e + 4p = 120      (ii)    (3 5 2 4)(e/p) = (165/120)               ½(4 −5 −23)(3 5 2 4)(e/p) = 1/2(4 −5 −23)(165/120)               (1 0 0 1)(e/p) = 1/2(60/30)      (e/p) = (30/15) 1 exercise book= 30 Ksh, 1 pen = 15 Ksh             2(30) + 15 = 60 + 15 = Ksh 75      (iii)    75x 36 = Ksh 2,700	M1    M1    A1    B1  B1    M1    M1      A1  M1  A1	For equating product of A and B to Zero.    For solving quadratic equation.  For values of x.    For equation 1  For Equation 2    For matrix equation.    Working using inverse to get required costs.      Cost of 2 exercise books and 1 pen
	Total	10
19	(i)    (ii)   (iii)	M1  M1    M1    M1      M1  A1    M1    M1    M1  A1	For varies  For introducing constant K  For value of K  For equation connecting X, Y and Z.    For making z subject and substituting values   of x and y.    For substituting % change in the formula.  For 1.2649  For 126.49%  Accuracy to 1 d.p
	Total	10
20	(a)  (i) OQ = r + p            OT = r + 1/5p (b)  (i)                (ii)                  (iii)                          mr + mp = r(1 − n) + p(4/5n + 1/5 )             m = 1 − n or m = 4/5 n + 1/5             1 − n = 4/5n + 1/5             4/5 = 14/5 n             9/5n = 4/5              n = 4/9             m = 1 − 49 = 59 → m = 5/9                 OS:SQ = 5 : 4	B1  B1    B1      B1  B1    M1    M1    A1      B1	For equating the two values of OS.    For working out to get the scalars.    For n = 4/9  For m = 5/9
	Total	10
21	(a)  (b) (i)        [− 1 0 0 1][1 4 1 2  1 2  3 2] =        [− 1 − 4 1 2  − 1 − 2 3 2]       Coordinates of A1B1C1D1       (ii) [ 1 0 0 −1][− 1 −4 1 2 − 1 − 2 3 2] =             [ − 1 − 4 − 1 − 2 −1 − 2 − 3 − 2 ]       Coordinates of A11B11C11D11       A11( −1, −1) B11(−4, −2) C11(−1, −3) D11(−2, −2)      (iii)            [ 0 −1 − 10][− 1 − 4 − 1− 2 − 1 ]             = [1 2 1 2  3 2  1 2] (c)          Coordinates of A111B111C111D111        A111(1, 1) B111(2,4) C111(3, 1) D111(2,2)        [a b c d] [1 4 1 2 1 2 3 2] = [1 2 1 4 3 2 1 2]         a + b = 1        4a + 2b = 2 Solving a = 0, b = 1         c + d = 1        4c + 2d = 4 Solving c = 1, d = 0       [a b c d] = [0 1 1 0]       Description; Reflection on line y = x	B1      B1        B1        B1    B1    B1  B1      B1      B1      B1	For ABCD      For A1B1C1D1      For A11B11C11D11      For A111B111C111D111  For diagram of A1B1C1D1    For diagram of A11B11C11D11      For diagram of A111B111C111D111      For working for a,b,c,d  For matrix [0 1 1 0]    For correct description of the matrix.
	Total	10
22	(a)     x  −5  −4  −3 −2   −1  0 1  2  y = x3 + 4x2 − 5x − 5  −5  15 19  13  3   −5  −5  9    (b)  Solutions to equation x3 + 4x2 −5x −5 = 0 are on line  Y = 0  These are x = −4.8 ± 0.1, 1.4 ± 0.1, −0.7±0.1 (c) (i)  Solution to x3 + 4x2 −x −4 = 0      (ii) y = x3 + 4x2 − 5x − 5           0 = x3 + 4x2 − x − 4           Y= − 4x −1           X = 1, −1, −4	B2            P1    C1  S1    L1            B2        B2	For all the 6 correct  B1 for 3 correct            For plotting all points correctly.  For a smooth curve.  For uniform scale on both y and x   axis.  For line y = 4x − 1            For any two correct values of x      For any two correct values of x
	Total	10
23	(a) (i)               16,200                  x       (ii)              16,200                 x + 3  (b) 16,200 − 16,200 = 60            x           x + 3       Becomes       60x2 + 180x − 48,600 = 0       Simplifying        X2 + 3x – 810 =0       X= −3 ± 9 − 4x1x − 8102                             2       X = − 3 ± 572                     2        X = 27      Or x = −30(ignore)  (c)             Discount = Ksh 60          Original Price = 16,200 = Ksh 600                                        27         60 x 100 = 60 %        600	B1  B1    M1    M1    M1    A1  B1    M1  M1    A1	For correct expression.  For correct expression.    For the expression   16,200 − 16,200 = 60        x           x +3  For forming quadratic equation from  above expression.  For the method used for working out.   Either formula or any other.  For x = 27  For − 30 which is ignored.    For discount = Ksh 60  For working out original price.  For correct percentage discount.
	Total	10
24	(i)   Diagram on last page (ii)   Length of AC = 7.8 ± 0.1 (iii)  Locus L1 on the diagram (iv)  Locus L2 on the diagram (v)   Locus L3 on the diagram (vi)  Length of O to L3 = 4.6 ± 0.1 (vii)  Locus L4 on the diagram (viii) Length of AD = 5.0 ± 0.1	B1  B1  B1  B1  B1  B1  B1  B1  B1  B1	<ABC drawn without using protractor. For Line AC Δ ABC complete  For AC = 7.8 ± 0.1 For locus L1 on the diagram For Locus L2 on the diagram For Locus L3 on the diagram For Length of O to L3 Locus L4 on the diagram AD = 5.0 ± 0.1
	total	10