Mathematics Paper 1 Questions and Answers - Lainaku II Joint Mock Examination 2023

Instructions to candidates

• This paper consists of TWO sections I and II.
• Answer ALL questions in section I and any five from section II.
• Show all the steps in your calculations giving your answers at each stage in the spaces below each question.
• Marks may be given for correct working even if the answer is wrong.
• Non-programmable silent electronic calculation and KNEC Mathematical tables may be used.

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

1. Without using mathematical table or a calculator, evaluate: (3 mks)
        36 – 8x -4 – 15 ÷ −3
          3x -3 + -8 (6 - (-2)
2. Simplify leaving your answer as a simple fraction (3 mks)
3. Solve the inequality −3x + 2 < x + 6 ≤ 17 – 2x and write down the integral values satisfying the inequality. (3 mks)
4. A triangle whose area is 16cm² has its shortest side as 12cm. Find the length of the shortest side of a similar triangle whose area is 36cm² (3mks)
5. Solve for x given that 5^{2x + 2} – 20 x 5^2x = 625 (3mks)
6. The line y = mx + 6 makes an angle of 78⁰ 58¹ with the x – axis. Find the co-ordinates of the point where the line cuts the X – axis. (3mks)
7. 3g of metal A of density 2.7g/cm³ is mixed with 1.6cm³ of metal B of density 3.2g/cm³
   Determine the density of the mixture (3mks)
8. A two digit number is such that when the digits are reversed, the value of the number increases by 36. If the sum of the unit digit and twice the tens digit is 16, find the number. (3mks)
9. On Monday the currency exchange rate was
                    1 Euro (E) = Kshs.95.65
                    1 US dollar($) = Ksh.76.50
   A gentle man Tourist decided to exchange half of his 2400E into Dollars.
   Calculate to 2 decimal places the number of dollars he received. (3 marks)
10. In the figure given below, AC is an arc of a circle centre B. Angle ABD = 60°, AB = BC = 7cm and CD = 5 cm.
                                                                            
    Calculate (3 d.p)
    1. The area of triangle ADB (2mks)
    2. The area of the shaded region. (2mks
11. If cos A = − 40 and A is obtuse, find without using tables or calculators
                      41
    1. Sin A (2mks)
    2. Tan A (1mk)
12. A matatu left town X at 9.37a.m. towards town Y at an average speed of 78km/h. At 9.47a.m. A car left the same venue at an average speed of 84.5km/h. Determine the time when the car caught up with the matatu . (3mks)
13. Using a set square, a ruler and a pair of compasses, divide the given line into five equal parts. Measure the length of one part. (3 marks)
                                     
14. Simplify the expression.    x² + 14x + 49 (3mks)
                                                   x² − 49
15. The acceleration of a particle in ms-2 is given by the expression Acc=3t –4
    Given that t=0 sec V=3m/s and S=0 metres
    Find:
    1. an expression for velocity Vms^-1 (2 mark)
    2. An expression for distance S metres from a fixed point O. (2 marks)
16. A salesman earns a basic salary of Kshs.19, 000 per month. In addition, he earns a commission of 5% for all sales above ksh.20,000. In February 2023, he sold goods worth 115,000. Calculate his total earnings that month . (3mks)

SECTION II (50 MARKS)
Answer only five questions

17.  A triangle with A(-4, 2), B(-6, 6) and C(-6, 2) is enlarged by a scale factor -1 and centre (-2, 6) to produce triangle A¹ B¹ C¹. Triangle A¹ B¹ C¹ is then reflected in the line y = x to give triangle A¹¹ B¹¹ C¹¹
    1. Draw triangle ABC, A¹ B¹ C¹ and A¹¹ B¹¹ C¹¹ and state the co-ordinates of A¹ B¹ C¹ and A¹¹ B¹¹ C¹¹ (6mks)
    2. If triangle A¹¹ B¹¹ C¹¹ is mapped onto A¹¹¹ B¹¹¹ C¹¹¹ whose co-ordinates are A¹¹¹(0,−2), B¹¹¹(4, −4) and C¹¹¹ (0, −4) by a rotation. Find the centre and angle of rotation (4mks)
18. A cylindrical water tank of diameter 5m and height 1.8m is supplied with water by pipe P of internal radius 2.5cm. Water flows through this pipe at the rate of 50m per minute. A drainage pipe Q can empty the full tank in 8 hours.
    1. Calculate the time in hours that pipe P alone would take to fill the empty tank (4mks
    2. The tank is initially half full. Water flows through pipe P into the tank for 2 hours. The drainage pipe Q is then opened and both pipes left running.
       Determine how long it will take to fill the tank (6 mks)
19. Ken and Jane cycle to school 20km away. Jane cycles at 2km/h faster than Ken and reaches there half an hour earlier. Given that the speed of Ken is xkm/hr. Find in terms of x
    1. The time taken by Ken. (1 mk)
    2. The time taken by Jane. (1 mk)
    3. Form an equation in x. (1 mk)
    4. Solve the equation in (a) above and find the speed of Ken and Jane. (7 mks)
20. The table below shows some paired values of X and Y for a known curve.
    

    X	0.0	0.2	0.4	0.6	0.8	1.0
    Y	0.0	0.4	1.6	3.6	6.4	10.0

    By establishing how x and y relates.
    Estimate the area under the curve for the interval 0 < X< 1 using
    1. The mid – ordinate rule with five mid – ordinates. (4mks)
    2. The trapezium rule with five Trapezia. (2mks)
    3. If the exact area is ¹⁰/₃ square units.
       Calculate the percentage error in the two estimates. (4mks)
21. The following are speeds in m/s of the first 50 vehicles at a police check point.
    

    Speed   in m/s                              x	No. of  vehicles   f                             f(x)
    10 - 19  20 – 29  30 - 39  40 - 49  50 - 59  60- 69  70 – 79  80 - 89  90- 99  100 -109	3  1  2  5  6 11  9  8  3  2

    
    
    1. Calculate the mean speed (4mks)
    2. Draw on the same axes using this information
       1. The Histogram (3 mks)
       2. The Frequency polygon (1 mk)
    3. Calculate the median speed. (2mks)
22. A trader sold an article at sh.4800 after allowing his customer a 12% discount on the marked price of the article. In so doing he made a profit of 45%.
    1. Calculate
       1. The marked price of the article. (2 marks)
       2. The price at which the trader had bought the article (2 mks)
    2. If the trader had sold the same article without giving a discount. Calculate the percentage profit he would have made. (3 mks)
    3. To clear his stock, the trader decided to sell the remaining articles at a loss of 12.5%. Calculate the price at which he sold each article.
       (3 mks)
23. The diagram below shows a parallelogram OPQR. Point M divides line PQ in the ration 2:3 PR and OM intersect at N. Given that OP = p and OR = r
                                                                 
    1. Express the following vectors in terms of   (2 mk)
                                                      
    2. Given that    
       By Expressing ON in two different ways find the ratio in which N divides PR. (8 mks)
24. The diagram below shows the graph of a moving matatu from Nakuru to Naivasha.
                                              
    1. Find the acceleration of the matatu. (2mks)
    2. Find the deceleration of the matatu (2mks)
    3. Calculate the distance the matatu travelled while accelerating. (2mks)
    4. Calculate the distance the matatu covered while traveling at an acceleration of 0m/s2 (2mks)
    5. Find the distance between the two Towns. (2 mks)

MARKING SCHEME

	SECTION A
1.	36 + 32 + 5      3x −3 + −8 (8)         73         − 73   = − 1	M1  M1   A1	Simplification of num  Simplification of deno
		03
2.		M1  M1  A1	Removal of fractional powers.  Removal of negative power and simplim
		03
3.	−3x + 2 < x + 6    −4 x < 4       x > −1     x + 6 ≤ 17 – 2x     3x ≤ 11      X ≤ 11/3     −1< x ≤ 3 2/3     Integral values of x={ 0, 1, 2, 3}	B1  B1  B1
		03
4.	A.S.F = 16:36      A.S.F = 4/9      L.S.F = √4/9                = 2/3             2/3 = 12/x       = 12 x 3              2       = 18 cm	B1  M1  A1	For  L.S.F    Equating the ratios
		03
5.	Let U=52X    52X(52) - 20(52X) = 625    25U - 20U = 625    U=125    5U = 625    5        5        U = 125    52X = 53    2x = 3     X = 3/2     x = 1.5	M1    M1          A 1	Expressing in the same base
		04
6.	Gradient = Tan 78° 581                 = Tan 78.97°  Gradient = 5.130    y = 5.13x + 6    At x –axis, y = 0    0 = 5.13x + 6    X = − 1.17    Co-ordinate (−1.17, 0)	B1  M1  A1	Gradient  Substitution for y=0  Co- ordinates
		04
7.	Metal A: Volume =   mass                                    density                               =   3g                                         2.7 g/cm3                              = 1.111cm3   Metal B: Mass = d x v                           = 1.6 x 3.2                           = 5.12g   Total mass = 5.12 + 3 (mixture)                           = 8.12g   Total volume = 1.111 + 1.6                        = 2.711   Density =   8.12                     2.711    = 2.995g/cm3	M1  M1  A1	For both vol &mass
		04
8.	Let no.be = xy  10x + y is the value of the number  ⇒ reversing digits, the new value is 10y + x  ∴ 10y + x – (10x + y) = 36  10y + x – 10x – y = 36  9y – 9x = 36  9(y – x) = 36  y – x = 4…..(i)   y + 2x = 16 …..(ii)  y – x = 4  y + 2x = 16  −3x = −12  x = 4  and y – 4 = 4  y = 8  ∴ number is 48	M1  M1  A1	For both equations✔ ✔elimination of one variable or equivalent
		03
9.	½ of 2400E = 1200E  In ksh. = 1200E x 95.65  = Ksh.114,780  Number of dollar = Kshs.114,780                                            76.50  = 1500.39 dollars ( to 2 d.p)	M1  M1  A1	For conversion.
		03
10.	Area of triangle ADB is ½ x 7 x 12 sin 60°                                7 x 6 sin 60°               = 36.373cm2 (to 3 d.p)   Area of unshaded sector  60 x 22 x 7 x 7 = 25.6667 360    7 Shaded area = 36.373 − 25.667 = 10.706cm2 (to 3 d.p)	M1  A1  M1  A1	Exp for the area
		04
11.	Sin A = 9/41               Tan A = − 9/40 (the ans must be –ve)	BI  B1  B1	Sin Ratio  Accept 0.2195  Tan Ratio  Accept − 0.225
		03
12.	9.47 – 9.37 = 10min  60min = 78km/hr             = 10 x 78                60  Dist. apart = 13km  R.S = (84.5 – 78) = 6.5km/hr Time taken = 6.5 Time to Catch up = 9.47a.m. + 2hrs                = 11.47 a.m.	M1  M1      A1	Exp for Dist apart
		03
13.	Length of one part = 2.5cm	B1  B1  B1	Drawing a line at an acute angle  Subdividing the line  Accept 2.4cm or 2.6cm
		03
14.	Num = (x + 7) (x +7)   Den = (x + 7) ( x – 7)     (x + 7) (x +7)     (x + 7) ( x – 7)     x + 7     x - 7	M1  M1  A1	Num simplified  Deno simplified
		03
15.	dv = 3t – 4   dt V= ∫3t – 4dt V= 3t2 − 4t + c        2 Since V = 3 when t = 0 Then 3(0)2 − 4(0) + c = 3    2 c = 3 V= 3t2 − 4t +3        2 S =  = t3 – 2t2 +3t + k    2 = S = 0, when t = 0    k = 0 s = t3 – 2t2 + 3t       2	M1         A1       M1     A1
		04
16.	Extra sales = Kshs. 115,000                                 − 20,000                          Kshs. 95,000   Commission earned = 5  x 95,000 = 4,750                                     100   Total earning = 4,750 + 19,000                         = Kshs. 23,750	M1  M1  A1	Exp for the commission
		03
17.	(a)  B1 for object drawn        B2 1st image drawn        B1 coordinates of 1st image        B2 for 2nd image & its coordinates given   (b)  B1 for the 1st bisector       B1 for the 2nd bisector       B1 for the centre of rotation C(5,−1)       B1 for the angle of rotation =180° or equivalent
		10
18.	Rate of water flow = 22/7 x 2.5 x 2.5 x 5000 = 98,214.28571cm3 Volume of tank = 22/7 x 2.5 x 2.5 x 1.8            = 35.35714286 m3 35.35714286 x 1000000            98214.28571 = 360mins = 6hrs Half full ⇒ pipe has worked for 3hrs Additional ⇒ 2 hours Total hours = 2 + 3                   = 5 hrs Fraction of Tank without water = 1 – 5/6                                = 1/6 of tank Both taps working together               = 1/6 – 1/8           = 1/24 of tank 1/24 of tank filled in 1hr 1/6 of tank = ?    = 1/6 x 24/1 x 1    = 4hrs	M1  M1  M1  A1  M1  M1  M1    A1  M1    A1	Rate of flow    vol of tank      Division    NB  Follow through for other   possible alternative   methods
		10
19.	Ken’s speed = x km/h.  Distance = 20km. Time taken by Ken = (20/x)hrs Time taken by Jane = (20/x+2)hrs 20/x – 20/x+2 = ½ 20x2 (x+2) – 20 x 2 x x = x(x + 2) 20 x 2(x + 2) – 20 x 2 x X = x2 + 2x 40x + 80 – 40x = x2 + 2x x2 + 2x – 80 = 0 x2 - 8x +10x – 80 = 0 (x + 10)(x – 8) = 0 Either x = -10 or x = 8 ∴ Ken’s speed = 8km/h and Jane’s speed=10km/h	B1  B1  B1  M1  M1  M1  M1  A1  B1  B1	For the correct eqn.  Accept equivant  Removal of fractions and   partial factorisation  For the eqn  Removal or fractions  Quadratic eqn equated to 0  Partial simplification  Factorization equated to 0  Accept equivalent   methods
		10
20.	x and y are related by the eqn y = 10x2    X  0.0  0.2   0.4   0.6   0.8   1.0   Y  0.0  0.4   1.6   3.6   6.4   10.0  Area = 0.2(0.1 + 0.9 + 2.5 + 4.9 + 8.1)         = 3.3 square units Trapezium rule Area = ½ x 0.2 ( 0.0 + 10.0 ) + 2 ( 0.4 + 1.6 + 3.6 + 6.4 ) = 0.1 (10) + 2 (12) = 0.1 x 34 = 3.4 square units  % error in mid ordinate rule  10 − 33 x 100   3     10                  10          3  = 1% % error in trapezium rule  10 − 17 x 100   3     5                  10          3   =  2%	B2  M1  A1    M1  A1  M1    A1  M1    A1	All mid values✔  B1 for any 3✔  ✔substitution
		10
21.	Speed   Mid/points(x)   class  f   fx     c.f  10-19   14.5   9.5-19.5  3  43.5   3   20-39   24.5  19.5-29.5  1  24.5   4   30-39   34.5  29.5-39.5  2  69   6   40-49    44.5  39.5-49.5  5  222,5  11  50-59   54.5  49.5-59.5  6  327  17  60-69   64.5  59.5-69.5  11  709.5  28  70-79  74.5  69.5-79.5  9  670.5  37  80-89  84.5  79.5-89.5  8  676  45  90-99  94.5  89.5-99.5  3  283.5  48  100-109  104.5  99.5-109.5  2  209 50   Σf= 50  Σfx= 3235     x = mean speed = Σfx  Σf  = 3235 = 64.7 m/s        50                 (c) Median speed will be given by 25th vehicle.       Median = 59.5 + 8 x 10 =                                     11                   = 66.7727m/s	M1     M1     M1     A1 S1     B2   B1   M1       A1	✔values for all   the mid pts  for all the fx   values✔  Correct linear   scale  All bars  Polygon
		10
22.	S.P. = 88%     88% =shs.4800     M.P.= shs 4800/88 x 100     M.P. = sh.5454.54 Accept M.P= Shs.5454.55  B. P.=145% = 4800  B.P. = 48000 x 100                    145          = shs 3310.34             Shs 3310.35 Profit= Shs.5454.50- 3310.30 percentage Profit = 5454.54 – 3310.34 x 100                                               10               3310.34                   = 0.6477 x 100                   = 64.77%  87.5 x 3310.30  100 = shs 2,896.55	M1   A1   M1   A1 M1 M1   A1 M1 MI A1	Follow through              For the profit
		10
23.	PR = PO + OR          OM = OP + PM          ON = KOM              ON = OP + hPR                K = 1 – h ……………………….(i) 2/5K = h………………………...(ii) Substituting for h in (i) K = 1 – 2/5K K = 5/7 h = 2/5 (5/7) h = 2/7 PN:NR=2:5	B1       B1   M1 M1     M1   M1 M1   A1 B1   B1	For equating   eqn1 &2  For extracting   the eqns.
		10
24.	ACC = 15 – 0                 20           = 0.75m/s2 Dece = 0 – 15                 20           = − 0.75m/s2 Area = 1 x 20 x 15             2           = 150m Area = 20 x 15         = 300m Area = 1 (20 + 60) X 15             2          = 600m	M1  A1  M1  A1  M1  A1  M1  A1  M1  A1
		10