Mathematics Paper 2 Questions and Answers - Mwakican Joint Pre Mock Examination 2021

Instructions to candidates

1. Write your name, admission number and class in the spaces provided above.
2. Sign and write the date of examination in the spaces provided above.
3. The paper contains two sections: Section I and Section II.
4. Answer All the questions in section I and strictly any five questions from Section II.
5. All answers and working must be written on the question paper in the spaces provided below each question.
6. Show all the steps in your calculations, giving your answers at each stage in the spaces below each question.
7. Marks may be given for correct working even if the answer is wrong.
8. Non-programmable silent electronic calculators and KNEC mathematical tables may be used, unless stated otherwise.

SECTION A (50 MARKS)
Answer all the questions in this section

1. Use logarithm table to evaluate. (4 mks)
   ∛0.52 × 0.312
          2.12²
2. 200 cm³ of acid is mixed with 300 cm³ of alcohol. If the densities of acid and alcohol are 1.08g/cm3 and 0.8 g/cm³ respectively, calculate the density of the mixture.(3 mks)
3. The coordinates of P and Q are P(5, 1) and Q(11, 4) point M divides line PQ in the ratio 2 : 1. Find the magnitude of vector OM. (3 marks)
4. The table below shows income tax rates in a certain year.
   

   Monthly income in Ksh	Tax rate in each Ksh
   1 – 9680	10%
   9681 – 18800	15%
   18801 – 27920	20%
   27921 – 37040	25%
   Over 37040	30%

   In that year, a monthly personal tax relief of Ksh. 1056 was allowed. Calculate the monthly income tax paid by an employee who earned a monthly salary of Ksh 32500. (4 mks)
5. Make w the subject of the formulae. (3mks)
   2x = √2w +8
              3w-5
6. A line passes through points (2, 5) and has a gradient of 2.
   
   1. Determine its equation in the form y=mx+c. (2mks)
   2. Find the angle it makes with x-axis. 1mk
7. A quantity P is partly constant and partly varies as the cube of Q. When Q=1, P=23 and when Q =2, P= 44. Find the value of P when Q = 5. (3mks)
8. The vertices of a triangle are A(1, 2) , B(3, 5) and C(4, 1). The co-ordinates of C’ the image of C under a translation vector T are (6, -2).
   
   1. Determine the translation vector T. (1mk)
   2. Find the co-ordinates of A’ and B’ under the translation vector T. (2mks)
9. Expand (1 -x)⁴ using the binomial expansion. (1mk)
   Use the first three terms of the expansion in (a) above to find the value of (0.98)⁴ correct to nearest hundredth. (2mks)
10. Find the centre and radius of a circle with equation:
    χ² + y² - 6χ + 8y – 11 = 0 (3mks)
11. Two grades of coffee one costing sh.42 per kilogram and the other costing sh.47 per kilogram are to be mixed in order to produce a blend worth sh.46 per kilogram in what proportion should they be mixed. (3mks)
12. Pipe A can fill an empty water tank in 3 hours while pipe B can fill the same tank in 5 hours. While the tank can be emptied by pipe C in 15 hours. Pipe A and B are opened at the same time when the tank is empty. If one hour later pipe C is also opened. Find the total time taken to fill the tank. (4 mks.)
13. Simplify the expression: (3mks.)
        (9t²- 25a²)    
    6t²+ 19at+15a² 
14. A business bought 300 kg of tomatoes at Ksh. 30 per kg. He lost 20% due to waste. If he has to make a profit 20%, at how much per kilogram should he sell the tomatoes. (3mks.)
15. Evaluate without using a Mathematical table or a calculator. (2mks)
    Log₆216 +[Log42 - Log6] ÷ Log49
16. Given that the ratio x: y = 2:3, find the ratio (5x-2y)∶ (x+y) (3 mk)

SECTION II (50mks)
Answer only five questions in this section in the spaces provide

17. Draw the graph of y= x^3+2x^2-5x-8 for values of x in the range -4≤x≤3 (5mks)
    

    x	-4	-3	-2	-1	0	1	2	3
    x3	-64							27
    2x2
    -5x
    -8
    y	-20

    
    By drawing suitable straight line on the same axis, solve the equations.
    1. x³+2x²-5x-8=0 (1mks)
    2. x³+2x²-5x-7=0 (2mks)
    3. 3+3x-2x²-x³=0 2mks
18. A transformation represented by the matrix maps the points A(0, 0), B(2, 0), C(2, 3) and D(0, 3) of the quad ABCD onto A¹B¹C¹D¹ respectively.
    
    1. Draw the quadrilateral ABCD and its image A¹B¹C¹D¹. (3mks)
       
    2. Hence or otherwise determine the area of A¹B¹C¹D¹. (2mks)
    3. Another transformation maps A¹B¹C¹D¹ onto A¹¹B¹¹C¹¹D¹¹. Draw the image A¹¹B¹¹C¹¹D¹¹. (2mks)
    4. Determine the single matrix which maps A¹¹B¹¹C¹¹D¹¹ back to ABCD. (3mks)
19. In the figure below (not drawn to scale) AB = 8cm, AC = 6cm, AD = 7cm, CD = 2.82cm and angle CAB = 50°.
    
    Calculate (to 2d.p.)
    1. the length BC. (3 marks)
    2. the size of angle ABC. (3 marks)
    3. size of angle CAD. (3 marks)
    4. Calculate the area of triangle ACD. (2 marks)
20. Three variables P, Q and R are such that P varies directly as Q and inversely as the square of R.
    
    1. When P = 18, Q = 24 and R = 4.
       Find P when Q = 30 and R = 10. (3mks)
    2. Express P in terms of Q and R. (1mk)
    3. If Q is increased by 20% and R is decreased by 10% find:
       1. A simplified expression for the change in P in terms of Q and R. (3mks)
       2. The percentage change in P. (3mks)
21. A surveyor recorded the following information in his field book after taking measurement in metres of a plot.
    

    	To E
    720 to F 240 G	1000 880 640 480 400 200	320 to D 600 to C 40 to B
    	From A

    1. Sketch the layout of the plot. (4 mks.)
    2. Calculate the area of the plot in hectares. (6mks)
22. A line L passes through points (-2, 3) and (-1,6) and is perpendicular to a line P at (-1,6).
    
    1. Find the equation of L. (2 mks)
    2. Find the equation of P in the form ax + by = c, where a, b and c are constant. (2 mks)
    3. Given that another line Q is parallel to L and passes through point (1,2) find the x and y intercepts of Q. (3 mks)
    4. Find the point of intersection of lines P and Q. (3 mks)
23. The figure below shows a square ABCD point V is vertically above middle of the base ABCD. AB = 10cm and VC = 13cm.
    
    Find;
    1. the length of diagonal AC (2mks)
    2. the height of the pyramid (2mks)
    3. the acute angle between VB and base ABCD. (2mks)
    4. the acute angle between BVA and ABCD. (2mks)
    5. the angle between AVB and DVC. (2mks)
24. The diagram below represents a conical vessel which stands vertically. The which stands vertically,. The vessels contains water to a depth of 30cm. The radius of the surface in the vessel is 21cm. (Take π=²²/₇).
    
    1. Calculate the volume of the water in the vessels in cm³ (3mks)
    2. When a metal sphere is completely submerged in the water, the level of the water in the vessels rises by 6cm.
       Calculate:
       1. The radius of the new water surface in the vessel; (2mks)
       2. The volume of the metal sphere in cm³ (3mks)
       3. The radius of the sphere. (3mks)

MARKING SCHEME

No	Working
1	No Log 0.52 0.312 2.122 T.716 T.4942 T.212 (0.3263) x 2 10 x 10-1 → 3.305 x 10-1 → 0.3305
2	Mass of acid = 200cm3 x 1.08g/cm3 =216g Mass of alcohol = 300cm3 x 0.8g/cm3 =240g Total volume of the mixture = 200 + 300 = 500cm3 Density of the mixture = Total mass                                     Total volume =216g + 240g       500cm3 = 456g      500cm3 = 0.912g/cm3
3
4	Taxable income = Ksh32500 Income              Tax(sh) 9680 x 10/100 = 968 + 9120 x 15/100 = 1368 + 9120 x 20/100 = 1824 + 4580 x 25/100 = 1145 Total tax payable = 5305 Tax due = Tax payable - Relief = sh 5305 - 1056 = sh4249
5	2x = √2w +8            3w-5 (2x)2 = 2w +8              3w - 5 4x2(3w - 5) = 2w + 8 12x2w - 20x2 = 2w + 8 12x2w - 2w = 8 + 20x2 w = 8 + 20 2       12x2 - 2
6a)	y - 5 = 2 x - 2 y - 5 = 2x - 4 y = 2x + 1
6b)	θ = Tan-1(gradient) = Tan-1 2 = 63.43
7	p = a + bQ3 23 = a + b -44 = a + 8b -21 = -76 b = 3 23 = a + 3 → a = 20 P= 20 + 3Q3  When Q = 5 P = 20 + 3(5)3 = 395
8
9	(1 - x)4 = 1(1)4(-x)1 + 4(1)3(-x)1+6(1)2(-x)2 + 4(1)3(-x) + 1(1)4(-x)4 = 1 - 4x + 6x2 - 4x3 + x4 1 - x = 0.98 → x = 0.02 (0.98)4 = 1 - 4(0.02) + 6(0.02)2-4(0.02)2 + (0.02)4 = 1 - 0.08 + 0.0024 - 0.000032 =0.922 ≈0.902
10	x2 + y2 - 6x + 8y - 11 = 0 x2 - 6x + (-6/2)2 + y2 + 8y + (8/2)2 = 11 + (-3)2 + (4)2 (x -3)2 + (y +4)2 = 36 (x -3)2 + (y + 4)2 = 62 Center → (3, -4)  Radius = 6 units
11	Let the ratio be x:y in kg respectively Cost of the mixture = sh(42x + 47g) Total mass of mixture = (x + y)kg Cost per kg of the mixture = Total cost of the mixture                                                     Total mass sh46= 42x + 47y              x + y 46x + 46y = 42x + 47y 4x = y x = 1  → x:y = 1:4 y     4
12	Rate of work of each pipe A = 1/3  B = 1/5 C = 1/15  Rate of work of A and B = 1/3 + 1/5 = 8/15 per hour Work done in 1 hr= 8/15 x 1 = 8/15 of the volume Volume still empty = 1 - 8/15 =7/15  Rate of work of A, B & C = 1/3 + 1/5 - 1/15  =7/15  Time taken to fill =7/15 ÷ 7/15 = 1hr Total time = 1 + 1 = 2hrs
13	Numerator  9t2 - 25a2 = (3t + 5a)(3t - 5a) Denominator 6t2 + 19at + 15a2 6t2 + 10at + 9at + 15a2 2t(3t + 5a)+3a(3t + 5a) (2t + 3a)(3t + 5a) Both combined  (3t + 5a)(2t - 5a)  = (3t - 5a)  (2t + 3a)(3t + 5a)    (2t + 3a)
14	Buying price = 300kg x sh30 per kg = sh9,000 After the loss = 80/100 x 300kg = 240kg 9000 = 100% sp = 120% sp =9000 x 120 = sh10,800             100 sp per kg =10800                    240 =ksh45
15	Log6216 +[Log42 - Log6] ÷ Log49 Log6216 +[Log(42/6)]÷ Log49 Log6216 + Log7 ÷ Log7 Log2216 + Log 7                 2Log7 Log6216 + ½ Log6216 = x   216 = 63 = 6x  x = 3 3 + ½ = 3½
16	x:y = 2:3 x  =  y  = k 2      3 x  = k    x =2k 2 y  = k    y =3k 3 5x - 2y = 5(2k)-2(3k)  x + y         2k + 3k =10k - 6k        5k = 4k     5k 5x - 2y  = 4  = 5x - 2y:x + y = 4:5  x + y       5
17
18
19	a) a2 = b2 + c2 - 2bcCosA a2 = 62 + 82 - 2(6)(8)Cos56º = 36 + 64 - 96(0.6428) a = √100 - 61.71 = √38.29 a = 6.188cm b)  b     =  a      sinB     sinA    6  = 6.188  sinB     sin50 sinB = 6sin50             6.188 B = Sin-1(0.7428) = 47.97º c) a2 = d2+c2 -2dcCosA 2.822 = 72+62 -2(7)(6)CosA 7.9524 = 85 - 84CosA cosA =85 - 7.9524                 84 A =Cos-1(0.9172) =23.48 d)Area = ½dcSinA =½ x 7 x 6 x Sin23.48 = 8.367cm3
20	a) P x Q           R2  P = k Q          R2  18 = k 24    k = 18 x 42            42                 24 k = 12 when Q =30 and R=10 P = 12 Q            R2  = 12 x 3           102 = 3.6 b)P = 12 Q                R2  c) P = k1.2Q    P = k1.48 Q/k2        (0.9R)2 Pnew = 1.48P d) P = 1.48(k Q )                     R2 P = 148 (k Q )       100     R2  P increases b 48%
21	a) 1cm rep. 100m b)Area 1 = ½ x 400 x 240 = 48 000m2 Area 2 = ½ x 240 x(240 + 720) = 115 200m2 Area 3 = ½ x 360 x 720 = 129 600m2 Area 4 = ½ x 120 x 320 = 19 200m2 Area 5 = ½ x 400 x (600 + 320) = 184 000m2 Area 6 = ½ x 280 x(400 + 600) = 140 000m2 Area 7 = ½ x 200 x 400 = 40 000m2  Total Area = 48 000 +115 200 + 129 600 + 19 200 + 184 000 + 140 000 + 40000 =676 000m2  = 67.6ha
22	a) Gradient of L = M =  6 -3        -1-(-2) = 3  = 3    1  y - 3  = 3  x + 2 y - 3 = 3x + 6 y = 3x + 9 b) Gradient of P x 3 = -1 Gradient of P = -1/3 y - 6  = -1  x + 1     3  3y - 18 = -x -1 x + 3y = 17 c) Gradient of Q = 3 y - 2  = 3 x - 1  y - 2 = 3x - 3 y = 3x - 1 d) Solve P & Q simultaneously x + 3y = 17 -3x + y = -1 3x + 9y = 51 -3x + y = -1 10y = 50 y = 5 x + 3(5) = 17 x = 2 Point of intersection (2,5)
23	a) AC =√102 + 102 = √200 = 14.14cm b) √132 - 7.072 = √119.0151 = 10.91cm c) G.θ = 7.07                13 θ = Cos-1( 7.07 )                   13 =57.25º d) Tan x = ( 10.91 )                      5 x = Tan-1 ( 10.91 )                     5 = 65.38 e) Tan xº =    5                     10.91 xº = Tan-1(5/1.91) = 24.62º = 2xº = 49.24
24	a) Volume of water =1/3 x 22/7 x 212 x 30 = 13,860cm3  b i) h  = r        H    R 30  = 21      R=21 x 36 36      R               30 =25.2cm ii) New volume = 1/3 x 22/7 x 212 x 36 = 23950.08cm3 iii) 4/3 x 22/7 x r3 = 10 090.08 r3 =10 090.08 x 21               88 r = √10 090.08 x 21               88 = 13.40cm