Mathematics Paper 2 Questions and Answers - Momaliche Joint Pre Mock Exams 2022

QUESTIONS
SECTION I

1. Solve the quadratic equation by completing the square method.
   x² – 7x + 10 = 0 (3mks)
2. Simplify √5- √3 by rationalizing the denominator. (2mks)
                √5+ √3
3. Grade x rice costs shs 75 per kg and grade y rice costs shs 50 per kg. The two grades are mixed in the ratio such that the blend costs sh 70 per kg. Find the ratio in which grade x rice was mixed with grade y rice. (3mks)
4. Given that A = ∜d-c²g make c the subject of the formula. (3mks) 
                            b+c²f
5. Solve the equation 2 sin (3x + 60) = 1 for 0º ≤ x ≤ 180º (3mks)
6.      
   1. Expand and simplify (2 – x)⁸ up to 4th term. (2mks)
   2. Use the simplified expression in (a) above to estimate the value of (1.99)⁸ giving your answer correct to 4 significant figures. (2mks)
7. Solve the equation 2 + log₃ (2x -7) = log₃(5x - 7) . (3mks) 
8.      
   1. Construct a triangle ABC in which AB is 2.2cm, BC is 3.5cm and angle ABC is 60º(2mks)
   2. A point D moves such that it is on the same side of BC as A. Construct the locus of D such that the area of ΔABC = area of ΔBCD. (2mks)
9. The equation of a circle is x² + y² – 6x + 8y – 11 = 0. Find the coordinates of the centre of the circle and its radius. (3mks)
10. The length of a rectangle is 8.3 cm and its width is 5.45cm. Calculate
    1. The relative error in area of the rectangle (3mks)
    2. The percentage error in area (1mk)
11. A triangle xyz whose area is 15.3cm² mapped onto Δx¹y¹z¹ by a transformation whose matrix is m = . Find the area of Δx¹y¹z¹ .(3mks)
12. Solve the triangle (3mks)
    
13. Fifteen men working at the rate of 9 hours per day take 20 days to complete a job. Find the number of days 27 men would take to complete the same job working at the rate of 10 hours a day. (3mks)
14. The 5th term of an arithmetic sequences is 71 and the 7th term is 59. Find
    1. The first term and the common difference. (2mks)
    2. The 10th term (2mks)
15. Use matrix methods to solve the simultaneous equations. (3mks)
    3x – 4y = 2
    6x + y = 13
16. Factorise: 2x² – x – 10 (2mks)

SECTION (II) (50MARKS)

17. The following table shows the distribution of marks obtained by 50 students of a certain school.
    

    Marks	45-49	50-54	55 - 59	60- 64	65 - 69	70-74	75- 79
    No. of students	3	9	13	15	5	4	1

    1. State the modal class (1mk)
    2. By using an assumed mean of 62, calculate
       1. The mean (5mks)
       2. The standard deviation (4mks)
18. The figure below is a triangle OAB, where OA = a and OB = b. A point R divides AB in the ratio 2:5 and a point T divides OB in the ratio 1:3. OR and AT intersect at D.
    
    
    1. Find in terms of a and b
       1. BT (1mk)
       2. OR (2mks)
       3. AT (1mk)
    2. Given that AD = KAT and RD = hRO where k and h are scalars. Find the values of k and h, hence express AD in terms of a and b (5mks)
19. John travels to work by either boda-boda or by tuk-tuk. If he travel by tuk-tuk on every one day, there is a probability of 0.75 that he travels by a boda boda the following day. If he travels by a boda boda on any one day, then he travels by tuk-tuk the following day with a likelihood of ⁵/₆ . There is a chance of ²/₃ that he travels by tuk-tuk on Tuesday.
    1. Draw a tree diaagram to illustrate the possible outcomes in 3 days. (2mks)
    2. Find the probability that he travel by;
       1. Boda-boda on Wednesday (2mks)
       2. Tuk-tuk on Wednesday (2mks)
       3. Boda – boda on Thursday (3mks)
       4. Tuk-tuk on Thursday (1mk)
20.      
    1. Fill the table below for the curves given by y = 3sin(2x + 30º) and y = cos2x for x value in the range 0º ≤ x ≤ 180º (2mks)
       

       x

       0º

       15º

       30º

       45º

       60º

       75º

       90º

       105º

       120º

       135º

       150º

       165º

       180º

       Y = 3 sin (2x + 30)

       1.50

       2.60

       -2.60

       1.50

       Y =cos 2x

       1.00

       0-0.87

       -0.50

       1.00

    2. Draw the graph of y = 3 sin (2x + 30º) Y = 3sin (2x + 30º) and y = cos 2x on the same axes. (4mks)
       x-axis 1cm rep 15º
       y-axis 1cm rep 0.5 units
    3. Use your graph to solve the equation 3sin (2x +30º) = cos 2x (2mks)
    4. Determine the following from your graph
       
       1. Amplitude of y=3sin(2x+30º) (1mk)
       2. The period of y=3sin(2x+30º) (1mk)
21. An arithmetic progression has the first term as a and the common difference as d.
    1. Write down in terms of a and d the 3rd, 9th and 25th term of the progression (1mk)
    2. The progression is increasing, and the 3rd, 9th and 25th terms form the first three consecutive terms of a geometric series. If the sum of the 7th and twice the 6th term of arithmetic progression is 78, calculate
       1. The first term and the common difference of the AP. (6mks)
       2. The sum of the first nine terms of the AP. (3mks)
22. The figure below is a right rectangular based pyramid VABCD where AB =5cm,
    BC =7cm and VC =VB=VA=VD=13cm and O is a point on the base of the pyramid vertically below V.
    Calculate
    1. The length of AC (2mks)
    2. VO the height of the pyramid. (2mks)
    3. The angle between the edge VB and the plane ABCD (3mks)
    4. The angle between the planes VBC and ABCD (3mks)
23. Three quantities L, M and N are such that L varies directly as M and inversely as the square root of N.
    1. Given that L = 2250 when M = 450 and N = 64, write down an equation connecting L, M and N. (4mks)
    2. If M decreased by 16% and N increased by 44%, calculate the percentage change in L. (3mks)
    3. In soccer competition, the number of goals (G) scored in a penalty shoot-out is partly constant and partly varies as the skill (S) of the player. Given that G = 8 when S = 2 and G = 12 when S =4, find the value of G when S = 6. (3mks)
24. The table below shows income tax rates
    

    Monthly taxable pay (k£)	Rate of tax ksh per £
    1- 435 436 – 870 871 – 1305 1306 – 1740 Excess over 1740	2 3 4 5 6

    A company employee earns a monthly basic salary of Ksh 28,000. He is also entitled to the following monthly allowances: house allowance of Ksh 9000, a medical allowance of sh 2000 and a commuter allowance of shs 1480.
    1. Calculate his total income tax. (5mks)
    2. He is entitled to a personal tax relief of Ksh 1056 per month. Determine the net tax. (1mk)
    3. If he received a 50% increase in his total income, calculate the corresponding percentage increase on the income tax. (4mks)

MARKING SCHEME

1. x² - 7x + 10 = 0
   x²-7x = -10 
   x² - 7x + ⁴⁹/₄ = -10 + ⁴⁹/₄
   (x - ⁷/₂)² = ⁹/₄
   x - ⁷/₂ = ± ³/₂
   X = ⁷/₂ ± ³/₂
   Either x = ⁷/₂ + ³/₂ = ¹⁰/₂ = 5
   or
   x = ⁷/₂ - ³/₂ = ⁴/₂ = 2
2. 75x + 50y = 70
       x + y
   75x + 50y = 70x +70y
   5x = 20y
   ^x/_y = ²⁰/₅ = 4
   x:y = 4:1
3. √5 - √3 x √5 - √3 
   √5 + √3   √5 - √3 
   (√5 - √3)( √5 - √3)
   5 - √15 + √15 - 3
   5 - √15 - √15 + 3 = 8 - 2√15
              5 - 3                2
   = 4 - √15
4. A = ⁴√d - c²g
             b + c²f
   A⁴ = d - c²g
           b + c²f
   A⁴b + A⁴c²f = d - c²g
   A⁴c²f + c²g = d - A⁴ b
   c²(A⁴ + g) = d - A⁴ b
   c² = d - A⁴ b 
            A⁴+ g
   c = ± √d - A⁴ b 
               A⁴+ g
5. 2sm(3x+60 = 1
   (3x+60) = 0.5
   3x+60 = 30
   3x = -30
   x = -10
   3x + 60 = 180 -30 = 150
   3x = 150 - 60
   X = 30
   3x+60 = 360 +30 = 390
   3x = 390 - 60 = 330
   x = 110º
   3x + 60 =510
   3x = 450
   x = 150º
6.        
   1. (2 - x)⁸ = 2⁸ - 2⁷x+ 2⁶x² - 2⁵x³
      = 256 - 1024x + 1792x² - 1792x²
   2. (1.99)⁸ = (2-0.01)⁸
      = 256 - 1024 (0.01)+ 1792(0.01)² - 1792(0.01)³
      = 256- 10.24 +0:1792- 0.001792
      = 245.937408
7. 2 + log₃(2x-7) = log₃(5x - 7)
   2log₃³ + log₃(2x - 7) =log₃(5x - 7)
   log₃9 + log₃(2x - 7) = log₃(5x - 7)
   log₃(9(2x - 7)) = log₃(5x - 7)
   9(2x - 7) = 5x - 7
   18x - 63 = 5x - 7
   13x = 56
   x = 4⁴/₃
8. 
   
   Area of ΔABC = ½ x 2.2 x 3.5sin60
   3.334
   ΔBCD = ½ x 3.5 x h = 3.334
   h = 1.9cm
   locus of Δ
9. x² + y² -6x + 84 - 11 = 0
   x² - 6x + y² + 8y = 11
   x² 4 - 6x + 9 + y² + 8y + 16 = 11 +9+16
   (x - 3)² + (y + 4)² = 36 = 6²
   (x - a)² + (y - b)² = r²
   Centre (3, -4)
   Radius - 6 units.
10. 8.3 ± 0.05cm by 5:45 ± 0.005 cm
    1. Max. Area = 8.35 x 5.455
       = 45.549250 cm² 
       Min Area = 8.25 x 5.445
       = 44.92125 cm²
       Actual area = 8.3 x 5.45 = 45.235 cm²
       AE = 45.54925 - 44.92125
                             2
       = 0.314
       R.E = 0.314
                45.235
       =0.006941527578
    2. Percentage error = RE x 100
       = 0.006941527578 x 100
       = 0.6941527578%
       ≈ 0.6942%(4sf) 
11. Area of Δxyz = 15.3 cm²
    Area scale factor = Determinant of matrix
    ASF = 3 x 2 - (-1x - 2)
    = 6 - 2 = 4
    Area of ΔX'Y'Z = 4 x 15.3
    = 61.2cm²
12. 
     9.4  =  16:6  
    SinB  Sin110
    SinB = 9.4Sin110 = 0.5321
                    16.6
    β = 32.1°
    θ = 180-32.1 =147.9°
    x² = 9.4²+10.8² - 2 x 9.4 x 10.8 cos110°
    = 88.36 + 116.64 - 203.04 Cos110
    = 88:36+ 116.64 + 203.04 x 0.3420
    = 274.43968
    x = 16.566cm(3d.p)
13. M   H   D
    15  9   20
    27  10  ?
    No. of days = 15 x 20 x 9
                             27 x 10º
    = 10 days
14. Tn = a + (n - 1)d
    1. T5 = a +(5-1) d = 71 (i)
       T7 = a +(7-1) d = 59 (ii)
       a + 4d = 71
       a + 6d = 59
       -2d = 12
       d = -6
       a + 4x - 6 = 71
       a = 95 .
    2. T10 = a5 + (10-1) x -6
       = 95 - 54
       = 41
15. 3x - 4y = 2
    6x + y = 13
    
16. 2x² - x - 10
    2x² - 5x + 4x - 10
    = x(2x - 5) + 2(2x - 5)
    =(2x - 5)(x + 2)
17.      
    1. Modal class 60-64
    2.        
       

       Class	x	f	d = x - 62	d2	fd	fd2
       40-49	47	3	-45	225	-45	675
       50-54	52	9	-10	100	-90	900
       55-59	57	13	-5	25	-65	325
       60-64	62	15	0	0	0	0
       65-69	67	5	5	25	25	125
       70-74	72	4	10	100	40	400
       75-79	77	1	15	225	15	225
       		48≈50			∑fd = -120	∑fd2= 2650

       1. Mean
          x = A + ∑fd
                      ∑f
          = 62 + -120
                      50
          = 62 - 2.4 = 59.6
       2. Standard deviation 
          √2550 - (-120)²
              50       50 
          = √47.24
          = 6.873
18.      
    1.          
       1. BT = ¾BO = -¾b
       2. OR =  5  OA +  2  OB
                  2+5        2+5
          = ⁵/₇a + ²/₇b
       3. AT = AO + OT
          = -a + ¼b/¼b - a
    2. AB = K(¼b-a) = ¼kb - ka
       AD = AR + RD
       = ²/₇AB + h(-⁵/₇a - ²/₇b)
       = ²/₇(b - a) + (-⁵/₇ha - ²/₇hb)
       = -²/₇a - ⁵/₇ha + ²/₇b -²/₇hb
       = (-²/₇ - ⁵/₇h)a + (²/₇ - ²/₇h)b
       k = -²/₇ - ⁵/₇h → k =²/₇ + ⁵/₇h
       ¼k = ²/₇ - ²/₇h
       ¼(²/₇ + ⁵/₇h) = ²/₇ - ²/₇h
       h = ⁶/₁₃
       k = ⁸/₁₃
       AD = ⁸/₁₃ (b-a) = ²/₁₃b - ⁸/₁₃a
19.      
    1.     
       
    2.      
       1. P(TB or BB)
          =(²/₃ x ¾) + (¹/₃ x ¹/₆)
          =½ + ¹/₁₈ = 9 +1 = ¹⁰/₁₈
                              18
       2. P(TT or BT)
          =(²/₃ x ¼) + (¹/₃ x ⁵/₆)
          = ²/₁₂ + ⁵/₁₈ = 6 +10 = ¹⁶/₃₆
                                  36
       3. P(TTB or TBB or BTB or BBB)
          =(²/₃ x ¼ x ¾) + (²/₃ + ¼ x ¹/₆) + (¹/₃ x ⁵/₆ x ¾) + (¹/₃ x ¹/₆ x ¹/₆ )
          = ⁶/₄₈ + ⁶/₇₂ + ¹⁵/₇₂ + ¹/₁₀₀
          = ¹⁸⁴/₄₃₂
       4. 1 - ²³/₅₄ = ³¹/₅₄
20.      
    1. 
       

       x	0º	15º	30º	45º	60º	75º	90º	105º	120º	135º	150º	165º	180º
       Y = 3 sin (2x + 30º)		2.60	3.00		1.50	0	-1.50	-2.60	-3.00		-1.50	0
       Y =cos 2x		0.87	0.50	0	-0.50		-1.00	-0.87		0	0.50	0.87

    2.     
       
    3. x = 84º or 174º 
    4.    
       1. Amplitude 3
       2. Period 180º
21.        
    1. T3 = a + 2d
       T25 = a + 24d
    2.      
       1. a +2d, a +8d, a + 24d
          a + 8d = a + 24d
          a + 2d    a + 8d
          (a + 8d)(a + 8d) = (a + 24d)(a + 2d)
          a² + 16ad + 64d² = a2 + 26ad + 48d²
          64d2 = 48d² = 26ad - 46ad
          16d² = 10ad
          10a = 16d
          a = 1.6d
          a + 6d + 2(a + 5d) = 78
          3a + 16d = 78
          4.8d + 16d = 78
          20.8d = 78
          d = 3.75 
          a = 6
       2. S9 = ⁹/₂ [2 x 6 + (8 - 1)¹⁵/₄]
          = ⁹/₂[12 + 30]
          = 189
22.            
    1. AC = √5² + 72
       = √74
       = 8.602cm
    2. VO = √13² - 4.301²
       = √150.501
       = 12.27
    3. BD = AC = 8.602
       BO = ½BD = 4.301
       ∠VBO = θ
       
       Cosθ = 4.301
                     13
       θ = Cos^-10.3308
       = 70.68º
    4. Midpoint of BC
       VM = √MO² + OV²
       = √2.5² + 12.27
       = √156.8029
       = 12.52
       
       Cos x =  2.5  
                   12.52
       x = 78.48
23.      
    1. L = aM
             √N
       2250 = 450a
                   √64
       a = 2250 x 8 = 40
                450
       L = 40M
             √N
    2. M1 = 0.84M , N1 = 1.44N
       L1 = 0.84aM = 0.84am
               √1.44N     1.2√N
       = 0.7L
       %change = (0.7 - 1)L x 100%
                               L
       = -0.3 x 100% = -30%
    3. G = a+bs
       8 = a + 2b (i)
       12 = a + 4b (ii)
       -4 = 2b 
       b = 2
       a = 8 - 4 = 4
       G = 4 + 2 x 6
       = 4 + 12
       = 16
24.           
    1. Total income tax 
       Taxable income = 28000 + 900 + 2000 + 1480
                                                       20
       = 2024
       Tax 
       435 x 2 = 870
       435 x 3 = 1305
       435 x 4 = 1740
       435 x 5 = 2175
       Rem  284 x 6 = 1704 
       Total income tax = 7794
    2. Net tax = 7794 - 1056
       = 6738
    3. Increase in income 
       = 150 x 2024 = 3036
          100
       Total income tax = 870 + 1305 + 1740 + 2175 + 1296 x 6
       = 6090 + 7776
        = 13866
       % increase in income tax = 13866 - 7794 x 100
                                                          7794
       = 77.906%