Mathematics Paper 2 Questions and Answers - Mathioya Mock Exams 2022

• The paper contains two sections: Section I and 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 calculators and KNEC Mathematical tables may be used, except where stated otherwise.

SECTION 1 ( ANSWER ALL QUESTIONS IN THIS SECTION)   

1. The dimensions of a rectangle are measured and given as 6.9cm by 3.06cm.   Calculate to 4 s.f the percentage error in the area.  (3mks)
2. Make P the subject of the formula     (3mks) 
   
3. Express as a surd and simplify   (3mks)
   1+cos30°
   1−sin60°
4. Use completing square method to solve     (3mks)
   8x² + 6x − 9 =0
5. The diagram below shows a circle centre O.  AP is the tangent to the circle.  Angle OPA is 23°.  Find the length of the tangent.   (4mks)  
   
6. Solve for the value x if  (x+5) − x − 3 = ²/₃       (3mks)
7. A variable N varies as the square of r and inversely as the square root of P.   What is the percentage change in N if r is increased by 15% and P is decreased by 36%   (3 mks)
8. The equation of a circle x² + y² + 6x − 10y – 2=0.   Determine the co-ordinate of the centre of the circle and its radius (3mks)
9. Find the equation of the tangent at point (3,1) to the curve y = x² – 4x +4       (3mks)
10. Solve for  in the equation 6 cos 2θ − sinθ − 4 = 0  in the range 0°≤θ≤180°     (3mks)
11.  
    1. Expand and simplify the expression (1+2x)⁶                           (2mks)
    2. Use the first four terms of the expression in (a) above to find the approximate value of (1.02)⁶  (2mks)
12. A tea dealer mixes two brands of tea, x and y to obtain 35kg of the mixture worth Ksh62 per kg.  If brand X is valued Ksh68 per kg and brand y at Ksh.53 per kg.   Calculate the ratio, in its simplest form, in which the brand x and y are mixed   (2mks)
13. Find the compound interest on Ksh.21,000 in 3 years of at rate of 20% p.a compounded semi-annually (3mks)
14. Two taps A and B, when opened at the same time can fill a tank in 3 hours 36 minutes.  Tap A working alone takes 3 hours longer than tap B to fill the tank.  How many hours does it take tap A alone to fill the tank   (4mks)
15. The points P and Q are 7.4cm apart.   By construction locate the locus R such that PR=2.5cm and angle PRQ=90°   (3mks)
16. Solve the simultaneous equation below     (3mks)
    2x – y = 3
    x² – xy= − 4 

SECTION II ( ANSWER ONLY FIVE QUESTIONS) 

17. An arithmetic progression of 41 terms is such that the sum of the first five terms is 560 and the sum of the last five terms is −250.   Find;
    1. The first term and the common difference (5mks)
    2. The last term  (2mks)
    3. The sum of the progression  (3mks)
18. The table below show the income tax brackets for a certain year 
    

    Income in Ksh p.m	Rate %
    1 - 8,000 8,001   - 16,000 16,001  -24,000 24,001 – 32,000 32,001   - 40,000 40,001 and above	10% 15% 20% 25% 30% 35%

    Mr Bundi earns a monthly salary of Ksh45,000.  He gets a house allowance of Ksh13,000 and a commuter allowance of Ksh6,000.   He is entitled to a family relief of Ksh1166 per month.  Find;
    1. Mr Bundi’s taxable income in Ksh per month (2mks)
    2. Total tax payable per month in Ksh. by Mr. Bundi   (5mks)
    3. Mr Bundi’s net salary per month if the following deductions are also made monthly.
       NHIF of Ksh.1500
       WCPS of 2%  of the basic salary.                                                             (3 mks)
19. PQRSV is a right pyramid on a horizontal square base of side 10cm.  The slant edge are all 8cm long. 
     
    Calculate correct to 2 d.p
    1. The height of the pyramid (2mks)
    2. The angle between
       1. Line VP and the base PQRS      (2mks)
       2. Line VP and line RS      (2mks)
       3. Planes VPQ and the base PQRS     (2mks)
    3. Volume of the pyramid  (2mks)
20. A bag contains 5 red, 4 white and 3 blue beads.  Two beads are selected at random one at a time without replacement.
    1. Draw a tree diagram and list the probability space (3mks)
    2. Find the probability that;
       1. The last bead selected is red (2mks)
       2. The beads selected were of the same colour  (2mks)
       3. At least one of the selected beads is blue       (3mks)
21.  
    1. Complete the table below for the functions y = sin(x+30) and y = 2cos x      (2mks)
       

       x°	0	30	60	90	120	150	180	210	240	270	300	330	360
       Sin(x+30)°	0.50		1.0			0	−0.50			0.87
       2 cos x	2.00			0	−1.0								2.0

    2. On the same axes, draw the graph of y = sin(x+30°) and y = 2 cos x for  0°≤x≤360° (5 mks)
    3. Using the graph;
       1. Solve the equation sin(x+30°) – 2cos x=0     (2 mks)
       2. Find sin(x+30°) when x=145°    (1 mk)
22. The position of two towns are A(30°S,  20°W) and B (30°S, 80°E),
    Find;
    1. The difference in latitudes between the two towns (1mk)
    2. The difference between A and B along the parallel of latitude correct to 2 decimal places in;
       1. Kilometres (take ∏ = ²²/₇ and radius of the earth = 6370km)  (3mks)
       2. in nautical miles (2mks)
    3. Find the local time in town B when it is 1.45p.m in town A .  (4mks)
23. In the figure below (not draw to scale)   AB=8cm. AC=6cm, AD=7cm CD=2.82cm and angle CAB=50°
    
    Calculate  (to 2d.p)
    1. the length BC   (3mks)
    2. the size of angle ABC  (3mks)
    3. Size of angle CAD   (3mks)
    4. Calculate the area of triangle ACD    (2mks) 
24. A particle moves in a straight line. It passes through point O at t=O with velocity V = −4m/s. The acceleration a m/s² of the particle at time t seconds after passing through O is given by a = 10t + 1
    1. Express the velocity V of the particle at time t seconds in terms of t (3mks)
    2. Find V when t=3    (1mk)
    3. Determine the value of t when the particle is momentary at rest    (3mks)
    4. Calculate the distance covered by the particle between t=2 and t=4  (3mks)

MARKING SCHEME

1. Max Area = 6.95 × 3.065 = 21.30175
   Min Area = 6.85 × 3.055 = 20.92675
   Actual Area = 6.9 × 3.06 = 21.114
   21.30175 × 20.92675 × 100%
                   2
                   = 0.8880
2. x² = yp − y²
            p − i
   px² − x²i = yp − y² 
   px² − yp = x²i − y²
   p(x² − y) = x²i − y² 
   p = x²i − y²
         x² − y
3.  
   
   cos 30 = ^√3/₂
   sin 60 = ^√3/₂ 
   
   
   
4.  x² + ¾x − ⁹/₈ = 0
   x² + ¾x = ⁹/₈
   x² + ¾x + ⁹/₆₄ = ⁹/₈ + ⁹/₆₄
   (x + ³/₈)² = ⁹/₈ + ⁹/₆₄
   x + ³/₈ = ± ⁹/₈ 
   x = ⁻³/₈ + ⁹/₈ = ¾
   or
   x = ⁻³/₈ − ⁹/₈ = 1½
5.  
   
   5 × 3 = 2 × x
    x = 7.5
   7.5 + 2 = 4.75
        2
   Radius = 4.75
   AP =  4.75  
            tan 23
        = 11.19cm
6. log₈(x+5) − log₈(x−3) = ²/₃ log₈8
   log ₈  = ²/₃ 
   x + 5 = 8²/₃
   x − 3
   x + 5 = 4
   x − 3
   x + 5 = 4(x−3)
   x + 5 = 4x − 12
   5 + 12 = 4x − x
   17 = 3x
   x = 5²/₃ 
7. N α  r²
           p
   N = kr²
           P
   New r = 115 = 1.15
                 100
   New P =  64   = 0.64
                  100
   N = (1.15 − 1) × 100
         √0.64
   0.6531 × 100= 65.31%
8. x² = 6x + y² − 10y = 2
   Completing square ......
   x² + 6x + (⁶/₂)² + y² − 10y + (⁻¹⁰/₂)² = 2 + (⁶/₂)² + (⁻¹⁰/₂)²
   (x + 3)² + (y − 5)² = 36
   (x − h) + (y − k) = r²
   h = −3
   k = 5
   Centre = (−3, 5)
   Radius = 6
9.  y = x² − 4x + 4
   ^dy/_dx = 2x − 4
   At x = 3
   Gradient = 2(3) − 4
                  = 2
   At (3, 1)
   y−1 = 2
   x−3
   y − 1 = 2(x − 3)
   y = 2x − 5
10. 6(1−Sin²θ) − sinθ − 4 = 0
    6 − 6sin²θ - sinθ − 4 = 0
    − 6sin²θ − sinθ + 2 = 0
    6sin²θ + sinθ − 2 = 0
    Let sinθ be y
    6y² + y − 2 = 0
    6y² + 4y − 3y − 2 = 0
    2y(3y + 2) −1(3y + 2) = 0
    (2y−1)(3y+2) = 0
    y = ½ or ⁻²/₃    
    30°, 150°
11.  
    1. 1 + 6(2x) + 15(2x)² + 20(2x)³ + 15(2x)⁴ + 6(2x)⁵ + (2x)⁶ 
       1 + 12x + 60x² + 160x³ + 240x⁴ + 192x⁵ + 64x⁶ 
    2. (1.02)⁶ = 1 + 0.02)⁶
       1 + 12(0.01) + 60(0.01)² + 160(0.01)³ 
          = 1.12616
12. 68x + 53y  = 62
       x + y
    68x + 53y = 62(x+y)
    68x + 53y = 62x + 62y
    6x = 9y
    ^x/y = ^3/₂ 
    x:y = 3:2
13. A = P(1 + ^r/₁₀₀)ⁿ  
    A = 21000(1 + ¹⁰/₁₀₀)⁶
    A = 21000(1.1)⁶ = 37202.78
    Compound Interest = 37202.78 − 21000
             = Sh. 16202.78
14. Tap B takex x hrs
    Tap A takes (x+3)hrs
    Both (¹⁸/₅) hrs
      1   +  1 = 5 
    x+3     x    18
    x + x + 3 =   5  
      x(x+3)       18
    36x + 54 = 5x² + 15x
    5x² −21x − 54 = 0
    x = 21 ±  √(441 + 1080)
                             10
    = 6 or  −1.8
    x = 6
    A = 6 + 3 = 9hrs
15.   
    
    Length = 7.4cm
    Radius = 2.5 from P
    Two tangents
16. y = 2x − 3
    x² − x(2x − 3) = −4
    x² − 2x² + 3x + 4 = 0
    −x² + 3x + 4 = 0
    x² −3x − 4 = 0
    x² − 4x + x − 4 = 0
    x(x− 4) + 1(x−4) = 0
    (x + 1) or ( x− 4) = 0
    x = − 1 or 4
    y = 2(−1) − 3   or  y  = 2(4) − 3
    = −2 −3                    = 8 − 3
    = −5                         = 5
17.   
    1. Sn = ⁿ/₂(2a + (n−1)d)
       n = 5
       560 = ⁵/₂(2a + 4d)
       ²/₅ × 560 = 2a + 4d
       224 = 2a + 4d
       112 = a + 2d ......(i)
       Cast 5 terms :
       41^st term = a + 40d
       40^th term = a + 39d
       39^th term = a + 38d
       38^th term = a + 37d
       37^th term = a + 36d
                total = 5a + 190d
       5a + 190d = − 250 ......(ii)
       Solve (i) and (ii) simultaneously
       a + 2d = 112
       a + 38d = − 50
       −36d = 162
       d = −4.5
       a + 2(4.5) = 112
       a − 9 = 112
       a = 121
    2. Last term = a + 40d 
                      = 121 + 40(−4.5)
                      = 121 − 180
                     = − 59
    3. Sn = ⁿ/₂ (2a+(n−1)d)
       S₄₁ = ⁴¹/₂(2×21 + (41−1)−4.5)
       S₄₁ = ⁴¹/₂(242 − 180)
       S₄₁ = ⁴¹/₂(62)
       S₄₁ = 41 × 31
       S₄₁ = 1271
18.         
    1. Taxable income = Basic slary + all alowances
       = 45000 + 13000 + 6000
       = 64000
    2. 1st tax band = 8000 x 10/100 = 800
       2nd tax band = 8000 x 15/100 = 1200
       3rd tax band = 8000 x 20/1000 = 1600
       4th tax band = 8000 x 25/100 = 2000
       5th tax band = 8000 x 30/100= 2400
       6th tax band = 24000 x 35/100 = 8400
       Gross tax = 16 400
       Net tax = 16, 400 - 1166
       = 15 234
    3. Net pay = Taxable income - all deductions
       = 64 000 - (15 234 + 1500 + 900)
       = 64 000 - 17 634
       = Ksh 46 366
19.          
    1. √0 = √(82 - 7.071)2
       = √64 - 50
       = 3.74
    2.     
       1. Cos θ = 7.071/8
          = 27.89
       2. Cos θ = 6/8 
          51.32
       3. Tan θ = 3.74/5
          = 36.80
    3. 1/3 x 100 x 3.74
       = 124.7
20.                  
    1.   
       
    2.     
       1. P(RR) or P (WR) or P(BR)
          (5/12 x 4/11) + (4/12 + 5/11) + (3/12 x 5/11)
          5/33 + 5/33 + 5/44
          = 5/12
       2. P(RR) or P(WW) or P(BB)
          (5/12 X 4/11) + (4/12 X 3/11) + (3/12 + 2/11)
          = 5/33 + 1/11 + 1/22 = 19/66
       3. P(RB) or P(WB) or P(BR) or P(BW) or P(BB)
          (5/12 X 3/11) + (4/12 X 3/11) + (3/12 X 5/11) + (3/12 X 4/11) + (3/12 X 2/11)
          5/44 + 1/11 + 5/44 + 1/11 + 1/22
          = 5/11
21.   
    
22.     
    
    1.    
       
       20 + 80 = 100
    2.        
       1. 100/360 X 2 X 22/7 X 6370 X Cos 30
          = 9632.13 km
       2. 100 X 60 X Cos 30
          = 5196.15 nm
       3. Time diffreence = 100 x 4 = 6 2/3
                                          60
          = 6 hrs 40 minutes
          time at - B = 13.45 + 6hrs 40 mins
          = 2025 hrs
          = 8.25 pm
23.           
    1. a² = b² + c² - 2bc Cos A
       a² = 6² + 8² - 2 x 6 x 8 x Cos 50
       a² = 36 + 64 - 96 (0.64.28)
       a =√38.29
       a = 6.187891402
       a = 6.19
    2.      b        =       a      
         SinB            Sin A
       Sin B =   6Sin 50  
                      6.19
       B= 47.95
    3. a2 = d2 + c2 - 2dc CosA
       2.822 = 72 + 62 - 2 x 7 x 6 x Cos A
       7.9524 = 85 - 84 Cos A
       - 77.0476 = -84Cos A
       Cos A = 0.91723
       A° = 23.48°
    4. Area = 1/2 dc Sin A
       = 1/2 x 7 x 6 Sin 23.48
       = 8.37cm²
24.        
    1. v = ∫(10t + 1)dt
       V = 5t² + t + c
       When t= 0 v = -4 hence -4 = 5 (02) + o + c
       c= -4
       V = 5t² + t - 4
       
       
    2. At  t = 3
       V = 5(3)2 + 3 - 4
       = 44 m/s
    3. At rest V=0
       5t² + t - 4 = 0
       5t² + 5t - 4t - 4 = 0
       5t (t + 1) - 4 (t + 1)= 0
       (5t - 4)(t + 1)= 0
       t = 4/5 or t = -1 (ignore)
       t = 4/5 or 0.8 seconds
    4. S = ⁴₂∫(5t² + t - 4)
       [5t³/3 + t²/2 - 4t]42
       [5(4)³/3 + (4)²/2 - 4(4)] - [5(2)³/3 + (2)²/2 - 4(2)]
       = 98.67 - 7.33
       = 91.34m