Mathematics Paper 2 Questions and Answers - Moi Kabarak High School Mock 2020/2021

Instructions to candidates

• The paper contains two setions:Section I and Section II
• Answer ALL the questions in Section I and strictly any five questions from Section II
• Show all the steps in your calculations, giving your answers at each stage.
• 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 unless stated otherwise.

SECTION I (50 marks)

1. Use logarithms    (4 marks)
   
2. Make x the subject of the formula  (3 marks)
   
3. Without using tables or calculators evaluate  (3 marks)
          3        
   2−3 sin 60°
4. Expand (x+¹/_2x)⁴ Hence give the constant term  (3 marks)
5. Y varies partly as square at x and partly inversely as square root of x. When x=4,, y=13 and when x=9, y=79. Find the relationship between the variables  (3 marks)
6. The second, fourth and the sixteenth term of an increasing arithmetic progression are consecutive terms of a geometric progression, If the first term of the A.P is −3, find the common ratio of the G.P. (3 marks)
7. Solve for θ given that:    (3 marks)
   2 cos ½θ =⁻⁴/₅ for −360° ≤ θ ≤ 360° 
8. Find the centre and the radius of a circle whose equation is (3 marks)
   ½x² − x + ½y² + y − ⁷/₂ =0
9. Complete the given table, for 0° ≤ x ≤ 180°  (1 mark)

   1. x	0	30	45	90	135	150	180
      y=sin 2x	0		1	0	−1		0

   2. Draw the graph of y=sin2x, hence  find the value of Sin 2x=0.4 (3 marks)
10. Use the matrix method to solve the following pair of simultaneous equations  (3 marks)
    2n+3m=12
    4m−2n=5
11. Mr. Omondi paid a tax of Kshs.2512 in the month of July. In that month he received a house allowance of Kshs. 5000 and a tax relief of shs. 1056. Use the table below to find his monthly basic salary in shilings.  (3 marks)
    

    Income in $ per month	Rate %
    1-484 485-940 941-1396 1997-1852 Excess over 1852	10 15 20 25 30

12. The marked price of a laptop is Kshs. 60,000. A customer buys it at shs. 73500 on hire purchase terms. If the deposit paid is Kshs. 7500 and an interest of 22½% p.a is charged on the balance. Calculate the number of months it takes to clear the balance. (3 marks)
13. A man uses 100m of fencing to enclose a rectangular area using a wall on one side. Find the maximum possible area that can be enclosed. (3 marks)
14. A triangular prism has cross section ADE as shown. BC//DE, BC=4cm and DE-6cm. If the prism has a length of 10cm and density of 1.32g/cm³. Calculate the mass of the prism of cross-section ABC. Given that the area of trapezium BCDE=36cm².  (3 marks)
    
15. Solve for t          (3 marks)
    (log₁₀t)²=3 − 2log₁₀t      
16. Find the percentage error, a+b given that a=3.0, b=4.24 and c=2.  (3 marks)
                                             c

SECTION II

17.                                →
    In the triangle below OA =a and OB=b. M is mid-point of Ab and N is a point on Ob such that ON=¹/₃OB. AN and OM intersect at P
             
    
    1. Express the following vectors in terms of a and b
       
       1. AB  (1 marks)
       2. OM   (1 marks)
       3. AN   (2 marks)
    2. If OP=tOM and AP=sAN, express OP in two different ways hence find the values of t and s. (5 marks)
    3. State the ratio AN:NP  (1 marks) 
18. The table below shows height in cm of form 4 students in Westlands Secondary School.
    

    Height(cm)	150-154	155-159	160-164	165-169	170-174
    Frequency(f)	18	16	24	28	14

    
    
    1. Using an assumed mean of 162, calculate;
       1. mean height    (3 marks)
       2. Standard deviation    (3 marks)
    2. if the heights of each student increases by 5cm. Write down
       1. The new mean    (1 marks)
       2. The new standard deviation  (1 marks)
    3. Calculate the 30^th percentile (2 marks)
19. Two identical bags A and B contained identical balls. Bag A has 5 yellow balls and 2 red balls. Bag B has 4 yellow balls and 3 red balls. Two balls are drawn from each bag without replacement.
    1. Draw a tree diagram to illustrate the above information  (2 marks)
    2. Find the probability that;
       1. Both balls are yellow  (2 marks)
       2. The first ball is red and the second yellow   (2 marks)
       3. A red and yellow balls are picked  (2 marks)
       4. At least one ball is red   (2 marks)
20. Using a ruler and a compass only;
    1. Draw triangle PQR in which PQ=7.5cm QR=5cm and angle PQR=60°  (2 marks)
    2. By shading the unwanted regions, find the position of S where S and Q are on opposite sides of PR such that:
       1. Angle PSR ≤ 60°  (2 marks)
       2. PS ≥ SR   (2 marks)
       3. PS ≥ 5cm     (2 marks)
       4. Area of triangle PSQ ≤ 7.5cm²    (2 marks)
21. Use the graph to answer the question below
             
    
    1. Using trapezium rule with 7 ordinates calculate the shaded area. (3 marks)
    2. Using mid ordinate rule with 6 strips, estimate the area of the shaded region.   (2 marks)
    3.  
       1. Calculate the actual area of the shaded region    (3 marks)
       2. Find the percentage error in using mid ordinate rule   (2 marks)
22. The positions of towns X and Y are given to the nearest degree as X(40°N, 124°E) and Y(40°N,56°W)
    1. Longitude difference   (1 marks)
    2. Find the distance in nautical miles between x and y along a parallel of a latitude.    (3 marks)
    3. Find the distance XY via the north pole in nautical mile  (3 marks)
    4. A plane flying at 200 knots leaves y at 2.00p.m. on Monday what time and day does it arrive at x via great circle.  (3 marks)
23. A manufacturer wishes to mix two brands of drink so that the ingredient per litre of the mixture is at least 18 units of iron, 14 units of calcium and 20 units of alcohol. The ingredient per litre of the brand one is 4 units of iron, 2 units of calcium and 2 units of alcohol. The ingredient per litre of brand two is 2 units of iron, 2 units of calcium and 4 units of alcohol. One litre of brand one costs shs.10 and one litre of brand two costs shs.14.
    1. Form inequalities representing the above information   (4 marks)
    2. Represent the inequalities on the grid provided    (3 marks)
    3. From the graph, determine the minimum cost per litre of the new brand of drink obtained by mixing brand one with brand two.  (3 marks)
24. The variables y and x are connected by the equation y=axⁿ
    

    x	1	2	3	4	5	6
    y	4	22.6	62.4	128	224	352

    
    
    1. Write the equation in the form of y=mx+c  (1 mark)
    2. Draw a straight line graph on the grid provided ad estimate the value of a and n (7 marks)
    3. Find the value of y when x=2.1   (2 marks)

Marking scheme

1.  
   
             =0.14096
2. 4x² = x² − yp
               2+y
   8x² + 4x²y = x² −yp
   x²(8 + 4y −1) =   −yp   
        8 + 4y −1     8+4y−1
   
   
3.    3        ×   2+3√3
   2−3√3               2 
           2        2+3√3
                            2
   6+9√3       6+9√3
           2                2
   4 − 27           −¹¹/₄   
          4   
   −6−9√3
             2  
        ¹¹/₄  
4. (x + ¹/2x)⁴ = x⁴ + 4(x³)¹/_2x + 6(x²)(¹/_2x)² +4x(¹/_2x)³ + (¹/_2x)⁴   
                    = x⁴ + 2x² + 1.5 + ¹/_2x² + ¹/_16x⁴ 
              Constant term 1.5 
5. y=mx² + c 
                √x
   13=16m + ^c/₂
   79 =81m + ^c/₃ 
   26 =32m + c
   237= 243m+c
   −211=−211m
         m=1
   26=32+c
     c=−6
     y= x² −  6
                 √x 
   
6. a+d, a+3d,a+15d
   −3 +15d  = −3 + 3d
   −3 + 3d        −3+d
   9+
    9−45d −3d + 15d² = 9 − 18d + 9d²
     −30d = −6d²
           d = 5
    −3 + 75 =  72
     −3  +15    12
          r=6
7. 2 cos ½θ = ⁻⁴/₅
           ½θ = 113.58, −113.58
         θ =227.16  or −227.16
8. x² −2x + y² + 2y −7 =0
   (x −1)² + (y+1)² =7+1+1
   (x −1)² + (y+1)² =9
   Centre(1,−1)
   radius 3 units
9.  

   1. x	0	30	45	90	135	150	180
      y=sin 2x	0	0.866	1	0	−1	−0.866	0

   2. 
      
      13.5° or 75°
10.  
    
    
11. Gross tax = (2512 + 1056)
                               20
                   =178.4
       484 × ¹⁵/₁₀₀ = 48.4
       456 × ¹⁵/₁₀₀ =68.40
        0.2x =61.6
             x=308
      (484 + 456+ 308) × 20
         24,960 − 5,000
          =ksh. 19,960
12. A=73500 −7500 = Ksh.66,000
    P=60000−7500=Ksh.52,500
    66,000−52,500(1+^20.5/₁₀₀)ⁿ
     1.2571 =  (1.225)ⁿ
             n=log 1.2571
                  log 1.225
            n=1.13yrs
          ∴ n= 13 months
13. Let length be x
    w=100-2x
             2
    A=x(50 − x)
    dA =50−2x
    dX
           50−2x =0
                x=25
    max=25 × 25
          =625m²
14. L.S.F =²/₃
    A.S.F =⁴/₉
      s=36cm²
     4×36 =28.8cm²
        5
    Vol =28.8 ×10 =288cm³
      1cm³ =1.32
        288 × 1.32
         =380.16g
15. let log₁₀t =x
    x² =3 − 2x
    x² + 2x − 3 = 0
    (x+3) (x−1)=0
    x=1 or −3
    log₁₀t = 1 or −3
      10¹ = t
      10⁻³= t
      t=10 or 0.00
16. a+b =3.0+4.24
      c          2.0
    Max =3.05+4.245
                 1.95
    Min = 2.95 + 4.235
                  2.05
    Err=(3.741 − 3.505) × 100
                 3.625
       =6.5%
17.  
    1.  
       1.  AB b − a
       2. OM ½a + ½b
       3. AN ¹/₃b − a
    2.  →
       OP =t(½a + ½b)
       →
       OP = a + s(¹/₃b − a)
       ½t = 1 − s ....(i)
       ½t = ¹/₃s ....(ii)
       ½t = 1 − ³/₂t
       ⁴/₂t = 1
        t ½=
         ¼ = ¹/₃s
           s=¾
    3. AN:NP
          4:−1
18.   
    1.  
       1. Mean height
          

          x	d	fd	fd2	c.f
          152 157 162 167 172	−10 −5   0    5   10	−180 −80  0 140 140	1800 400 0 700 1400	18 34 58 86 100

          x̄ =Σfd + A
                Σf
             =²⁰/₁₀₀ + 162 =162.2
       2. Standard deviation
          
              √42.96 = 6.554
    2.  
       1. New mean
          162.2 + 5
           =167.2
       2. new standard deviation
           =6.554
    3. 30^th percentile
       154.5 + ¹²/₁₆ × 5
       154.5 + 3.75
             =158.25
19.  
    1. 
       
    2.  
       1. Both balls are yellow
          P(AYY) or P(BYY)
          (½ × ⁵/₇ × ⁴/₆) = (½ × ⁴/₇ × ³/₆)
          ⁵/₁₂ + ¹/₇ = ⁸/₂₁
       2. First ball is red, second ball yellow
          P(ARY) or P(BRY)
          (½ × ²/₇ × ⁵/₆) + (½ × ³/₇ × ⁴/₆)
          ⁵/₄₂ + ¹/₇ = ¹¹/₄₂
       3. P(AYR) or P(ARY) or P(BYR) or P(BRY)
          (½ × ⁵/₇ × ²/₆) + (½ × ²/₇ × ⁵/₆) + (½ × ⁴/₇ × ³/₆) + (½ × ³/₇ × ⁴/₆)
          ⁵/₄₂ + ⁵/₄₂ + ¹/₇ + ¹/₇ = ¹¹/₂₁
       4. At least one ball is red
          P(AYR) or P(ARY) or P(BYR) or P(BRY) or P(ARR) or P(BRR)
          (½ × ⁵/₇ × ²/₆) + (½ × ²/₇ × ⁵/₆) + (½ × ⁴/₇ × ³/₆) + (½ × ³/₇ × ⁴/₆) + (½ × ²/₇ × ¹/₆)+(½ × ³/₇ × ²/₆)
          ⁵/₄₂ + ⁵/₄₂ + ¹/₇ + ¹/₇ + ¹/₄₂ + ¹/₁₄ =¹³/₂₁
20.  
    1. 
       
    2.  
       4. Area of triangle PSQ ≤ 7.5cm²
          A= ½ × 7.5 × 2
          =7.5cm²
21.  
    1. 
       

       y0	y1	y2	y3	y4	y5	y6
       −4	−3.75	−3	−1.75	0	2.25	5

       ½ × ½ ( 4+5) + 2(3.75 +3+1.75+0+2.25)
       ¼ × 30.5
       =7.625 units²
    2. 
       

       	y1	y2	y3	y4	y5	y6
       x	0.25	0.15	1.25	1.75	2.25	2.75

       
       A=½ (3.9375+3.4375+2.4375+0.9375+(0.0625+3.5625)
           ½ × 15.375 
            =7.6875 sq units
    3.  
       1. Actual area of shaded region
          
          (⁸/₃−8) + ((9−12) − (⁸/₃−8))
          −¹⁶/₃+ ⁷/₃ 
          = 7¹/₃ sq. units
       2. Percentage error
          0.0208 × 100
             ²²/₃
          =0.2836%
22.  
    1. Longitude difference
       124 + 56 =180°
    2. 180 × 60cos40
          =8,237.28nm
    3. Distance XY
       100 × 60=6000nm
    4. t= 6000
            200
       =30hrs
       time difference  180×4 =12hrs
                                  60
       Time at X =2.00a.m
23.  
    1. let brand one be x and brand two be Y
       4x +2y≥18, 2x+2y≥14, 2x+4y≥20, x70, y70
    2. 
       
    3. (10 × 2) + (14×5)
       =Kshs. 90    
24.  
    

    x	1    0	2  0.301	3  0.477	4  0.602	5 0.699	6 0.778
    y	4 0.602	22.6  1.354	62.4 1.795	128  2.107	224   2.35	352 2.547

    
    
    1. log y =log a + n log x    
    2.