Mathematics Questions and Answers - Form 2 Mid Term 2 2022

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INSTRUCTIONS TO CANDIDATES

  1. Write your name and index number in the spaces provided at the top of this page.
  2. This paper consists of two sections: Section l and Section II
  3. Answer all questions in section l and any five questions from Section II.
  4. Show all the steps in your calculations, giving your answers at each stage in the spaces below each question.
  5. Marks may be given for correct working even if the answer is wrong.
  6. Non- programmable silent electronic calculators and KNEC Mathematical tables may be used.


QUESTIONS

SECTION I(30 MARKS)
Answer ALL Questions from this section in the spaces provided

  1. Use logarithm tables ONLY, evaluate to 4 significant figures  1 auygdad(4mks)
  2. The sum of three consecutive odd integers is 219. Determine the first three such integers (3 mks)
  3. A Kenyan company received US Dollars 100,000.The money was converted into Kenya shillings in a bank which buys and sells foreign currencies as follows:
                                   Buying                              Selling
                         (in Kenya shillings)          (in Kenya shillings)
    1 US Dollar            77.24                                77.44
    1 Sterling Pound   121.93                              122.27
    1. Calculate the amount of money, in Kenya shillings, the company received. (2 mks)
    2. The company exchanged the Kenya shillings calculated in (a) above, into sterling pounds to buy a car from Britain. Calculate the cost of the car to the nearest sterling pound. (2 mks)
  4. A solid cone of height 12cm and radius 9 cm is recast into a solid sphere. Calculate the surface area of the sphere. ( 4 marks)

  5. The G.C.D of two numbers is 12 and their L.C.M is 240. If one of the numbers is 60, find the other number. (2mks)
  6. Solve for x in the equation: 6x2-13x+6=0 (3mks)
  7. Simplify without using tables 4 cos 450 sin 600 (2mks)
  8. Use mathematical tables only to evaluate 11.45 sin 38.3 (3mks)
  9. If the area of a regular nonagon is 185.1cm2. What is the length of each side? (3mks)
    SECTION II (30 MARKS)
    Answer ANY THREE Questions from this section in the spaces provided
  10. A helicopter flies from Kaptiony due south for 300km. It then flies on a bearing of 2550 for 350km. From there it flies on a bearing for 0400 for 400km.
    1. Draw an accurate diagram showing the journey of the helicopter using a scale of 1:5000000. (5mks)
    2. From your diagram, find the distance and bearing of Kaptiony from the final position of the .helicopter. (2mks)
    3. Given that the helicopter flies at a steady speed of 200kmh-1, find how long the whole journey took. (3mks)
  11. The figure below shows two intersecting circles with centres P and Q and radius 5cm for the small one and 6cm for the big one. AB is a common chord of length 8cm. Calculate;
    11 ddadada
    1. the length of PQ (1 mark)
    2. the size of;
      1. angle APB (2marks)
      2. angle AQB (2 marks)
    3. the area of the shaded region(5 marks)
  12. Below are the measurements of a wheat field using a baseline XY recorded in metres.
     

    Y

     
     

    240

     

    TO R 60

    190

     
     

    180

    75 TO Q

     

    150

    50 TO P

    TO S 100

    120

     
     

    100

    100 TO N

    TO T 30

    50

     
     

    20

    20 TO M

     

    X

     
    1. Using a scale of 1cm represents 20m. Sketch the map of the wheat field. (4mks)
    2. Find the area of the field in hectares. (4mks)
    3. If the cost of one hectare is sh65, 000 find the cost of the wheat field. (2mks)
  13. The figure below shows a glass in form of a frustum of a cone whose top and bottom diameter of 7cm and 3.5cm respectively. Its depth is 10cm. Taking π=22/7,
    13 sdsdsds
    Calculate;
    1. Its total surface area. (5 marks)
    2. Its capacity. (5 marks)


MARKING SCHEME

SECTION I(30 MARKS)
Answer ALL Questions from this section in the spaces provided

  1. Use logarithm tables ONLY, evaluate to 4 significant figures  1 auygdad(4mks)
  2. The sum of three consecutive odd integers is 219. Determine the first three such integers (3 mks)
    • let the numbers be a, b and c respectively such that:

      b =a + 2
      c = a + 2 + 2 = a + 4
      a + (a+2) + (a+4) = 219
      a + a + 2 + a + 4 = 219
      3a + 6 = 219
      3a = 213
      a = 71
      b = 73 
      c = 75


  3. A Kenyan company received US Dollars 100,000.The money was converted into Kenya shillings in a bank which buys and sells foreign currencies as follows:
                                   Buying                              Selling
                         (in Kenya shillings)          (in Kenya shillings)
    1 US Dollar            77.24                                77.44
    1 Sterling Pound   121.93                              122.27
    1. Calculate the amount of money, in Kenya shillings, the company received. (2 mks)
      • 7, 724, 000 (student should show working)

    2. The company exchanged the Kenya shillings calculated in (a) above, into sterling pounds to buy a car from Britain. Calculate the cost of the car to the nearest sterling pound. (2 mks)
      • 63, 171.669
        to the nearest pound:
        £ 63, 172
        (student should show working. 
        NOte: € - is for Euro)
  4. A solid cone of height 12cm and radius 9 cm is recast into a solid sphere. Calculate the surface area of the sphere. ( 4 marks)
    • S.A of cone = S.A of the sphere
      S.A of cone = 2πr2 + πdh 
      = (2 x 22/7 x 9 x 9) + (22/7 x 18 x 12)
      =3564/7 + 4572/7
      =8136/7 = 1,162.2857cm2
  5. The G.C.D of two numbers is 12 and their L.C.M is 240. If one of the numbers is 60, find the other number. (2mks)
    • G.C.D = 12
      let number be x
      60   x
      30  x/2
      3 15   x/4
      5 x/12 
      x/12  1
      2 x 2 x 3 = 12 (The GCD),
      hence x is not divisible by 5.
      x/12 
        1 1
        2 x 2 x 3 x 5 x x/12 = 240   
      12 x 5 x x/12 = 240
      5x = 240 
      x= 240/5
      x= 48
       
  6. Solve for x in the equation: 6x-13x + 6 = 0 (3mks)
    • 6x2 - 4x - 9x + 6 = 0
      2x(3x - 2) -3(3x-2)= 0
      (2x - 3)(3x -2)= 0
      2x - 3= 0 hence 2x = 3, hence x = 2/3
      3x - 2= 0 hence 3x = 2, hence x = 3/2


  7. Simplify without using tables 4 cos 450 sin 600 (2mks)
    • 4 Cos (360 + 90) Sin (360 + 240)
    • 4 Cos 90 Sin (240 - 180  - 3rd quadrant hence value is negative
    • 4 Cos 90 - (Sin 60)
      4 (0)- (0.86603)
      = 0
  8. Use mathematical tables only to evaluate 11.45 sin 38.3 (3mks)
    • 11.45 x 0.62251
      7.1277395
  9. If the area of a regular nonagon is 185.1cm2. What is the length of each side? (3mks)
    • interior angle = 140
      It forms iscosceles traing;les with angele of 70, 70 and 40.
      S is the length of the equal sides ( lines from the edges of the nanogon to the center) ,
      hence total area:
      area = 9  x 1/2 x s2 x Sinθ
      9 x 1/2 x s2 x sin 40 = 185.1
      s2   185.1 x 2     
                9 x sin 40
            370.1          
      9 x 0.64279
      = 370.1/ 5.78511
      s2= 63.9745 approx = 64
      s2 = 64
      s = 8cm 
      Each of the 9 triangles can be split in th middle by a perpendicular line, to give a triangle with angles of 20, 70 and 90.
      to find length of the lower side:
      x = 8 Sin 20
      x =2.73616
      2x =  5.47232 cm
      Length of side = 5.47232 cm



    • SECTION II (30 MARKS)
      Answer ANY THREE Questions from this section in the spaces provided
  10. A helicopter flies from Kaptiony due south for 300km. It then flies on a bearing of 2550 for 350km. From there it flies on a bearing for 0400 for 400km.
    1. Draw an accurate diagram showing the journey of the helicopter using a scale of 1:5000000. (5mks)
    2. From your diagram, find the distance and bearing of Kaptiony from the final position of the .helicopter. (2mks)
    3. Given that the helicopter flies at a steady speed of 200kmh-1, find how long the whole journey took. (3mks)
  11. The figure below shows two intersecting circles with centres P and Q and radius 5cm for the small one and 6cm for the big one. AB is a common chord of length 8cm. Calculate;
    11 ddadada
    1. the length of PQ (1 mark)
      7.472cm
    2. the size of;
      1. angle APB (2marks)
        • Sin θ = 4/5 
          Sin -1 0.8 = θ = 53.13 x 2
          = 106.26  ± 1
      2. angle AQB (2 marks)
        • Sin θ = 4/6
          Sin -1 2/3 = θ = 41.8
          41.8 x 2 = 83.6 ± 1
    3. the area of the shaded region(5 marks)
      •   (83.6/360 x 22/7 x 6 x 6) -  (2 x 1/2 x 4.472 x 4) = 26.27428 - 17.888 = 8.38628
      • (106.26/360 x 22/7 x 5 x 5) - (2 x 1/2 x 3 x 4)
        23.1916666667 - 12 = 11.1916666667
        8.38628 + 11.1916666667 = 
        19.5779466667
        (allow for erros of margin/ rounding off etc)
  12. Below are the measurements of a wheat field using a baseline XY recorded in metres.
     

    Y

     
     

    240

     

    TO R 60

    190

     
     

    180

    75 TO Q

     

    150

    50 TO P

    TO S 100

    120

     
     

    100

    100 TO N

    TO T 30

    50

     
     

    20

    20 TO M

     

    X

     
    1. Using a scale of 1cm represents 20m. Sketch the map of the wheat field. (4mks)
    2. Find the area of the field in hectares. (4mks)
    3. If the cost of one hectare is sh65, 000 find the cost of the wheat field. (2mks)
  13. The figure below shows a glass in form of a frustum of a cone whose top and bottom diameter of 7cm and 3.5cm respectively. Its depth is 10cm. Taking π=22/7,
    13 sdsdsds
    Calculate;
    1. Its total surface area. (5 marks)
      glass has open upper side, therefore:
      TSA = area of bottom circle + area of middle part
      bottom circle = 22/7 x 1.75 x 1.75= 9.625 cm2
      middle part lengths:
      22/7 x 3.5 = 11
      22/7 x 7 = 22
      Area = (2 x 1/2 x 5.5 x 10) + (11 x 10) 
      = 55 + 110 = 165
      165 + 9.625 = 174.625cm2
    2. Its capacity. (5 marks)
      Vol = 1/3 x 22/7 x 10 ( r12 + r22 + (r1 x r2))
      Vol = 1/3 x 22/7 x 10 ( 3.52 + 72 + (3.5 x 7))
      Vol = 1/3 x 22/7 x 10 ( 12.25 + 49 + 24.5)
      Vol = 1/3 x 22/7 x 10 ( 85.75)
      Vol = 1/3 x 22/7 x 857.5
      898.333333333 cm3

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