Mathematics Questions and Answers - Form 2 Term 3 Opener Exams 2023

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INSTRUCTIONS TO CANDIDATES
  • This paper consists of two sections: Section I and Section II.
  • Answer all questions in section I and Section 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.
  • KNEC Mathematical tables may be used.

SECTION I (50 Marks)

Answer all the questions in the spaces provided

  1. Find the possible values of x in the equation.                                                   (3 marks)
    9x = 27(2x+12)
  2. Express F2MathT3O2023Q2 in the form a/b where a and b are constants hence find the difference between a and b.           (3 marks)   
  3. A trader buys 200 items at a total cost of shs 6000 .She sells 150 of them at a profit of 25% and the remainder at a loss of 8% .
    1. Calculate the amount of net profit she made                                                  (2marks)
    2. Express the net profit as a percentage of the initial cost price of all items     (2 marks)
  4. The width of a room is 8m less than length. Find the measurement of the room if its Perimeter is 48m. Hence calculate the area of the room.                 (3 marks)
  5. Use tables to evaluate √1.875+2.4232 leaving your answer correct to 2 decimal places.                                (2 marks)
  6. Given that sinθ=0.8 where  is an acute angle ,find without using mathematical tables or calculator 
    1. Cos  θ                                               (2marks)
    2. tan⁡(90−θ)                                (1mark)
    3. tan θ                              (1mark)
  7. The currency exchange rates at a given bank in Kenya are as follows.
     Currency  Buying   Selling 
     1 Sterling Pound  135.50   135.97 
     1 US dollar  72.23  72.65
    A tourist arrived in Kenya with 5000 US dollars which he converted to Kenya shillings upon arrival. He spent Ksh.214, 500 and converted the remaining to Sterling pounds. How many pounds did he receive?                               (3marks)
  8. Use reciprocal tables and cube tables to evaluate                            (3 marks)
    5.23     4      
                 0.7052
  9. Find the surface area of the solid below                                                                  (3 marks)
    F2MathT3O2023Q9
  10. Without using a calculator or mathematical tables simplify. (3 marks)
    36 − 8 × − 4 −15  ÷ − 3
       3× − 3 + − 8(6 −( −2)
  11. An isosceles triangle has congruent sides of 20 cm. the base is 10 cm. determine the area of the triangle.       (3 marks)
  12. Two similar containers have capacities of 1000 litres and 1728 litres respectively .Given that the smaller container has an area of 750 cm3.Calculate the surface area of the larger container.                       ( 3marks)
  13. The circle whose length is 2.2m subtends an angle of 60° at the centre. Calculate the area of the minor segment of the circle.
    (Take  = 22/7)                  (3 marks) 
  14. Find the coordinates of the point of intersection of the lines 2x − 3y + 5 = 0 and   2y − x = −3.               (3 marks)
  15. Given that  (x+60)° = cos⁡(2x)°find tan (x+60)°                                   (3marks)
  16. A piece of land is to be divided into 20 acres or 24 acres or 28 acres for farming and Leave 7acres for grazing. Determine the smallest size of such land.                                (3 marks)

SECTION II

  1. A triangle A(1,1), B(2, 4) and C(4, 2) undergoes:
    1. Reflection in the line y =° x. Obtain the coordinates of its image A’B’C’ after the above transformation and show both the object and the image on a Cartesian plane.                                 (3 marks)
    2. On the same axis, draw A’’B’’C’’, the image of A’B’C’ after a negative quarter turn about the origin and state the coordinates of A’’, B’’ and C’’.                                           (2 marks)
    3. Describe the transformation that maps triangle A’’B’’C’’ onto triangle ABC.        (2 marks)
    4. Draw the line x = −3 and draw triangleA’’’B’’’C’’’, the image of triangle A’’B’’C’’ under a reflection in the linex = − 3. State the coordinates of triangleA’’’B’’’C’’’.      (3 marks)
  2. A line l1 passes through the point −2,3 and  −1, 6
    1. Find the equation of l1 in the form y = mx+c                                               (3marks)
    2. Find the equation of l2 which is perpendicular to line l1 at −1, 6 in the form  ax + by − c = 0 Where a,b and c are constants            (3marks)
    3. Given that another line l3 is parallel to l1 and passes through point (1 ,2)
      1. Find the equation of  l3 in the form y = mx + c                                  (2marks)
      2. Find the x-intercept of line l3                                                        (2marks)
  3. The figure below shows two circles of radii 10cm and 8cm and centres O1 and O2 respectively. The common chord AB=12cm  If O1O2 is a perpendicular bisector of AB use (π=3.142)
    F2MathT3O2023Q19
    1. Calculate correct to 2 decimal places
      1. angle AO1B                                                                                             (1mark)
      2. Angle AO2B                                                                                               (1 mark)
      3. Area of triangle AO1B                                                                         (2marks)
      4. Area of triangle AO2B                                                                                (2marks)
    2. The area of the shaded part  to 4 significant figures                                        (4 marks)
  4.  
    1. The price of an article was raised by 25% and a month later the new price was lowered by 20%. What was the new price if the original price was Ksh. 750?                     (3 marks)
    2. Water flows from a tap at the rate of 27cm3/second into a rectangular container of length 60 cm, breadth 30 cm and height 40 cm. If at 5.00 p.m. the container was half full, what will be the height of water at 5.08 p.m.?                          (4 marks)
    3. Akinyi sold a mobile phone costing Ksh. 4000 at a profit. She earned a commission of 10% on the profit. Find the amount earned if she made a profit of 40%.                    (3 marks)
  5. A salesman is paid a salary of Sh. 10,000 per month .He is also paid a commission of sales above Sh 100000 .In one month he sold goods worth Sh. 500,000 .If his total earnings that month was shs 46,000
    1. Calculate the rate of commission                                                                  (4marks)
    2. If the rate of commission in (a) above was reduced to 7.65 % and his monthly salary increased by sh. 1000 , find his total new earning that month for selling the same amount of goods .                                                                                                            (4marks)
    3. The marked price of TV set in shop is sh.40000 Mueni bought the TV at 10% discount calculate the amount of money she paid for the TV.                    (2marks)     

MARKING SCHEME

 NO  WORKING   MARKS 
 1. 32x = 36x+36
2x = 6x + 36
2x − 6x = 36
−4x = 36
x = −9
 M1

 M1

 A1
 2. r        = 12.42727 ........ (i)
10r    = 124.27272 ....(ii)
100r   = 1242.72727 ..... (iii)
1000r = 12427.27272 ...... (iv)
Subtract (ii) from (iv)
1000r = 12427.27272
    10r = 124.27272 − 
   990r = 12 303
r = 12 303
         990
a = 12 303, b = 990 
a − b = 12 303 − 990 = 11 313
 



 M1




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 3.
  1.  25  × 4500 =
    100 
    paid = sh 1125
      8  × 1500 = shs 120
    100
    M.P = 1125 − 120 = sh 1005
  2. 1005 × 100 =
    6000
    =163/4%
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 4. let width = x
length =x +8
p = 2(x+8+x) = 48
4x = 48 − 16
x = 8cm
length =16cm width = 8cm
Area = 16×8 = 128cm2


 M1

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 A1
 5. √1.875 = 1.3693
2.4232 = 5.871
⇒ 1.3693 + 5.871 = 7.2403
≅ 7.24 (2 d.p)
 M1


 A1
 6.
  1. x2 = 52 − 42
    x = 3
    cos θ = 3/5  
  2. Tan (90 − θ) = 3/4
  3. Tan θ = 4/3
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 A1
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 7. Ksh.5000 x 72.23 = Ksh.361,150
Remainder after expenditure
361,150 = 214,500 = Ksh.146,650
No. of £ = 146650 × 1
                    135.97
= £1078.55

 M1

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 A1

 8.
5.23 = 140.61
     4     = 4 × 10 × 0.1418 = 5.672
0.7052
⇒ 140.61 + 5.672 = 146.282


 M1

 M1A1

 9.
1/2 × 7 × 24 × 2 = 128cm2
25 × 4 = 100cm2
7 × 4 = 28cm2
24 × 4 = 96cm2
total surface area = 168 + 100 + 28 + 96
= 392cm2
 M1

 M1


 A1
10.
 N
36 − 8× − 4 −15 ÷ −3
36 − 8 × − 4 + 5
36 + 32 + 5 = 73
D
3 × −3 + −8(6−(−2)
−9 − 64 = −73
 N/D =  73   = −1
          −73



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 A1

 11. s = 0.5(10 + 20 + 20) = 25
A = √(25(25−10)(25−20)(25−20))
   = √(25×15×5×5)
   = 96.82 cm2 (4 s.f)
 M1
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 A1
 12.
V.S.F = 1728 = 216
             1000    125
L.S.F= ³√(216/125)
= 6/5
A.S.F= (6/5)2 = 36/25
Area of larger container = 36/25 × 750 = 1080 

 M1



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 A1

 13 L = /360 × 2r
2.2 =  60  x 2 x 22 x r
         360           7
r = 2.2 x 360 x 7
           60 x 44
r = 2.1
 60  × 22 × 2.12  − 1/2 × (2.1×2.1× sin 60)
360     7
 2.23 − 1.9095 = 0.4004

 M1




 M1

 A1
 14.
2x − 3y = −5 .... (i)
2y − x = −3 ........(ii)
From (ii), x = 2y + 3
2(2y+3) − 3y = −5 
4y + 6 − 3y = −5 
y = −11
x = 2 (−11) + 3 = −19
Coordinates are (−19, −11)

 M1

 A1 both values of x & y

 

 

 B1

 15. x + 60 + 2x = 90
3x = 30
x = 10°
Tan 70 = 2.747
 M1

 B1
 A1
 16.
 2  20   24   28 
 2  10   12   14 
 2  5   6   7 
 3  5   3   7 
 5  5  1  7
 7  1  1  7
   1  1  1
23 × 3 × 5 × 7 = 840
840 + 7 = 847




M1 for the table

 

 

 

 

 B1
 A1

 17.
  1. A'(−1, −1), B'(−4, −2) and C'(−2,−4)
  2. A''(−1, 1), B''(−2, 4) and C''( −4,2)
  3. Reflection on the y axis (x=0)
  4. A'''(−5, 1), B'''(−4, 4) and C'''( −2,2)
    F2MathT3O2023Ans17
B1
B1 reflection on y=-x
B1 ∆A'B'C'
 
B1 rotation
B1 ∆A''B''C''
B1 B1
 
B1
B1 reflection on x=-3
B1 ∆A'''B'''C'''
 18.
  1. (−2 ,3) (−1 ,6)
      6 − 3    = 3
    −1−(−2)
    3 = y−3 
           x+2
    y − 3 = 3x + 6
    y = 3x + 9
  2. m2 = −1/3
    1/3 = y − 6
                x+1
    − x − 1 = 3y − 18
    − x −3y + 17 = 0
    x + 3y − 17 = 0
  3.  
    1. 3 = y−2 
            x−1
      y − 2 = 3x − 3
      y = 3x − 1
    2. y = 3x − 1
      3x − 1 = 0
      3x = 1
      x = 1/3
  M1



 M1

 A1

 B1
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 19.
  1.  
    1. sin-1 0.6 = 36.87
      angle AO1B = 2×36.87 = 73.74°
    2. sin-1(6/8) = 48.59
      Angle AO2B = 2 × 97.18°
    3. 1/2 × 10 × 10 × sin73.74 
          =48.00cm2
    4. 12 × 8 × 8 × sin97.18
           =31.75cm2
  2. (73.14 × 22 ×10×10) − [1/2 × 10 × 10 sin73.74]
       360      7
    64.376 − 48.00 = 16.376cm2
    (97.18 × 22 × 8 × 8)  − [1/2 × 8 × 8 sin97.18]
       360      7
    52.297 − 31.749 = 22.548cm2
    shaded region = 16.376 + 22.548
                            = 38.92cm2


 B1 observe the condition 

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 20.
  1. Original price ⇒ 750=100%
    After increase  125  × 750 = 937.5 shillings
                           1000
    After reduction ⇒ 80% = x
    Price after decrease ⇒ 937.5×80 = 750 shillings
                                              100
  2. Rate =   Volume   
                time taken
    Volume of water that flows in 8 mins = 27 × 8 × 60
                                                               = 12 960 cm3
    Volume of water as at 5.00 p.m = 0.5×40×60×30
                                                       =36 000cm3
    Volume as at 5.08 p.m
    . = 36 000 + 12 960 = 48 960 cm3
    Height of water = 48 960 = 27.2 cm
                                 60×30
  3. Selling price = 140  × 4000 = Sh. 5600
                            100
    Profit = 5600 − 4000 = Ksh. 1600
    Commission earned =  10  × 1600 = Sh. 160
                                        100


 M1

 M1A1


 M1
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 A1
 

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 21
  1. commission = 46000 − 10000
                         = ksh36000
    x/100 × 400,000 = 36000
    4000x = 3
             x = 9%
  2. 7.65 × 400,000=shs.30600
    100
    new salary = 1000 + 10000 = sh.11000
    Total earning = 30600 + 11000 = sh.41600
  3.  90 × 40,000 = sh.36000
    100

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