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

Share via Whatsapp
INSTRUCTIONS TO CANDIDATES
  • This paper consists of TWO sections.  Section 1 and section 11.
  • Answer all the questions in section 1 and only five from section 11.
  • Show all the steps in your calculations, giving your answers at each stage in the spaces provided below each question.
  • Marks may be given for correct working even if the answer is wrong.
  • Non-programmable silent electronic calculators and mathematical tables may be used.

SECTION I

Answer all the questions in this section.

  1. Use logarithms to evaluate      (3mks)
     1.46 x 183  
    0.97 x 0.041                                                                 
  2. Find the integral values of x which satisfy the inequalities. (3mks)
    2(2−x) < 4x – 9  ≤ x + 11
  3. Make v the subject of the formula   (3mks)
    4N =   M   + N     
            p+1/V                                                 
  4. Solve the equation
    9x + 3 = 243 (2mks)
  5. Mutua bought 8 pairs of trousers and six shirts at Sh. 4140. Had he bought twice as many shirts and half as many trousers, he would have saved Sh. 180. Find the cost of each item. (3 marks)
  6. Determine the equation of the perpendicular bisector of a line joining the points
    P(−2,−4) and Q(4,8) (3mks)
  7. The volume of a cube is 9.261cm3. Find
    1. the length of the cube                                   (1mk)
    2. the perimeter of one of it’s faces                                                                          (1mrk
    3. the total surface area of the cube                                                                              (2mks)
  8. Simplify leaving your answer in surd form.     (3mks)
    1+3√2
    1 − √3
  9. A play station can be purchased on cash at sh 30,000.  It can also be purchased on hire purchase terms at a deposit of sh 10,000  followed by twelve equal monthly installments of sh. 3200 per month.  Determine the percentage rate of interest per month.    (3mks)
  10. The floor dimensions of a rectangular tank were measured as 3.4cm by 2.7m.  calculate the percentage error in calculating the area of the floor of the tank.        (3mks)
  11. The mean mass of 18 form 4 boys is 64 kg. When their class teacher is added the mean becomes 65 kgs. What is the mass of the class teacher in kilograms. (3mks)
  12. The sum of four consecutive odd numbers is 88. Find the numbers. (3mks)
  13. Solve the equation    (3mks)
    4 cos(2x – 60) = 2 for the range 0° ≤x ≤300°     
  14. Given that log  y = 3.143 and log x = 2.425,evaluate
    4 log y3 – 3 log ∛x (3mks)
  15. Given OP = (4 5 ), OQ =(3 8 ), and OR =(7 −4 )
    1. Express in column form p – 2q – 3r (1mk)
    2. Express PQ and QR as column vectors, hence write QR in terms of PQ. (2mks)
  16. Triangle ABC is right angled at C.  BC = 8cm and AC = 6cm.  A perpendicular from C cuts AB at D.  Find the length CD. (4mks)

SECTION II

Answer only 5  questions from this section.

  1. Plot trapezium ABCD with vertices A(2,0), B(4,0),C(3,2) and D(2,2) (1mk)
    1. Find the area of trapezium ABCD. (1 mk
    2. ABCD is reflected in the x – axis to give the image Al BlCl Dl.  Plot the image and write down its co-ordinates.  (2mks)
    3. The image of AlBlClDunder a negative quarter turn about the origin is AIIIBIIICIIIDIII.  Plot this image and write down its co-ordinates. (2mks)
    4. Plot AIIIBIIICIIIDIII the image of  AIIBIICIIDII under an enlargement scale factor -½ and centre (O,O) (1mk)
    5. Describe fully a transformation that will map AIIBIICIIDII onto ABCD. (2mks)
  2. In triangle PQR,  PQ=20m,QR=30m  and angle PQR=70°.
    1. Calculate
      1. the area of triangle PQR. (2mks)
      2. the length PR (2mks)
      3. the angle QPR (2mks)
    2. T is the top of a 10m pole standing vertically on point  Q. Find
      1. length TP (1mk) 
      2. length TR (1mk)
      3. area of triangle TPR (1mk)
  3. A container in the shape of a frustrum of a cone has an internal base diameter of 18cm and internal top diameter of 28cm.  the container is open at the top and has a depth of 30cm as shown below.
    MathF32023ET2OP2Q19
    1. Calculate
      1. the slant height of the container (3mks)
      2. the capacity of the container in litres (3mks)
    2. The container is half-filled with water.  Calculate the surface area of the container that is in contact with water. (4mks)
  4. In triangle OAB, O(O,O) ,A(2,8) and B(10,2).  C is the midpoint of OA and E is the midpoint of AB
    1. Find
      1. the co-ordinates of C and E (3mks)
      2. the magnitude of CE and OB (3mks)
    2. Express the vector CE in terms of OB (1mk)
    3. Given that OM=OE + 2CE, find
      1. the co-ordinates of M (2mks)
      2. the co-ordinates of the image of M under a translation (−8 −2 )      (1mk)
  5. A married man earns a basic salary of sh. 10800 per month and is housed by his employer for a nominal rent of sh. 500 per month.  His taxable income is hence deemed to be 115% of his basic earning less the rent paid.  He is entitled to a family relief of sh. 1160 per month and a life insurance relief at a rate of 10% of insurance premium of sh. 600 p.m.
    1. calculate his taxable income in  £p.a. (1mk)
    2. calculate relief on his insurance in sh p.a (1mk)
    3. Using the income tax table below
      1. calculate the gross amount of tax he pays in sh p.a (4mks)
          Income in £ p.a    Rate in sh per £                              
         1- 2100  2  
         2101 -4200  3  
         4201- 6300  4  
         6301- 8400  5  
         Over 8400  6  
      2. Calculate his net tax in sh. per month                                                                             (2mks)
    4. Apart from tax,he is also deducted sh 280 for NHIF. Calculate his net earnings in sh per            month.                                                                                                                        (2mks)
  6. Marks scored by  students in a class were recorded as shown below.
    35   30   45   64   62   52   37   43   55   49
    60   39   54   37   42   59   62   48   57   32
    46   64   31   44   56   35   47   38   53   61
    46   59   45   51   49   47   40   45   36   46
    1. Complete the frequency table below with uniform classes as shown in the first class.
       Marks  30-34             
       No. of students              
    2. determine the modal class. (1mk)
    3. calculate
      1. the mean mark. (4mks)
      2. the percentage of students who scored between 45 and 54 marks.    (2mks)
      3. The simplified fraction of students who passed the exam if the pass mark was 45.  (1mk)
  7. The effort E required to lift a load L is partly constant and partly varies as L.  Use the table below for values of E and L to answer the questions that follow.
     L  2    4   6   8   10   12 
     E  5.5  6   6.5   7   7.5   8 
    1. Write an expression for E in terms of L. (1mk)
    2. Draw a graph of E against L (3mks)
    3. From the graph, determine
      1. gradient of the graph (2mks)
      2. the E-intercept of the graph. (1mk)
    4. Write the equation connecting E and L. (1mk)
    5. Find from the graph
      1. the effort  when the load is 7. (1mk)
      2. the load that can be lifted by an effort of 6,25. (1mk)
  8. In the figure below, O is the centre of the circle.  Chord AB of the circle is 8cm long.  CM is perpendicular to AB and CM=AB.
    MathF32023ET2OP2Q24
    Calculate
    1. the radius of the circle (3mks)
    2. the obtuse angle AOB (2mks)
    3. the shaded area (3mks)
    4. the unshaded area (2mks)

MARKING SCHEME

 1  MathF32023ET2OP2Ans1
103 × 6.719
= 6719

 All logs ✓M1

✓Addition and Subtraction M1

A1 (Allow 6718.9)

 2  4 − 2x < 4x − 9
 13 < 6x
 2.1667 < x
 4x − 9 ≤ x + 11
 5x < 20
 x ≤ 4
 x = 3, 4

 M1 separating inequalities

M1 solving each

A1

 3   4N − N =      M    
                  P + 1/y 
 3N (P + 1/y) = M
 3NP + 3N/Y = M
 3N/Y = M − 3NP
   Y =  3N 
          M − 3NP
 Follow through

 M1

 M1

 A1
 4    32x + 2 = 35
   2x + 2 = 5
         2x = 3
          x = 1.5
 M1

 A1 Allow  1½

 5   8t + 6s = 4140 .........(i)
  4t + 12s =3960.........(ii)
 multiply (ii) by 2
 8t + 24s = 7920
  8t + 6s = 4140   
       18s = 3780
           s = 210
 8t + 6(210) = 4140
             8t   = 2880
                 t = 360
 
 B1 for both



 M1 for eliminating one unknown


A1 for both

 6  Grad of L1 = 8 + 4 = 2
                      4 + 2
Grad of ⊥ = −½
Mid point of L1 ( −2+4−4+8)
                               2         2
                 =(1, 2)
 y − 2 = −½
 x − 1
2y − 4 = −x + 1
2y + x = 5
 

 B1



 M1


 A1
 7  a) 3√9.261 = 2.1cm
 b) 2.1 × 4 = 8.4cm
 c) Area of 6 faces 
     6 × 2.1 × 2.1 = 26.46cm2
 B1
 B1

 M1A1
 8  1+3√2 × 1 + √3
1 − √3     1 + √3
Num
 (1+3√2)  (1 + √3)
 = 1 + √3 + 3√2 + 3√6
 Den
 (1 − √3) ( 1 + √3)
 =  1 + √3 − √3 − 3 = −2
  1 + √3 + 3√2 + 3√6
           −2
  =  −1 − √3 − 3√2 − 3√6
                      2
 


 M1


 M1



 A1

 9  10000 + 12 × 3200 = 30000(1 + r/100)12
                     48400 = 30000 ( 1+ 0.01r)12
                    1.6133 = ( 1 + 0.01r)12
                 1 + 0.01r =  12√1.6133
                 1 + 0.01r = 1.0407
                       0.01r = 0.0407
                              r = 4.07%
 
 M1


 M1

 A1
 10  L.L   2.65, 3.35
 U.L  2.75, 3.45
Actual area = 2.7 × 3.4
                   = 9.18
 Min area =  2.65 × 3.35
                = 8.8775
 Max area = 3.45 × 2.75
                 = 9.4875
 Absolute error = ½(9.4875 − 8.8775)
                        = 0.305
 % error = 0.305 × 100
                  9.18
            = 3.322%
 





 M1 max & min


 M1


 A1
 11  Total mass of boys
  18 × 64 = 1152
 Boys + teacher = 1152 + x
   65  × 19 = 1235
  x + 1152 = 1235
            x = 1235 − 1152
               = 83
  
 M1


 M1

 A1
 12  Let the numbers be x, x+2, x+4, x+6
 x + x+ 2 + x + 4 + x + 6 = 88
           4x + 12 = 88
                   4x = 76
                    x = 19
 19, 21, 23, 25
 
 M1


 A1
 B1
 13  Cos(2x − 60) = 0.5
 Cos−1(2x − 60) = 60°
 2x − 60 = 60, 300, 420, 660
     x = 60, 180°, 240°, 360°
 
 B1
 B1 (for 60°, 180°)
 B1 (240°, 360°)
 14  4 log y3 − 3 log 3√x
12 log y − log x
 12 × 3.143 − 2.421
    = 35.295
 
 M1
 M1
 A1
 15  a)
 MathF32023ET2OP2Ans15a
b)
MathF32023ET2OP2Ans15b
 QR = −4PQ
 or PQ =  −¼QR
 



  B1



 B1 both or for PQ and PR



 B1
 16

 MathF32023ET2OP2Ans16AB = √(62 + 82) = 10

CD2 = 62 − x
       = 36 − x2
CD2 = 82 − (0 − x)2 
        = 64 − 100 + 20x − x2
36x2 − 64 − 100 + 20x − x2 
   36 = −36 + 20x
   72 =   20x
     x = 3.6cm
 CD = √(62 − 3.62
      = 4.8cm

 B1 both or for PQ and PR 

 B1


 



 M1 both eqns

 M1 equating and simplifying
 

 A1

 17  MathF32023ET2OP2Ans17  
 18




























 MathF32023ET2OP2Ans18
  1.   
    1. A = ½ × 20 × 30 sin 70 = 281.9m
    2. PR2 = 202 + 302 − 2(20× 30cos 70°)
      400 + 900 − 410.4
            = 889.6
      PR = 29.83m
    3. 29.83  =   30  
      Sin70°    Sinθ
      Sin θ = 30 Sin 70 = 0.9450
                      29.83
          θ   = 70.92°
  2.  
    1. TP = √(202 + 102)
            = 22.63m
    2. TR =  √(102 + 302)
           = 31.62m
    3. ½(22.36 + 31.62 + 29.83) = 41.91
      A = √(41.91(41.91−22.36)(41.91−31.62)(41.91−29.83))
        = √(41.91 × 19.55  × 10.29  × 12.08)
       = 319.1m2
 







 
 M1  A1

 M1


 A1

 M1

 A1



 B1

 B1

 M1

 A1
 19

 

  1.  
    MathF32023ET2OP2Ans19 
    1. 18/28 =     h    
                   h+30
       28h = 18h + 540
           h = 54cm
      L + L = √(842 + 142)
               = 85.16
      18/28   1      ∴ L = 54.75
                  85.16
      L = 85.16 − 54.75 = 30.41cm
    2. V = 1/3 × 142 × 22/7 × 84 = 17248.0
      V = 1/3 × 92 × 22/7 × 54 =      4582.3     
                                               12665.7cm3 
      Capacity = 12665.7 = 12.6657litres
                          1000
  2.   
    MathF32023ET2OP2Ans19b
    9/r = 54/69
        r = 11.5cm
    Base area = 22/7 × 92 = 254.6
    C.S.A 
    A = 22/7 × 11.5 × (54.75 + 15.205)
    A = 22/7 × 9 × 54.75 = 1548.6
                                          979.8
    T.S.A = 254.6 + 979.8 = 1234.4cm2



 







 M1




 M1


 A1
 M1 (both eqns)

 M1 (subtraction)

 A1











 M1

 M1


 M1(both)

 A1
 20  MathF32023ET2OP2Ans20  





 B1


 B1



 B1

 B1

 B1

 B1


 B1

 B1



 B1


 B1
 21
  1.  10800 × 12 = 115 =   £ 7452
        20               100
    500 × 12                = −    300  
         20                        £7152p.a
  2. 600 × 10/100 × 12 = Sh. 720p.a

  3.  Income £ p.a  Rate   Amount of tax
     1 - 2100
     2101 - 4200
     4201 - 6300
     6301 - 7152
     

     2 × 2100
     3 × 2100
     4 × 2100
     5 × 852

      4200
      6300
      8400
      4260       
     Sh 23160 p.a
    1. Gross tax = Sh. 23160 p.a
    2. 23160 − (720 + 1160 × 12)
      23160 − 14640 = Sh. 8520 p.a
      8520 = Sh. 710 p.m
        12
  4. 10800 − (280 + 710)
         = Sh 9810 p.m
  M1








 B1  M1 (for 852)
 


 M1

 A1

 M1
 A1
 22
  1.  
     Marks  No. of students   (f)  mid point (x)    fx
     30 - 34
     35 - 39
     40 - 44
     45 - 49
     50 - 54
     55 - 59
     60 - 64

     
     B1
        3
        7
        4
       11
        4
        5
        6     
     Σf = 40
         
        B1
        32
        37
        42
        47
        52
        57
        62

     
       B1

     96
     259
     168
     517
     208
     285
     372    
     Σfx
     = 1905
     B1

  2. Modal class 45 − 49
  3.  
    1. x̄ = 1905 = 47.625
              40
    2. 15/40 × 100 = 37.5%
    3. 26/40 = 13/20
 












 B1

 M1 A1

 M1 A1
 B1
 23
  1.   
    MathF32023ET2OP2Ans23
  2. E = K + hL
  3.  
    1. Gradient =  8 − 6 = 2/8 = 1/4    M1  A1
                       12 − 4
    2. E - intercept = 5   B1
  4. E = 5 + ¼L        B1
  5.  
    1. E = 6.75    B1
    2. L = 5     B1
 
 24

 MathF32023ET2OP2Ans24

  1. r = 8 − x
    (8 − x)2 =  x 2 + 42 
    64 − 16x + x2 = x2 +16
    64 − 16 = 16x 
             x = 3
    radius = 8 − 3 = 5cm
  2. Tan θ = 4/3
           θ = 53.13°
    ∠ AOB = 2 × 53.13
               = 106.26°
  3. 106.26 × 22/7 × 25 − ½ × 25 Sin 106.26     
       360
    23.19 − 12.0
    = 11.19
  4. 22/7 × 25 − 11.19
      = 78.57 − 11.19
          =  66.67                     
 









 M1

 M1


 A1


 M1
 A1
 M1 (both)

 M1 (subtraction)
 A1
 M1 (Area of circle)

 A1
Join our whatsapp group for latest updates

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


Tap Here to Download for 50/-




Why download?

  • ✔ To read offline at any time.
  • ✔ To Print at your convenience
  • ✔ Share Easily with Friends / Students


Get on WhatsApp Download as PDF
.
Subscribe now

access all the content at an affordable rate
or
Buy any individual paper or notes as a pdf via MPESA
and get it sent to you via WhatsApp

 

What does our community say about us?

Join our community on:

  • easyelimu app
  • Telegram
  • facebook page
  • twitter page
  • Pinterest