Mathematics Paper 1 Questions and Answers - Form 4 Term 2 Opener Exams 2023

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INSTRUCTIONS:

  • Answer all the questions in Section I and any FIVE in Section II.
  • Answer the questions on the spaces provided.


QUESTIONS

SECTION 1

  1. Without using calculators or mathematical tables, evaluate leaving your answer in standard form. (3 Marks)
      1.33×0.51   
    0.19×0.0017
  2. Three bells are programmed to ring after an interval of 15 minutes, 25 minutes and 50 minutes. If they all rang together at 6.45am, when will they next ring together? (3 Marks)
  3.  The volumes of two similar solid cylinders are 1920cm3 and 810cm3. If the area of the curved surface of the smaller cylinder is 300cm2, find the area of the curved surface of the larger cylinder. (4 Marks)
  4. Solve for x in the equation(32x )= 81×94(3 Marks)
  5. A class of 30 students uses 75 pencils in a term. If the number of students is reduced to 24, how many pencils are likely to be used in a term?(3 Marks)
  6. Given that sin (3x-35)o – cos (x+20)= 0 and x is an acute angle, find its value. (3Marks)
  7. Simplify 4x-20 (3 Marks)
                   5-x
  8. The shaded region in the figure below shows a section of a road on a roundabout. Calculate the area in cm2. (use π=3.142) (3 Marks)
    8 ayadad
  9. Solve the inequalities and represent your solution on a single number line.
    x-5≤3x-8<2x-3 (3 Marks)
  10. A tourist visits Kenya and changes $400 to Ksh. At the end of the holiday, she has only Ksh. 1450 left. How many dollars did she spend in the holiday if the exchange rate is as per the table below.

    Currency

    Buying  Ksh

    Selling Ksh

    1US dollar $

    79.25

    81.50

    (3 Marks)
  11. Use tables of squares, square root and reciprocal only to evaluate. (4 Marks)
    (0.06458)1/+(2/0.4327)2
  12. Every week, the age of people who attend a cinema is recorded. In a particular week the data was as shown in the table below.

    Age(x) in years

    0 ≤ x < 5

    5 ≤ x < 15

    15 ≤ x < 25

    25 ≤ x < 45

    45 ≤ x < 50

    No of people

    14

    41

    45

    70

    15

    On the grid provided, draw a histogram to represent the distribution.
    Scale: 1cm to represent 5 units on horizontal axis
                2cm to represent 1 unit on vertical axis. (4 Marks)
    12 adadad
  13. 13 adada Find the scalar m to satisfy the equation 5p + mq = c (3 Marks)
  14. Find the sum of interior angles of a regular polygon with 18 sides. (2 Marks)
  15. The figure below shows a triangle ABC not drawn to scale; D is a point on line AC. Given that BC=14cm, DC=7cm and ∠ABC=∠BDC. Find the length of AD (3Marks)
    15 Adadada
  16. A positive two digit number is such that the product of the digits is 24. When the digits are reversed, the number formed is greater than original number by 18. Find the number. (3 Marks) 

SECTION II
Answer any five questions in this section (50 Marks)

  1. Two aero planes S and T leave airport A at the same time, S flies on a bearing of 060° at 750km/h while T flies on a bearing of 210° at 900km/h.
    1. Using a suitable scale, draw a diagram to show the positions of aero planes after two hours. (4 Marks)
    2. Use your diagram to determine
      1. The actual distance between the two aero planes (2 Marks)
      2. The bearing of T from S (2 Marks)
      3. The bearing of S from T (2 Marks)
  2. On the grid provided, draw triangle ABC whose vertices are A (3, 4), B (1, 3) and C (2, 1).
    1. Draw ΔA ′B′C′ the image of ABC under a rotation of +90º about (0, 0). (2 Marks)
    2. Draw Δ A"B"C" the image of A ′B′C  under a reflection in the line y=x. (2 Marks)
    3. Draw Δ A""B""C"" the image of ∆ A"B"C" under a rotation of -90º about (0, 0). (2 Marks)
      graph papaers
    4. Describe a single transformation that maps ΔABC onto Δ A""B""C"". (2 Marks)
    5. Write down the equations of the lines of symmetry of the quadrilateral BB"A""A′ . (2 Marks)
  3. A bus and a matatu left voi for Mombasa ,240km away at 8.00am.they travelled at 90km/h and 120km/h respectively. After 20minutes the matatu had a puncture which took 30 minutes to mend.it then continued with the journey.
    1. How far from voi did the matatu catch up with the bus. (6 Marks)
    2. At what time did the matatu catch up with the bus? (2 Marks)
    3. At what time did the bus reach Mombasa? (2 Mark)
  4. A straight line L1 whose equation is 3y – 2x = -2 meets the x – axis at R
    1. Determine the coordinates of R. (2 Marks)
    2. A second line L2 is perpendicular to L1 at R. Find the equation of L2 in the form y = mx+c where m and c are constants. (3 Marks)
    3. A third line L3 passingthrough(-4, 1) is parallel to L1. Find
      1. The equation of L3 in the form y=mx+c where m and c are constants. (2 Marks)
      2. the coordinates of points S at which L3 intersects L2 (3 Marks)
  5. A group of people planned to contribute equally towards a water project which needed Ksh.2,000,000 to complete. However 40 members of the group withdrew from the project. As a result each of the remaining members were to contribute Kshs.2,500 more.
    1. Find the original number of members in the group (5 Marks)
    2. Forty five percent of the value of the project was funded by constituency. Development fund (CDF). Calculate the amount of contribution that would be made by eachof the remaining members. (3 Marks)
    3. Members contribution were in terms of labour provided and money contributed. If the ratio of the value of labour to the money contribution was 6:9. Calculate the total amount of money contributed by the members (2 Marks)
  6. The diagram below shows a circle ABC with AB=12cm, BC=15cm, and AC=14cm
    22 adaddada
    Calculate to 4 significance figures:
    1. The angle ACB (3 Marks)
    2. The radius of the circle (3 Marks)
    3. The area of the shaded region (4 Marks)
  7. The displacement of a particle S metres, t seconds after passing a fixed point O is given by S=3+2t-5t2
    Calculate:
    1. The displacement of the particle 2 seconds later (2 Marks)
    2. The time taken for the particle to return to O (2 Marks)
    3. The maximum displacement of the particle (3 Marks)
    4. The initial velocity of the particle (3 Marks)
    5. The acceleration of the particle after t seconds (1 Mark)
  8. The figure below is a right pyramid VEFGH with a square base of 8 cm and a slant edge of 20 cm. Points A, B, C and D lie on the slant edge of the pyramid such that VA=VB=VC=VD=10 cm and plane ABCD is parallel to the base EFGH.
    24 adadad
    1. Find the length of AB. (2 Marks)
    2. Calculate, correct to 2 decimal places:
      1. the length of AC; (2 Marks)
      2. the perpendicular height of the pyramid VABCD. (2 Marks)
    3. The pyramid VABCD was cut off. Find the volume of the frustum ABCDEFGH correct to 2 decimal Places. (4 Marks)


MARKING SCHEME

  1. 1.33 x 0.51 / 0.19 x 0.0017
      1.33 x 0.51   x 1000000
        0.19 x 0.0017    1000000
    = 133 x 51 x 100
           19 x 17
    = 19 x 7 x 17 3 x 100
               19 x 17
    = 7 x 3 x 100
    = 2100
    = 2.1 x 103
  2. = L.C.M of 15, 25 and 50

    2

    15

    25

    50

    3

    15

    25

    25

    5

    5

    25

    25

    5

    1

    5

    5

     

    1

    1

    1

    =2 x 3 x 5 x 5 = 150 minutes = 2 hours 30 minutes
    6.45 am + 2 hours 30 minutes = 9.15 am
  3. V.S.F = 1920 = 64
                  810     27
    LSF = 3√64/27 = 4/3
    ASF = (4/3)2 = 16/9
    9 → 300 cm2
    16 → ?
    = 16 x 300
            9
    = 5331/3 cm2
  4. (32x)3 = 81 x 94
    36x = 34 x 324
    36x = 34 x 38
    36x = 312
    6x = 12
    x = 2
  5. 30 students → 75 pencils
    24 students → ?
    = 24 x 75
          30
    = 60 pencils
  6. (3x - 35)º + (x + 20)º = 90º
    3x + x - 35º + 20º = 90º
    4x - 15º = 90º
    4x = 105º
    x = 26.25º
  7. 4x - 20 / 5 - x
    = 4(x - 5) / 5 - x
    = -4(5 - x) / 5 - x
    = -4
  8. 8 auydgua
  9. x - 5 ≤ 3x - 8 < 2x - 3
    x - 5 ≤ 3x - 8
    x - 3x ≤ -8 + 5
    -2x ≤ -3
    3x - 8 < 2x - 3
    3x - 2x < -3 + 8
    x < 5
    3/2 ≤ x < 5
  10. 1 us dollar $ → ksh. 79.25
    $400 → ?
    = ksh. (400 x 79.25)
    = ksh.31700
    Amount spent = Ksh. 31700 - Ksh. 1450
    = Ksh.30250
    = $ 30250 / 79.25
    = $381.70
  11.                   
    11 adada
  12.         
    graph adad
  13.                 
    13 adada
  14. Sum of interior angles of n sided polygon
    = 90º(2n - 4)
    = 90º (2(18) - 4)
    = 90º (32)
    = 2880º
  15. BC/DC = AC/BC
    14/7 = AC/14
    AC = 14 x 14 = 28 cm 
                   7
    AD = 28 - 7 = 21 cm
  16. Let the number mn
    m x n = 24
    when reversed nm the new value = 10n + m
    (10n + n) - (10m + n) = 18
    9n - 9m = 18 ......................(i)
    n = 24/m ............................(ii)
    9(24/m) - 9m = 18
    216 - 9m2 = 18m
    9m2 + 18m - 216 = 0
    m2 + 2m - 24 = 0
    m2 + 6m - 4m - 24 = 0
    m(m + 6) - 4(m + 6) = 0
    (m - 4)(m + 6) = 0
    m = 4 or -6
    m must be positive
    = 4
    n = 6
    The number is 24
  17.                                   
    1.                       
      17 ajhdha
    2.                        
      1. Actual distance  = 10.7 (±0.1) x 300 = 3210 km ± 30 km
      2. = 224º ± 1º   or S 44ºW
      3. 044º   or N 44ºE
  18.                      
    1. A'(-4, 3), B'(-3, 1) and C'(-1, 2)
    2. A''(-3, -4), B''(1, -3) and C''(2, -1)
    3. A'''(-4, -3), B'''(-3, -1) and C'''(-1, -2)
      C adadada
    4. its a reflection in the line y = -x
    5. Lines of symetry are y = 0 or x -axis and x = -1.5
  19.                   
  20.                   
    1. At x - axis y is 0
      3(0) - 2x = -2
      -2x = -2
      x = 1
      R (1, 0)
    2. Equation of L1 is y = 2/3 x -2/3
      Gradient of L1 = 2/3
      Gradient of L2 = -3/2
      Equation of L2 → y - 0 / x - 1 = -3/2
      2y = -3x + 3
      y = -3/2 x + 3/2
    3.                
      1. Gradient of L3 = 2/3
        Ans = y = 2/3x + 11/3
      2. S(-1, 3)
  21.                        
    1. Let initial number of members be x
      initial contrubution per member = 2000000
                                                                  x
      New contribution per member 
      =2000000
         x - 40
      2000000 - 2000000 = 2500
        x - 40            x
      250x2 - 1000000x = 80000000
      x2 - 40x - 32000 = 0
      x2 + 160x - 200x - 3200 = 0
      (x - 200)(x + 160) = 0
      x = 200 or -160 Therefore, x = 200
    2. 55/100 x 2000000
      = 1100000
      Contribution = 1100000 / 160 = Ksh.6875
    3. 9/15 x 1100000
      =Ksh. 660,000
  22.                                
  23.                            
    1. s = 3 + 2(2) - 5(2)2
      = -13m
    2. 3 + 2t - 5t2 = 0
      3 + 3t - 5t - 5t2 = 0
      (3 + 5t)(1 - t) = 0
      t = 3/5 or 1 ∴ t = 1 second
    3. At maximum displacement v = 0
      2 - 10t = 0
      t = 1/5
      s = 3 + 2 (1/5) - 5(1/5)2 = 31/5 metres
    4. v = 2 - 10 (0)
      = 2 m/s
    5. a = - 10 m/s2

 

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