MATHEMATICS - FORM 3 END TERM 1 EXAMS 2020

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Questions

  1. Evaluate and simplify without using a calculator. (3mks)
    1
  2. The sum of interior angles of a polygon is 1980o. Find the number of sides the polygon has. (2mks)
  3. Simplify as far as possible by rationalizing the denominator. (3mks)
    3
  4. Use table of reciprocal only to work out (3mks)
    4
  5. Solve 3x - 2≤ 5x - 6 < 2x + 12 and represent your solution on a number line. Hence state the integral values. (4mks)
  6. Evaluate without using mathematical tables. (3mks)
    2 log 5 - ½log 16 + 2 log 40
  7. Given that P=2.6 cm, Q=4.0 cm and R=7.8 cm. Find the percentage error in the expression. (3mks)
    P + Q
      R
  8. From a point 20m away on a level ground the angle of elevation to the lower window line is 29o and the angle of elevation to the top line of the window is 32o. Calculate the height of the window. (3mks)
  9. The size of an interior angle of a regular polygon is 156o. Find the number of sides of the polygon. (2mks)
  10. The number 5.81 contains an integral part and a recurring decimal. Convert the number into an improper fraction and hence a mixed fraction. (3mks)
  11. Simplify.   (3mks)
    2y2 - xy + x2
      2x2 - 2y2
  12. Given that Sin A= 4/5 , Cos B=5/12 A and B are acute angles. Without using tables calculate
    Sin B Cos A + Sin A tan B                                                                           (3mks)
  13. A two digit number is such that the sum of the digits is 11 where the digits are reversed the number exceed the original number by 9. Calculate the original number. (3mks)
  14. Two boys and a girl shared some money. The elder boy 4/5 got of it, the younger boy got 2/5 of the remainder and the girl got the rest. Find the percentage share of the younger boy to the girls share. (4mks)
  15. Solve the following simultaneous equations. (4mks)
    x2- xy=2
    x + y=3
  16. Use the table of squares, square roots and reciprocals to evaluate to 3 decimal places the question below. (4mks)
    16
  17. The diagram below shows two circles centre A and B which intersect a point P and Q. Angle PAQ=70o and <PRQ=40o and PA=AQ=8 cm.
    17
    Use the diagram to calculate to 2 d.p
    1. the length PQ   (2mks)
    2. The length PB   (2mks)
    3. Area of minor segment circle centre A. (2mks)
    4. Area of the shaded region (4mks)
  18. The following table shows the heights to the nearest centimeter of some maize plants in a research farm.

    Height (cm)

    80-84

    85-89

    90-94

    95-99

    100-105

    105-109

    110-114

    115-119

    Frequency

    5

    14

    16

    17

    24

    12

    11

    4


    1. State the modal class   (1mk)
    2. Find to 2 d.p
      1. Mean height     (4mks)
      2. The difference between the mean height and the median height. (5mk)
  19.  
    1. Three points A(0,4) B(2,3) and C(-2, -1) are vertices of a triangle. Find ;
        1. The gradient of AC     (1mk)
        2. The gradient of the perpendicular bisector of line AC   (1mk)
        3. The coordinates of the mid-point of line AC. (1mk)
    2.  
      1. the gradient of AB   (1mk)
      2. The gradient of the perpendicular bisector of lines AB     (1mk)
      3. The coordinates of the mid-point of AB           (1mk)
    3.  
      1. Find the equation of the perpendicular bisector of AC   (1mk)
      2. The equation of perpendicular bisector of AB   (1mk)
      3. Hence find the coordinates of the circumcentre of the triangle.(2mks)
  20. The position vectors of points A and B with respect to the origin O, are (-85) and (12-5) respectively. Points M and N are the mid points of AB and OA respectively.
    1. Find
      1. The coordinates of N and M   (3mks)
      2. The magnitude of NM. (3mks)
    2. Express vector NM in terms of OB (1mk)
    3. Point P maps onto P1 by a translation (-58) given that OP=OM +2 MN, find the coordinates of P1 (3mks)
  21. The information of a piece of land was entered in a field book as shown below.
    21
    1. Sketch the map of the land (3mks)
    2. Find the area of the land in hectares. (7mks)
  22.  
    1. Draw the graph of the function y=2x2 + 4x - 3 on the graph paper provided `for -4 ≤ x ≤ 2.5         (5mks)
    2. Use your graph to solve the equations
      1. 2x2 +4 x-3=0             (2mks)
      2. x2+ x-5=0     (3mks)
  23. The base of an open rectangular tank is 3m by 2.5 m and its height is 4m.
    1. Calculate
      1. The capacity of the tank in litres. (3mks)
      2. The total surface area in m2 of the tank. (2mks)
    2. An open cylindrical tank has an equal capacity and same height as the rectangular tank in (a) above. Calculate ;(correct to one decimal places)
      1. The radius of the cylindrical tank. (3mks)
      2. The total surface area, in m2, of the tank. (2mks)
  24. Transline bus left Nairobi at 8.00 and travelled Kisii at an average speed of 80km/h. Given that the distance between Nairobi and Kisii is 400km, calculate;
    1. The time the car arrived in Nairobi. (3mks)
    2. The time the two vehicles met. (3mks)
    3. The distance from Nairobi to the meeting point. (2mks)
    4. The distance of the bus from Kisii when the car arrived in Nairobi. (2mks)

Answers

  1.  16/5 + (1/4 x 7/2) - 31/16
    16/5 + 7/8 - 31/6 = -131/120  √1 Numerator
    8/3 - (7/5 x 3/4)+15/4
    8/3 -21/20 + 15/4 = 322/60   √1 Denominator
    -131/120 x 60/322 = -131/644 √1

  2. (2n-4)90 = 1980 √1
    2n-4=22
    2n=26
    n=13  √1
  3.  
    a3
  4. 3/0.6735 + 13/0.156
    3(1/0.6735) + 13(1/0.156)
    (3 x 6.735 x 101) + (13 x 1.56 x 101) √1
    3 x 1.485 + 13(6.41) √1
    4.455 + 83.3333 = 87.785 √1
  5. 3x - 2 ≤ 5x - 6 < 2x + 12
    3x - 2 ≤ 5x - 6
    3x - 5x ≤ -6 + 2
    -2x ≤ -4
    x ≥ 2

    5x - 6 < 2x + 12
    5x - 2x < 12 + 6
    3x < 18
    x < 6

    a5

    2,3,4,5
  6. 2log5 - 1/2log16 + 2log40
    log 52 - log 161/2 + log 402
    log(25x 402)
              4
    log 1000 = 4
  7. max quotient = 
    2.65 + 4.05 
        7.75
    =0.8645
    min quotient =
    2.55 + 3.95
       7.85
    =0.8280
    Working quotient = 
    2.6 + 4.0
        7.8
    =0.8462
    Error =
    0.8645 - 0.8280
         2
    =0.01825
    %error = 0.01825/0.8462 x 100%
    =2.157%

  8. x + y = 20tan32
    y = 20tan27
    a8
    x + 20tan27 = 20tan32
    x = 20tan32 - 20tan27
    x = 2.307
    height of the window = 2.307m
  9. Exterior <
    = 180-156
    =24o
    n = 360/24
    = 15 sides

  10. let n = 5.8111.... (i)
    10n = 58.1111....(ii)
    100n = 581.1111....(iii)
    subtract eqn (ii) from eq(iii)
    90n = 523
    n=523/90
    =573/90
  11.  2y2 - xy -x2
    2x2 - 2y2
    2y2 - xy - x2 = (2y2 - 2xy) + (xy - x2)
    = 2y(y-x) + x (y-x)
    = (2y+x)(y-x)
    2x2 - 2y2 = 2 (x-y)(x+y)
    (2y+x)(y-x) = -1(2y+x)(x-y)
    2(x-y)(x+y)      2(x-y)(x+y)
    = -2y-x
       2(x+y)

  12. a12
    sin A = 4/5
    sin B = 12/13
    cos B = 5/13
    cos A = 3/5
    Tan B = 12/5

    sinBcosA + sinAtanB
    12/13 x 3/5 + 4/5 x 12/5
    36/65 + 48/25 =
    900+3120
        625
    =4020/1625
    =2154/325
  13. x + y = 11 ... (i)
    (10y + x) - (10x + y) = 9 ... (ii)
    10y+x-10x-y=9
    9y-9x=9
    y-x = 1 ...(iii)
    x+y = 1
    -x+y  = 1
    2x=10
    x=5, y = 6
    the original number = 56

  14. let x be amount shared
    elder boy = 4/9x
    younger boy = 2/5(5/9x) = 2/9x
    girl = 1 - (4/9x + 2/9x)
     = 1/3x
    (2/9x ÷ 1/3x) x 100%
    =66.67%
  15. x2 - xy = 2
    x+y = 3, y = 3-x
    x2 - x(3-x) = 2
    x2 - 3x + x2 = 2
    2x2 - 3x - 2 = 0
    (2x2 - 4x)+(x-2)=0
    2x(x-2)+1(x-2)=0
    (2x+1)(x-2)=0
    x = -1/2 or 2
    when x = 2, y =1
    when x = -1/2, y = 31/2

  16.  a16
  17.  
    1. sin 35 = x/8
      x=8sin35
      PQ = 2 x 8sin35
      = 9.18cm
    2. sin 20 = 4.5886/PB
      PB = 4.5886/sin20 = 13.42cm
    3. 70/360 x 22/7 x 8 x 8 - ½x 8 x 8sin70
      =9.04cm2
    4. Area of minor segment centre B
      40/360 x 22/7 x 13.422 - ½x 13.422sin40
      =5.01
      Area of shaded region
      (½ x 82sin70 + ½ x 13.422sin40)-(9.04+5.01)
      =73.90cm2
  18.  
    1. 100-104
    2.  
      Height  f x fx cf
      80-84 5 82 410 5
      85-89 14 87 1218 19
      90-94 16 92 1472 35
      95-99 17 97 1649 52
      100-104 24 102 2448 76
      105-109 12 107 1284 88
      110-114 4 117 468 103
        ∑f=103   ∑fx = 10181  
      1. X∑fx/∑f = 10181/103
        =98.84
      2. median = 94.5+((103/2 - 35)/17)5
        =99.35

        99.35-98.84 = 0.66cm
  19.  
    1.  
      1. A(0,4)
        C(-2,-1)
        4+1 = 5/2 = 2.5
        0+2
      2. -2/5
      3. (0-2/2,4-1/2) = (-1,1.5)
    2.  
      1. M =
        3-4 =1/2
        2-0
      2. m = 2
      3. (0+2/2,4+3/2)
        (1,3.5)
    3.  
      1. y - 1.5 = -2/5(x+1)
        y = -2/5x + 1.1
      2. y-3.5 = 2(x-1)
        y = 2x+2.5
      3. y = -2/5x + 1.1
        y = 2x + 2.5
        solve simultaneously
        x = -0.58
        y= 1.33
        (-0.58,1.33)
  20.   
    1.  
      1.  
        a20ai
      2.  
        a20aii
    2.  
      a20b
  21.   
    a21
    1. 1/2 x 35 x 55 = 962.5
    2. 1/2 x 85 x 65 = 2762.5
    3. 1/2 x 50 x 60 = 1500
    4. 1/2 x 90 x 15 = 675
    5. 1/2 x 115 x 60 = 3450
    6. 1/2 x 40 x 30 = 600
    Total = 9950m2
    9950/10000 = 0.995ha
  22.  
    1.  
    2.  
      1. x = 0.6 or -2.6
      2. y = 2x2 + 4x - 2
        0 = 2x2 +2x -10
        y = 2x + 7
        x = -2 or 2.5
  23.  
    1.  
      1. V = 3 x 2.5 x 4
        =30 m2
        =30x1000
        =30,000l
      2. (3x2.5x1)+(2.5x4x2)+(3x4x2)
        =55.5
    2.  
      1. Πr2 = 3 x 2.5
        r = square root of (3x2.5)/11
        =1.545m
        =1.5m
      2. 22/7 x 1.52 + 22/7 x 2 x 1.5 x 4
        =7.071 + 37.716
        =44.787m2
  24.  
    1. T = 400/80 = 5hrs
      8:00AM
      5:00
      1:00PM
    2. 80 x 3/2 = 120km
      RS = 80 + 120 = 200km/h
      t = 120/200 = 3/5hrs
      =36min

      9:30AM
         36
      10:06AM
    3. 10:06
        8:00
      2:06
      time = 126/20
      S = 80km/h
      D = 126/60 x 80 = 168km
    4. t = 400/120 = 3hrs 20 min
      9:30
      3:20
      11:50

      13:00
      11:50
      1:10

      D = 70/60 x 80
      =280/3

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