Mathematics Paper 1 Questions and Answers - Form 3 Mid Term 2 Exams 2021

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Instructions

  • Answer all the questions in section I and ONLY Five in section II.
  • Show all the calculations in the spaces provided
  • KNEC mathematical tables and non-programmable calculators may be used.

SECTION I

  1. Evaluate without using a calculator. [1 Mark]
    Math F3 MT2 PP1 Q1 2021
  2. The equation of a straight line L1 is 3y+4x=12
    1. Find the gradient of L1 [1 Mark]
    2. The equation of another line L2 which is perpendicular to L1 and passes through (1,2) [2 Marks]
  3. Evaluate using mathematical tables only expressing your answer to 4 significant figures. [3 Marks]
        3      +  (0.7918)2      
    0.2311
  4. Given that:
    sin⁡ (3x−35)= cos (x+20). Find x [2 Marks]
  5. The size of an interior angle of a regular polygon is (3x)° while the exterior angle is (x−20)°. Find the number sides of the polygon [3 Marks]
  6. Three bells ring at intervals of 9 minutes, 15 minutes and 21 minutes. The bells will next ring together at 11.00pm. Find the time the bells had last rung together.  [3 Marks]
  7. Find all the integral values of x which satisfy the following inequalities [3 Marks]
    2(2−x)<4x−9<x+11
  8. At a party, every two people shared a plate of Ugali between them. Every 3 people shared a plate of soup and every 4 people shared a plate of meat. If 65 plates were used in total. How many people were there? [3 Marks]
  9. Find the value of x which satisfies the equation; [3 Marks]
    Math F3 MT2 PP1 Q9 2021
  10. Mary and John live 140km apart. Mary starts from her home at 7.00am and drives towards John’s home at 80km/hr. John starts at 7.30am and drives towards Mary’s home at 100km/hr. at what tome did they meet? [3 Marks]
  11. Two points P and Q have coordinates (−2, 3) and (1, 3) respectively. A translation maps point P to P1 (10,10).
    1. Find the translation vector [1 Mark]
    2. Find the coordinates of Q the image of Q under the translation. [1 Mark]
    3. Find the values of M and N if;
      mP − nQ = Math F3 MT2 PP11c Q1 2021[3 Marks]
  12. A Kenyan company received $100,000 US dollars. The money was converted into Kenya shillings in a bank which buys and sells foreign currencies as follows;
                                      Buying.               Selling
    1 US Dollar ($)              77.23                 78.11
    1 Sterling Pound (£)      121.04               122.93
    1. Calculate the amount of money, in Kenya shillings the company received[2 Marks]
    2. The company exchanged the Kenya-shilling 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 Marks]
  13. In the figure below, O is the centre of the circle. Angle OAB=30° and angle BAC = 23°. Find angle ABC. [3 Marks]
    Math F3 MT2 PP13 Q1 2021
  14. A number n is such that when it is divided by 27, 30 or 45, the remainder is always 3. Find the smallest value of n. [2 Marks]
  15. A particle accelerates uniformly from rest and attains a maximum velocity of 30m/s. after 16 seconds. It travels at this constant velocity for 20 seconds before decelerating to rest after another 8 seconds.
    Calculate the total distance travelled by the particle. [4 Marks]
  16. The figure below shows a rhombus PQRS with PQ=9cm and <SPQ=60°,S×Q is a circular arc center P.
    Math F3 MT2 PP16 Q1 2021
    Calculate the area of the shaded region correct to 2 decimal places. [3 Marks]

SECTION II
Answer any 5 Questions in this Section. (50 Marks)

  1. A salesman received a basic salary of sh. 50,000 a year together with a commission of 6% on the value of goods sold and a car allowance of sh. 2.50 per km.
    1. Find the total amount he received in a year in which he sells goods worth sh. 625,000 and travels 10,000km. [4 Marks]
    2. The next year he travels 12,000km and receives a total of sh. 134,000.
      1. Calculate the value of goods sold. [4 Marks]
      2. Calculate the percentage increases in the value goods sold. [2 Marks]
  2. The following measurements were recorded in a field book at a farm using XY=400m as the baseline.
     C60
     


     B100 
     A120
      Y 
    340
    300
    240
    200
    140
     80
      X
     120D
     100E
     160E 


    1. Using the scale of 1:4000 (1cm represents 40m) draw accurately the map of the farm. [4 Marks]
    2. Determine the actual area of the farm in hectares. [4 Marks]
    3. If the farm is on sale at Ksh. 80,000 per hectare, how much does the farm cost?
      [2 Marks]
  3. The minor arc PQ of a circle radius 21cm subtends an angle of 120° at the centre of the circle as shown below.
    Math F3 MT2 PP19 Q1 2021
    1. Find the area of the minor sector POQ [2 Marks]
    2. Find the perimeter of the minor sector POQ [3 Marks]
    3. The minor sector POQ is folded to form a right circular cone.
      Calculate:
      1. The radius of the cone. [3 Marks]
      2. The height of the cone. [2 Marks]
  4. A triangular piece of land ABC has sides AB=100m, BC=150m and AC=190m.
    Math F3 MT2 PP1 20 Q1 2021
    1. Calculate the area of the triangular piece of land ABC [2 Marks]
    2. Calculate the value of angle ACB. [3 Marks]
    3. A new piece of land ABCD is a trapezium with AD//BC whose area is three times that of triangle ABC, calculate the perimeter of ABCD. [5 Marks]
  5. Three business partners, Bela, Joan and Trinity contributed Kshs. 112, 000, Kshs. 128, 000 and Kshs. 210, 000 respectively to start a business. They agreed to share their profits as follows:
    30% to be shared equally
    30% to be shared in the ratio of their contributions
    40% to be retained for the running of the business.
    1. If at the end of the year, the business realised a profit of Kshs. 1. 35million
      Calculate:
    2. The amount of money retained for running the business at the end of the year.
      [1 Mark]
    3. The difference between the amounts received by Trinity and Bela. [6 Marks]
    4. Express Joan’s share as percentage of the total amount of money shared between the three partners. [3 Marks]
  6. In the figure below, O1 and O2 are the centres of the circles whose radii are 4 cm and 7 cm respectively. The circles intersect at A and B and angle AO1O2 = 60˚
    Math F3 MT2 PP1 22 Q1 2021
    Find by calculation; take π = 3.142
    1. The angle AO2O1 [1 Mark]
    2. The area of the quadrilateral AO1BO2 [4 Marks]
    3. The shaded area [5 Marks]
  7. Members of a certain group decided to raise sh. 225,000 to buy a plot of land, with each contributing the same amount. Before the due date for collection of the contribution, ten of the members withdrew from the project.
    1. Letting n represent the original membership of the group, show that the increase in contribution per member was 2250000 [4 Marks]
                                                  n(n-10)
    2. If the increase in contribution per person was sh. 1125, what was the original number of members in the group? [4 Marks]
    3. Calculate the percentage increase in the contribution per person caused by the withdrawal of the members. [2 Marks]
  8. In the figure below, O is the center of the circle. <AEB=50°, <EBC=80° and <ECD=30°
    Math F3 MT2 PP1 24 Q1 2021
    Giving reasons calculate
    1. <CDE [2 Marks]
    2. <DFE [2 Marks]
    3. Obtuse <COE [2 Marks]
    4. <ADE [2 Marks]
    5. <CAE [2 Marks]


MARKING SCHEME

  1. 17/7 − 11/6 = 102 − 7725/42
                             42
    525 × 61 = 5
    742    51     7
    2/3 of 9/43/2 − 8/7 = 21 −165/14
                                           14
    5/7 × 14/5 = 2
  2.  
    1. 3y/3 = −4/3x + 12/3
      y = −4/3x + 4
      G = −4/
    2. 4/3x = −1 
      x = ¾
      y − 2 = ¾
      x − 1
      4y − 8 = 3x− 3
      4y = 3x + 5
      y = ¾x + 5/4
  3.  3 ×      1    
             0.2311
    2.311 × 10−1
    0.4327 × 101
    = 4.327
      ×     3
       12.981
    (7.918 × 10−1)2
    62.70×10−2 = 0.6270
    12.981 + 0.6270
            = 13.608

  4. 3x − 35 + x + 20 = 90
     47 − 15 = 9
    4x = 105
    x = 26¼°
  5. 3x + x − 20 = 180
    4x = 200
    x = 50°
    exterior = 50 − 20 =  30°
    n = 360 = 360 = 12 sides
           ext      30
  6.  
    Math F3 MT2 PP1 ans 6 2021
    3 × 3 × 5 × 7 = 315 minutes
    = 5hrs 15mins
    11:00
    −5:15
      5:45
    =5:45pm
  7. 4 − 2x < 4x − 9
    13/66x/6
    21/6 < x
    4x − 9 < x +11
    3x/320/3
    x < 62/3
    21/6 < x < 62/3
    integral values; ( 3, 4, 5, 6)
  8.  
    Math F3 MT2 PP1 ans 8a 2021
    2 x 2 x 3 = 12
    12/2 = 6 plates of Ugali
    12/3 = 4 plates of soup
    12/4 = 3 Plates of Meat
    12 people = 13 plates 
                      65 plates
    12 x 65 = 60 people
        13
  9.  
    Math F3 MT2 PP1 ans9 2021
    4x2 = 12x − 9 
    4x2 −12x + 9 =0
    4x2 − 6x − 6x + 9 = 0
    2x(2x−3) −3(2x−3) =0
    (2x − 3)(2x−3) = 0
    x= 1.5

  10. Math F3 MT2 PP1 ans 10 2021
    ½ × 80 = 40km
    Relative speed = 80 + 100 = 180km/hr
    T= D/S100/180 = 33mins
    7.30
    + 33  
    8.03 am.
  11.  

    1. Math F3 MT2 PP1 ans 11a 2021
    2.  
      Math F3 MT2 PP1 ans 11b 2021
    3.  
      Math F3 MT2 PP1 ans 11c 2021
      3(−2m−n=−12)
      2(3m − 3n = 9) -−
          −6m − 3n  = − 36
       +   6m − 6    = 18    
                  − 9n   = −18
                          n = 2
      −2m =−10    m = 5
  12. 1 US Dollor Buying 77.23
    1. 100,000 = 7,723000
    2. 7,723,000 = 62824 pounds
       122.93

  13. Math F3 MT2 PP1 ans 13 2021
    120 +23 = 143°
    180 − 143 = 37°
    <ABC = 37°
  14.  
    Math F3 MT2 PP1 ans 14 2021
    270 + 3 =273
  15.  
    Math F3 MT2 PP1 ans15 2021
    D = Area under velocity time graph
    A = ½ ×16 × 30 = 240m
    B = 20 × 30 = 600m
    C = ½ × 8 × 30 =120m
    Total distance = 960m
  16. Area of rhombus
    ½ x 9 x 9 Sin 60 × 2
    = 81 sin 60
    =70.148 cm
    Area of sector
    3.142 × 9 × 9 × 60  = 42.417
                            360
    70.148 − 42.417
    = 27.731 cm2
  17.  
    1. 6/100 × 625000
      = 37500/=
      Shs 2.50 = 1km
                       10,000
                    = 25,000 
      Total = 50,000 + 37500 + 25000
              = Shs 112,500 
    2.  
      1. 134000 − 50,000 = 84,000 
        12000 x 2.5 = 30,000
        Commission = 84000 − 30,000
                          = 54000
        100 × x = 54000 × 100
          6     100             6
        Value of goods = 900,000/=
      2.   900,000
        625,000
           275,000
        275000 × 100 = 44%
        625000
  18.  

    1. Math F3 MT2 PP1 ans 18a 2021
    2. A = ½ × 1.5 × 1.5 = 1.125cm2
      B = ½ × 4 × 5 = 10cm2
      C = ½ × 5.5 × 1.5 = 4.125cm2
      D = ½ × 2 × 3 = 3cm2
      E = ½ × 4 × 5 = 10cm2
      F = ½ × 6.5 × 1 = 3.25cm2
      G = ½ × 5.5 × 1.5 = 4.125cm2
      H = ½ × 2.5 × 3  =3.75cm2
                                   39.375cm2
      1cm rep 40m
      1cm2 =1600m2
      39.375
      39.375 × 1600 = 63000m2
                                10000
                            = 6.3ha
    3. 6.3 × 80000 = Shs 504,000
  19.  
    1. πr2 θ/360
      A = 22/7 × 21 × 21 × 120/360
         = 462cm2
    2. πDθ/360
      22/7 × 12 × 120/360 = 44cm
      44+42=86cm
    3.  
      1. 44cm = 22/7D
        D = 7/22 × 44 =14/2 = 7cm
      2.   
        Math F3 MT2 PP1 ans 19c 2021
        √(212 − 72)
        √392
         = 19.8cm

  20. Hero's formula
    1. S = ½(100+190+150)
      A = √(280(220−150)(220−100)220−190))
      A = √(55.44 × 106)
         = 7. 446 × 103 = 7446m2
    2. <ACB
      Cos C = Math F3 MT2 PP1 ans 20b 2021
      Cos C = 48600 = 0.8526
                   57000
      C = Cos−1 0.8526
      <ACB = 31.5°
    3. 7446×3
      ABCD = 22338m2
      Area of ACD 
      22338 − 7446
       = 14892m2
      7446 = ½ × 160 × h
      7446 = 75h
        75      75
         h =99.28m
      ABCD
      = ½ × (150+AD) × 99.28 = 22338
      150 + AD = 450
                AD = 300m
      <CAD = 31.5°
      (CD)2 = 1902 + 3002 −2 × 190 × 300 cos 31.5
      (CD)2 = 36100 + 90000 − 97201
      (CD)2 = 28899
      CD = 170m
      Perimeter = 100+150+170+300
                     =720m
  21.  
    1. 40/100 × 1350000
       = 540,000
    2. 30/100 × 1350000
       =405000
      405000 = 135000each
           3
       B  :  J  : T
      112:128:210
       56 : 64 :105
      Bela = 56/225 × 405000 = 100,800
      Total(Bela) = 135000+100800
                      =235,800
      Trinity = 105/225 × 405000 = 189000
      Total = 189000 + 135000
              =324,000
    3. Difference = 324000 − 235800
                      = Shs 88,200
    4. (64/225 × 405000) + 135000
             = 250,200
      250,200 × 100
      810,000
      = 308/9%
  22.  
    Math F3 MT2 PP1 ans 22 2021
    1.  
      Math F3 MT2 PP1 ans 22a1 2021
      sin 60° = h/4 h= 3.464
      Math F3 MT2 PP1 ans 22a2 2021
      Sin θ = 3.464
                    7
      θ = Sin−1 = 29.66°
    2. ½ × 4 × 4 sin 120 = 6.93cm2
      ½ × 7 × sin 59.32 = 21.072
      =28cm2
    3. Unshaded area
      sector 1
      πr2 θ/360
      3.142 × 4 × 4 × 120/360  = 16.76cm2
      ∆O1AB = 6.93
      16.76 − 6.93 = 9.83cm2
      Sector 2
      3.142 × 7 × 7 × 59.32 = 25.37cm2
                                 360
      25.37 − 21.07 = 4.3cm2
      sum 
      9.83+4.3 = 14.13cm2
      Shaded area 
      28 − 14.13 = 13.87cm2
  23.  
    1. Original contribution = 225000
                                            n
      After 10 withdrew = 225000
                                      n−10
      225000 − 225000
        n−10          n       
      n(n−10)
      225000n −225000n+2250000
                     n(n−10)
      = 2250000
          n(n−10)
    2. 2250000 = 1125
       n(n−10)
      n(n−10) = 2000
      n2−10n−2000=0
      n2 − 50n + 40n − 2000 =0
      n(n−50) + 40(n−50)=0
      (n+40)(n−50) = 0
      n = 50 members
    3. 225000 = 4500
           50
      225000 = 5625
          40
      1125 × 100
      4500
      =25%
  24.  
    1. <CDE = 100°
                =Opposite angles in a cyclic quadrilateral BCDE
    2. <DFE = 30°
               = Angle sum of triangle DEF add up to 180°
    3. <COE = 160°
               = Radius subtends equal angles at the circumference.
    4. <ADE = 40°
                =Angle subtended by chord AE at the circumference.
    5. <CAE = 80°
               = Angle subtended by chord CE at the circumference.

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