Mathematics Questions and Answers - Form 2 Mid Term 2 Exams 2021

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Instructions to candidates.

  • The paper contain two sections: Section I and section II.
  • Answer All the questions in section I and strictly any five questions from section II
  • All answers and working must be written in the question paper in the spaces provided below each question.
  • Show all the steps in your calculations, giving your answers at each stage in the spaces below each question.
  • Non-programmable silent electronic calculators and KNEC mathematical tables may be used, except unless stated otherwise.

SECTION I (50 marks)
Answer all the questions in this section in the spaces provided

  1. Simplify:                  (3 marks) 
    F2 MT2 Math Q1 2021
  2. A plot in the shape of a rectangle measures 608 m by 264 m. Equidistant fencing posts are placed along its length and breadth as far apart as possible. Find:
    1. The distance between the posts (2 marks)
    2. The number of posts used (2 marks)
  3. A ship P of 180 km West of a port Q. Another ship R is at a distance of 90 km and on a bearing of 050° from P. A third ship S is due East of R and due north of Q. By scale drawing determine the bearing of S from P. (Use a scale of 1 cm for 30 km)     (4 marks) 
  4. Simplify the following by use of common factors:   (3 marks) 
    F2 MT2 Math Q4 2021
  5. A business woman bought 288 bananas at sh 10 for every 12. She sold all of them at sh 20 for every 18. What was her percentage profit? (4 marks)
  6. Solve the simultaneous equations (3 marks)
    3x − 2y = 7
    5x + y = 3
  7. When a piece of cloth was washed, it shrank. Its length decreased from 150 cm to 120 cm.
    1. In what ratio did it decrease? (1 mark)
    2. Suppose the width decreased in the same ratio. What is the new width if the original width was l.4 m? (2 marks)
  8. Given the following currency exchange rate, calculate to 3 significant figures the number of dollars that can be exchanged for 25 Sterling pounds.  (3 marks)
    1 US dollar ($) = Ksh 76.85
    1 Sterling pound (£) = Ksh 115.30 (3 marks) 
  9. A cylindrical tank whose diameter is 1.4 metres and height 80 cm is initially empty. Water whose volume is 492.8 litres is poured into the tank. Determine the fraction of the tank filled with water.    (Take π = 22/7     (4 marks)
  10. A man is now three times as old as his daughter. In twelve years time he will be twice as old as his daughter. Find their present ages. (3 marks)
  11. The number .5 81..contains an integral part and a recurring decimal. Convert the number into an improper fraction and hence into a mixed number. (3 marks)
  12. An article which is marked for sh 450 is sold to a customer for sh 393.75. What percentage discount is the customer allowed? (3 marks)
  13. On a certain map a road 20 km long is represented by a line 4 cm long. Find the area in hectares of a ranch represented by a rectangle measuring 2.8 cm by 1.6 cm on this map. (3 marks)
  14. Syengo spends one-third of his salary on food, one-quarter on rent, three-fifths of the remainder on transport and saves the rest. If he spends sh 1 800 on transport, find how much money he saves. (3 marks)
  15. The base of an open vessel is as shown in the figure below. The curved ends are semicircular and the height of the vessel is 18 cm.
    F2 MT2 Math Q15 2021
    Calculate the area of metal sheeting required to construct the vessel, allowing 10% for wastage. (4 marks)
  16. In the figure below PQ is parallel to RS. Angle QPT = 20º, angle TRS = 15°, angle PQY = 90°, RTY is a straight line.
    F2 MT2 Math Q16 2021
    Calculate:
    1. acute angle PTR
    2. Angle QYT

SECTION II (50 marks)
Answer any five questions in this section in the spaces provided.

  1.  
    1. Using a ruler and a pair of compasses only, construct triangle ABC in which BC 6 cm, _AB = 8.8 cm and angle ABC 22 ½°. (4 marks)
    2. Measure AC and angle ACB. (2 marks)
    3. Construct a circle that passes through A, B and C. (3 marks)
    4. What is the radius of this circle? (1 mark)
  2.  
    1. The angle of elevation of the top of a vertical tower from a point P is 30°. The angle of elevation of the top of the tower from another point Q which is nearer the foot R of the tower is 45°. The distance between P and Q is 20 metres and the points P, Q and R are on the same straight line on level ground. Using a scale of 1 cm to represent 5 m, draw an accurate scale drawing to represent the above information. (4 marks)
    2. Use your scale drawing to determine
      1. the height of the tower (2 marks)
      2. the distance QR (2 marks)
      3. the distance PR (2 marks) 
  3.  
    1. A Jua Kali artisan made an article and sold it to a wholesaler at a profit of 20%. The wholesaler sold the article to a retailer at a profit of 30%. The retailer finally sold the article to a customer at a profit of 50%.
      If it cost the artisan sh 500 to make the article, find how much the customer paid for it. (3 marks)
    2. A customer paid sh 1 560 for another article. Determine how much the wholesaler had paid for it. (3 marks)
    3. During a clearance sale the retailer reduced his prices by 10%. Find the percentage profit the retailer _made on an article which had cost the artisan sh 1 000 to make (4 marks)
  4.  
    1. A newly built classroom measuring 6.3 m long, 4.5 m wide and 3.2 in high is to be cemented on the floor and all inside walls. The classroom has one door measuring 1.85 m by 80 cm and four windows measuring 1.5 m by 70 cm each. Cementing materials cost sh 500 per square metre while labour costs 20% of the cost of cementing materials. Calculate to one decimal place, the total surface area to be cemented. (5 marks)
    2. the cost of cementing materials. (2 marks)
    3. the total cost of cementing the classroom. (3 marks)
  5. A train left Mombasa on Monday evening and travelled to Kisumu according to the travel time table below. The train arrived in Kisumu on Wednesday morning of the same week.
     Mombasa   dep.   1930h 
      Mtito Andei   arr.  0250h
     dep.  0335h 
      Nairobi  arr.  1050h
     dep.  1240h
     Nakuru    arr.  1900h
     dep.  2015h
     Kisumu  arr.  0900h
    1. Determine the time the train took to travel between
      1. Mombasa and Mtito Andei
      2. Mtito Andei and Nairobi
      3. Nairobi and Nakuru
      4. Nakuru and Kisumu (4 marks)
    2. Calculate the total time for the whole journey. (4 marks)
    3. Given that the railway road distance between Mombasa and Kisumu is 1 200 km, calculate the avenge speed for the whole journey. (2 marks)
  6. A rectangular sheet measuring 80 cm by 50 cm is 2 mm thick and is made of metal whose density is 2.5 g/cm3. A square of side 5 cm is removed from each corner of the rectangle and the remaining part folded to form an open cuboid.
    1. Calculate
      1. the area of metal which forms the cuboid. (2 marks) 
      2. the mass of the empty cuboid in kilograms. (4 marks)
    2. The cuboid is filled with water whose density is 1 g/cm3. Calculate the mass of the cuboid when full of water. (4 marks)
  7.  
    1. Copy and complete the tables (i) and (ii) below for the functions y = 7 − 3x and y= 2x − 8 respectively
      1. y = 7 – 3x
         x   −2   −1   0   1   2   3   4    5 
         y  13    7          −8 
      2. y = 2x − 8
         x   −4   −2   0   2   4   6   8    10 
         y  −16      −8       4     
    2. On squared paper and on the same grid draw the graph of y = 2x - 8 and y = 7 – 3x   (4 marks)
    3. What is the nature of the two graphs you have drawn? (1 mark)
    4. Use your graphs to solve the simultaneous equations. (1 mark)
      3x + y = 7
      2x − y = 8
  8. The table below represent a surveyor’s a field –book record for a piece of land.
       Metres to D   
         250  
         130  90 to C
     To E 60     100  
          40  80 to B 
       From A  
    Calculate the area of the field in hectares. (10 marks) 


MARKING SCHEME

  1. Numerator = 4/5 (2 + ¼ − 3/8) ÷ (5/2 × 3/16
                    = 4/5 × 17/8 ÷ 15/32
                    = 4/5 × 15/8 × 32/1516/5
    Denominator = 3/5 × 16/548/25 
    Expression = 16/5 ÷ 48/2516/5 × 25/485/3 = 12/3       (3 marks) 
  2.  
    1. Distance between posts is the HCF of 608 and 264      (2 marks)
      608 = 2 × 2 × 2 × 2 × 2 × 19
      264 = 2 × 2 × 2 × 3 × 11 
      HCF= 23 = 8
      Distance between posts = 8m
    2. Number of posts = 2(l + b)
                                       8
                              =2(608 + 264)
                                          8
                              = 1744 = 218
                                    8                          (2 marks)
  3.  
    F2 MT2 Math Ans3 2021  
    Bearing of S from P72°                        (4 marks)
  4. = 4a(c−4a) − b(c−4a) +4
                   c−4a
    =(4a−b)(c−4a) +4
             c−4a
    =4a−b+4                               (3 marks) 
  5. Buying price (BP) = 288 × 10 =Sh 240
                                  12
    Selling price (SP) = 288  =Sh 320
                                  18
    Percentage profit = SP − BP × 100
                                     BP
                             = 320 − 240 × 100
                                     240
                             =331/3%                         (4 marks) 
  6. 3x − 2y = 7 ..........(i)    ( mult.eqn (i) by 1
    5x + y = 3   ..........(ii)     and eqn. (ii) by 2
    ⇒3x − 2y = 7
     10x + 2   = 6
      13x       = 13
             x=1
    Subst. for x in eqn. (ii)
    5 × 1 + y = 3
                y = 3 − 5
                   = −2
    Hence x = 1, y= −2                            (3 marks)
  7.  
    1. In what ratio did it decrease? (1 mark)
      120 :150 ⇒ 5:4
    2. new width = 4/5 × 140cm
                     = 112cm or 1.12m        (2 marks) 
  8. Given the following currency exchange rate, calculate to 3 significant figures the number of dollars that can be exchanged for 25 Sterling pounds.
    1 US dollar ($) = Ksh 76.85
    1 Sterling pound (£) = Ksh 115.30
    Convert sterling pounds into Kenya shillings
    25 Sterling pounds = Kshs 25 × 115.30
    Now convert Ksh. into dollars = 25 × 115.30
                                                       76.85
                                               =$ 37.5
  9. Volume of tank = πr2h where r = 70cm, h = 80cm 
                           = 22/7 × 70 ×70 × 80 litres
                                              1000
                           = 1232litres
    Fraction filled = 492.8
                            1232
                         = 4928 
                            12320
                         = 2/                (4 marks)
  10. Let daughters age be y years ⇒ man's age is 3y years.
    In 12 years time: duaghter will be (y+ 12) yeras old
    and man will be (3y+ 12 ) years old
    ∴ 3y + 12 = 2(y + 12)
      3y + 12 = 2y + 24
    3y − 2y = 24 − 12 ⇒ y = 12                       
    Hence daughter's age is 12 years and man's age is 36years (3 marks)
  11.  
    F2 MT2 Math Ans11 2021
            Let r = 5.81818181.........(i)
    then 100 r = 581.818181.........(ii)
    Subtract eqn (i) from eqn. (ii)
                 99r = 576
                    r= 576
                         99
                     = 64/11 or 59/11                     (3 marks)
  12. Marked Price (MP)  = sh. 450
    Selling Price (SP) = sh. 393.75
    % discount = MP − SP × 100
                            MP
                     =450 − 393.75 × 100
                                450
                     = 56.25 × 100
                         450
                     = 12.5%
  13. 4cm on map represents 20km
    1cm on map represents 5km = 5000m
    2.8cm on map rep 2.8 × 5000 = 14000m
    1.6cm on map rep. 1.6 × 5000 = 8000m
    ∴ Area of ranch = 14000 × 8000m2
                          = 14000 × 8000
                                       104
                         =11200ha                                       (3 marks) 
  14. Fraction spent on food and rent = 1/3 + ¼ = 4+3
                                                                       12     12
    Remainder = 1 − 7/125/12
    Fraction spent on transport = 3/5 of 5/123/5 × 5/12
    Fraction saved = 5/12 − ¼ = 5 − 3 = 2/12 or 1/6
    ∴ ¼ of salary = sh.1800
       1/6  = ?
    1800 × 1/6 × 4/= 1200
    =Kshs. 1200                                        (3 marks)
  15. base area = 22/7 × 14 × 14 × ½ × 2 +40 × 28 = 1736cm2 
    Curved surface area = (22/7 × 28 × ½ + 40 + 22/7 × 28 × ½ + 40) × 30 (4 marks)
                                 =5040cm2
    total area = 1736 + 5040 + 1736 = 8512cm2
    Metal needed = 110 × 8512 = 9363.2cm2
                            100
  16.  
    1. Angle PTR = 15 + 20 = 35º
    2. ∠20 + 90 + ∠QYT + (180 − 35) = 360
      ∠QYT = 105°

  17.  
    1.  
      F2 MT2 Math Ans17a 2021
    2. AC = 4.1cm ± 0.1cm
      ∠ACB = 122 ± 1°
    3.  
    4. radius = 5.2 ± 0.1cm
  18.  
    1.  
      F2 MT2 Math Ans18a 2021
    2.  
      1. Height TR = 5.5 × 5 = 27.5 ± 0.5m                  (2 marks)
      2. Distance OR = 5.5 × 5 = 27.5 ± 0.5m              (2 marks)
      3. The distance PR = 11 × 5 = 55 ± 0.5m             (2 marks)
  19.  
    1. Wholesaler paid 120 × 500 = sh 600
                              100
      Retailer paid      130 × 600 = sh 780
                              100
      Customer paid   150 × 780 = sh 1170
                              100                                         (3 marks)
    2. Let the amount paid by wholesaler be x 
      Retailer paid 130/100x = 1.3x
      Cutomer paid 150/100 × 1.3x = 1.95x
      ∴ 1.95x = 1560
             x = 1560
                   1.95
               = Sh. 800             (3 marks)
    3. Without the slae customer would have paid        (4 marks)
      1000 × 120 × 130 × 150 = Sh.2340
                  100    100    100
      Less 10% reduction = 90/100  × 2340
      ⇒ Selling price (SP)     = Sh 2106
      Buying Price (BP) for retailer
                                       = 1000 × 120/100 × 130/100
                                       = Sh. 1560
      ∴ %profit  = 2106 − 1560 × 100
                              1560
                     =35%
  20.  
    1. Area of front & back walls = 6.3 × 3.2 × 2
                                           =40.32m2
                  Area of side walls = 4.5 × 3.2 × 2
                                            =28.2m2
                   Area of floor = 6.3 × 4.5
                                      = 28.35m2 
      Total area of floor and walls = 40.32 + 28.8 + 28.35
                                               = 97.47m2 
      Area of door = 1.85 × 0.8 = 1.48m2
      Area of windows = 1.5 × 0.7 × 4 = 4.2m2
      Total area not cemented = 1.48 + 4.2
                                          = 5.68m2
      ∴ Area to be cemented = 97.47 − 5.68
                                        = 91.79 = 91.8m2             (5 marks)

    2. Cost of cementing materials = 91.8 × 500
                                               = sh. 45,900        (2 marks)
    3. Cost of labour = 20% of sh 45900
                           =20/100 × 45900
                           = sh 9180
      Total cost of cementing = 45900 + 9180
                                        = sh. 55,080            (3 marks)
  21.  
    1.  
      1. Mombasa to Mtito Andei time
           = (2400 − 1930) + 2:50 = 4:30+2:50
                                                =7h 20 min
      2. Mtito Andei to Nairobi time
        =1050 − 0335 = 7h 15min
      3. Nairobi to Nakuru time 
        =1900 − 1240 = 6h 20min
      4. Nakuru to Kisumu time
        =(2400 − 2015) + 9:00 = 3:45 + 9
                                           = 12h  45min       (4 marks)
    2. Stoppage time at Mtiti Andei
      = 0335 − 0250 = 45min
      Stoppage time at Nairobi
      =1240 − 1050 = 1h 50min
      Stoppage time at Nakuru 
      =2015 − 1900 = 1h 15min
      Total stoppage time 
      =45min + 1h 50min + 1h 15min
      =3h 50min
      Travellinig time from Mommbasa to Kisumu
      =7h 20min + 7h 15min + 6h 20min + 12h 45min = 33h 40min
      Time for whole journey 
      = 33h 40min + 3h 40min 
      =37h 30min                                                              (4 marks)
    3. Average speed = Distance covered = 1200
                                   Time taken          37.5
                            =32km/h                              (2 marks)
  22.  
    1.  
      1. Area of whole sheet = 80 × 50 cm2
        Area of sheet removed = 5×5×4cm2
        ∴ area of remaining part
        =(80 × 50) − (5×5×4)
        =4000 − 100
        =3900cm2                    (2 marks)
      2. Volume of metal cuboid = 3900 × 0.2
                                          = 780cm3
        mass of empty cuboid = mass of metal 
                                        = volume × density
                                        = 780 × 2.5g
                                        = 780 × 2.5 kg
                                               1000  
                                       =1.95kg                 (4 marks)
    2. Dimensions of the cuboid are l = 80 − 10 = 70cm  
      w = 50 − 10 = 40cm, h=5cm
      Capacity of cuboid = 70 × 40 × 5cm3
      Mass of water =  volume × density
                          = 70 × 40 × 4 × 1g
                         = 70 × 40 × 4 ×1 kg
                                  1000
                         =14kg                    (4 marks)
  23.  
    1.   
      1.  
         x   −2   −1   0   1   2   3   4    5 
         y  13  10  7  4  1  −2   −5   −8 
      2.  
         x   −4   −2   0   2   4   6   8    10 
         y  −16    −12   −8   −4   0  4  8   12
    2. Scale used:
      Horizontal axis; 1cm rep. 2 units
      Vertical axis: 2cm rep 5 units  
      (4 marks)
    3. Both graphs are straight lines 1 marks)
    4. x=3, y = −2   (1 marks)
  24.  
    F2 MT2 Math Ans24 2021
    AP=40m
    AQ = 100m
    AR = 130m
    AD = 250m
    PB = 80m
    QE = 60m
    RC = 180m
    Area of ∆ APB                = ½ × 40 × 80         = 1600m2
    Area of ∆ AQE                = ½ × 100 × 60       = 3000m2
    Area of trapezium BPRC  = ½(80 + 4=100)90 = 8100m2
    Area of ∆ DQE                = ½ × 150 ×60        = 4500m2
    Area of ∆ DRC                = ½ × 120 × 100     = 6000m2
    By addition, area of ABCDE                            =23200m2                   
    ∴ area of field = 2.32ha

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