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

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INSTRUCTIONS

  • This paper consists of two sections. Answer all the questions in section 1 and any 5 questions in section 2.
  • KNEC mathematical tables and non-programmable calculators may be used when necessary

SECTION 1

Answer all the questions in this section

  1. Evaluate 36 –8×–4 –15÷ –3       (3mk)
                  –3×–3–8(–6+ –2) 
  2. Simplify a+b2a–b        (3mk)
                  2         3 
  3. Find the greatest number which divides 181 and 236 and leaves a remainder of 5 in each case (2mk)
  4. A rectangle measures 20cm by 15cm.If each dimension is increased by 2.5cm, by what percentage is
    1. the perimeter of the rectangle increased (2mk)
    2. the area of the rectangle is increased (2mk)
  5. The angle of elevation of the top of a tree from a point P on horizontal ground is 30º. From another point Q 8 metres from the base of the tree, the angle of elevation of the top of the tree is 48º.
    1. Calculate to one decimal place the height of the tree. (1mk)
    2. Calculate the distance between P and Q (2mk)
  6. Given that cos θ= –0.8070 , find θ for 0≤θ≤720 (3mk)
  7. A piece of wire 40cm is bent to form a right-angled triangle whose hypotenuse is 17cm long. Find the lengths of the other two sides of the triangle (4mk)
  8. Solve for x in log 5–2+log(2x+10)=log(x–4) (3mk)
  9. Solve the quadratic equation by completing of squares giving your answer to 3sf     (4mk)
    11x2 –13x+3=0 
  10. Rationalize the denominator and simplify      (4mk)
    4√5+3√2 
     2√2–√5
  11. Use a calculator to work out

    1. work out using calc     (1mk)

    2. solve using calc (1mk)
  12. A tourist from Kenya left for Ethiopia. He exchanged sh 9898 into Ethiopian Birr at the rate of 1 Eth.Birr=ksh7.95. He spent ¾ of the money he got and converted the balance back to Kenyan money at the rate of 1Eth.Birr=ksh7.98 Calculate what he finally got to 2dp (3mk)
  13. Simplify the expression    (3mks)
    simplifying an expression
  14. Three types of tea costing sh203,sh146 and sh197 per kg are blended in the ratio of 2:5:k. Find the value of k,if the blend when sold at sh221 per kg gives 30% profit    (3mk)
  15. A two digit number is such that 4 times the units digit exceeds the tens digit by1. If the digits are reversed, the number formed is decreased by 45. Find the number. (3mk)
  16. A triangular field has dimensions 21m by 52m by 47m.
    1. calculate the area of the field to the nearest m2. (2mk)
    2. calculate the length of a straight ditch dug from the largest angle meeting the opposite side at right angles. (1mk)

SECTION 2

Answer any 5 questions in this section

  1. A bookseller bought a number of cartons of books at a cost of ksh 57600 from Kagumo bookstore. Had he bought the same books from Kerugoya bookstore, it would have cost him ksh 480 less per carton. This would have enabled him to buy 4 extra cartons of books for the same amount of money. By taking x to be the number of cartons of books he actually bought;
    1. write an expression in x
      1. for the cost of each carton he bought at Kagumo bookstore (1mk)
      2. for the cost of each carton had he bought from Kerugoya bookstore (1mk)
    2. find the value of x (6mk)
    3. The bookseller later sold all the books he had bought each carton at ksh 720 more than he had paid for it. Determine the percentage profit he made (2mk)
  2. A cylindrical metal bar of diameter 14cm and length 2m is melted and moulded into spherical balls. In the process, 5% by volume of metal is lost and what remains makes balls of radius 3.5cm.
    1. calculate the volume of metal used to make the balls. (3mk)
    2. find to the nearest whole number the number of balls made (3mk)
    3. find the total surface area of the metal bar (2mk)
    4. find the total surface area of the balls made (2mk)
  3.  
    1. Draw the graph of y=(2x–3)(x–1) for the interval –2 ≤ x ≤ 4 (6mk)
    2. use your graph to solve
      1. 2x2–5x+3=0 (1mk)
      2. 2x2=3x+2 (3mk)
  4.  
    1. Plot triangle ABC with coordinates A(1, 1), B(3, 1) and C(1,3 ) (1mk)
    2. Plot A′B′C′ the image of ABC under an enlargement scale factor 2 centre A and write down it’s coordinates (2mk)
    3. Plot A′′B′′C′′ the image of A′B′C′ under a reflection in the line x+y=0 and write down it’s coordinates (3mk)
    4. A′′B′′C′′ is then reflected in the line y=0 to give A′′′B′′′C′′′.Give the coordinates of A′′′B′′′C′′′ (2mk)
    5. Describe fully a rotation that maps A′′′B′′′C′′′ onto A′B′C′ (2mk)
  5. Three businessladies Wanjiku, Muthoni and Njoki decided to buy a lorry. The .marked price of the lorry was 2.8million shillings. The dealer agreed that the ladies could pay a deposit of 60% of the marked price and the rest to be paid within a year.The ladies raised the deposit in the ratio of 3:2:5 respectively. At the end of the year the lorry had realized 2.08million shillings which the three shared in the ratio of their contribution. However, they were required to contribute for the balance of the lorry from these earnings again in the ratio of their original contributions.
    1. calculate amount to be paid as deposit (1mk)
    2. how much did each contribute to pay for the deposit? (3mk)
    3. how much did Njoki receive at the end of the year? (1mk)
    4. calculate the total amount Muthoni and Njoki contributed to pay for the balance. (3mk)
    5. how much money did Wanjiku remain with after paying her share of the balance? (2mk)
  6.  
    1. Make a table and draw the graph of y= sinx–cosx for x in the range 0°≤x≤360°, with y values to 2dp and x intervals of 30° (6mk)
    2. use the graph to find the value of y when
      1. x=75° (1mk)
      2. x=255° (1mk)
    3. find the values of x for which y= –0.9 (2mk)
  7. The figure below is a segment of a circle centre O radius r units.CM is the perpendicular bisector of AB. 
    figure showing segment of a circle
    Given that CM=1cm and AB=2√3cm,
    1. calculate the radius of the circle centre O from which the segment was cut (3mk)
    2. calculate the angle that chord AB subtends at the centre of the circle (2mk)
    3. hence calculate
      1. the length of arc ACB (2mk)
      2. the area of the segment AMBC (3mk)
  8. A rectangular sheet of metal which measures 120cm by 0.8m is 1.5mm thick and is made of material whose density is 2.2 g/cm3. From each of the four corners of the rectangle, a square of side 10cm is cut off and the remaining part folded to form an open cuboid.
    1. calculate
      1. the capacity of the cuboid in cm3 to the nearest whole number (3mk)
      2. the mass of the empty cuboid in kg to the nearest whole number (3mk)
    2. the cuboid is filled with a liquid whose density is 0.75g/cm3. Calculate the mass in kg of the cuboid when full of the liquid (2mk)
    3. calculate the mass of metal lost in kg (2mk)

Marking Scheme

SECTION 1

Answer all the questions in this section

  1. Evaluate 36 –8×–4 –15÷ –3       (3mk)
                  –3×–3–8(–6+ –2) 
    numerator 36 + 32 + 5 = 73            M1
    denominator 9 − 8(−) = 9+64=73    M1
      Ans  73/73 =1                                A1
  2. Simplify a+b – 2a–b        (3mk)
                  2         3
    3(a+b) −2(2a−b)      M1
               6
    3a + 3b − 4a +2b     M1
               6
    =5b − a  or  −a+5b   A1
          6                6
  3. Find the greatest number which divides 181 and 236 and leaves a remainder of 5 in each case (2mk)
    GCD of 181−5 and 236−5
     =GCD  M1
     The number is 11     A1
  4. A rectangle measures 20cm by 15cm.If each dimension is increased by 2.5cm, by what percentage is
    1. the perimeter of the rectangle increased (2mk)
      Original perimeter  = 2(20+15) =70
                   Increase = 4 × 2.5 =10                 M1
                        %Inc = 10/70 × 100 =14.29%   A1
    2. the area of the rectangle is increased (2mk)
      Original area  =20×15=300
      Area with increase =(2.5 +20) (2.5 + 15) = 393.75
      Increase= 93.75                                                     M1
           %inc= 93.75 ×100 =31.25%                              A1
                       300
  5. The angle of elevation of the top of a tree from a point P on horizontal ground is 30º. From another point Q 8 metres from the base of the tree, the angle of elevation of the top of the tree is 48º.
    1. Calculate to one decimal place the height of the tree. (1mk)
      angle of elevation
      Tan 48 = h/8
             h = 8 tan 48 =8.885      B1
    2. Calculate the distance between P and Q (2mk)
      Tan 30 = 8.885                   M1
                      PF
            PF = 8.885 = 15.39
                    tan30
           PQ = 15.39 − 8 =7.389   A1
  6. Given that cos θ= –0.8070 , find θ for 0≤θ≤720 (3mk)
    θ = cos−1 (−0.8070)
    Acute θ =36.20°
    2nd & 3rd quad
    θ = 180 − 36.20 = 143.8°
    θ = 180 + 36.20 =216.2°
    Beyond 360°  θ = 143.8 + 360 = 503.8°
                         θ = 216.2 + 360 = 576.2°
  7. A piece of wire 40cm is bent to form a right-angled triangle whose hypotenuse is 17cm long. Find the lengths of the other two sides of the triangle (4mk)
    right angled triangle
    a+b = 23.....(i)
    a2 + b2 =289....(ii)              M1
    Substitute (i) in (ii)
    a2 + (23 − a)2 =289
    a2 + 529 − 46a + a2 =289
    2a2 − 46a + 240 =0
      a2 − 23a +120 =0              M1
      a2 − 8a − 15a + 120 = 0    M1 -Attempt to solve correctly
    a(a − 8) −15(a−8) = 0
                  a=15 or 8
                  b=8  or 15              A1
  8. Solve for x in log 5–2+log(2x+10)=log(x–4) (3mk)
    log 5 − 2 log10 + log(2x+10) = log(x−4)       
    Log (5/102  × (2x +10) ) =log(x−4)                
    x/10 + ½ = x − 4
    −9/10x = −9/2 
            x = −9 ×  10 
                    2     −9
               =5
  9. Solve the quadratic equation by completing of squares giving your answer to 3sf     (4mk)
    11x2 –13x+3=0 
    x2 − 13/11 = −3/11 
    x2 −13/11 + (−13/11)2 = −3/11 + (−13/11)2

    (x − 13/11)2−3/11 169/121
                       =136/121
    x − 13/11 =±√136/121
    either x = √136/121 + 13/11
               =11.66 +13 =2.242
                    11      11
    Or x = −√136/121 13/11
               =11.66 +13 = 0.1218
                           11
  10. Rationalize the denominator and simplify      (4mk)
    4√5+3√2 
     2√2–√5
    (4√5+3√2)(2√2+√5)
     (2√2–√5) (2√2+√5)
    Num
    8√10+20+12+3√10

    =32+11√10
    den
    8+2√10 − 2√10 − 5

    =3
    Ans = 32+11√10
                    3
  11. Use a calculator to work out

    1. work out using calc     (1mk)
      =0.6816   B1

    2. solve using calc    (1mk)
      =0.2030   B1
  12. A tourist from Kenya left for Ethiopia. He exchanged sh 9898 into Ethiopian Birr at the rate of 1 Eth.Birr=ksh7.95. He spent ¾ of the money he got and converted the balance back to Kenyan money at the rate of 1Eth.Birr=ksh7.98 Calculate what he finally got to 2dp (3mk)
    1 Eth. Birr =sh 7.95
         ?          =sh 9898
    =Birr 1245.03                M1
    Bal after spending
    =¼ × 1245.03 = 311.26   M1
    1 Eth. Birr =sh7.98
        311.26  =    ?   
        =311.26 × 7.98
        =sh 2483.80              A1
  13. Simplify the expression    (3mks)
    simplifying an expression
    num  (3t−5a)(3t+5a)         M1
    den 6t2 +9at+10at+15a2
        =3t(2t+3a) +5a(2t+3a)
        =(3t+5a)(2t+3a)           M1
    (3t−5a)(3t+5a)  = 3t−5a
    (3t+5a)(2t+3a)     2t+3a    A1

  14. Three types of tea costing sh203,sh146 and sh197 per kg are blended in the ratio of 2:5:k. Find the value of k,if the blend when sold at sh221 per kg gives 30% profit    (3mk)
    130% =221
    100% = ?
    =22100 =sh 170
         130
      2   (203) +  5   (146)  +  K   (197)  =170
    7+k              7+k              7+k
    406+730+197k =170(7+k)
          1136+197k =1190+170k
                     27k = 54
                          k=2
  15. A two digit number is such that 4 times the units digit exceeds the tens digit by 1. If the digits are reversed, the number formed is decreased by 45. Find the number. (3mk)
    let the number be xy
    4y − x =1......(i)
    10x+y−(10y+x) =45
    10x+y−10y−x =45
     9x−9y =45......(ii)        M1
    From (i) x=4y−1, subst in (ii)
    9(4y−1)−9y =45
    36y−9−9y =45
       27y = 54
           y=2                      M1
    x = 4 × 2−1
       =8−1 =7
    The no. is 72                     A1
  16. A triangular field has dimensions 21m by 52m by 47m.
    1. calculate the area of the field to the nearest m2. (2mk)
      area of triangular field
      Area
      S=½(47+21+52) =60
      A= √(60(60−52)(60−47)(60−21))
        = √(60×8×13×39) =493.3m2
        =493m2
    2. calculate the length of a straight ditch dug from the largest angle meeting the opposite side at right angles. (1mk)
      ½ × 52 ×h =493  
         h= 493 × 2 = 18.96m
                  52

SECTION 2

Answer any 5 questions in this section

  1. A bookseller bought a number of cartons of books at a cost of ksh 57600 from Kagumo bookstore. Had he bought the same books from Kerugoya bookstore, it would have cost him ksh 480 less per carton. This would have enabled him to buy 4 extra cartons of books for the same amount of money. By taking x to be the number of cartons of books he actually bought;
    1. write an expression in x
      1. for the cost of each carton he bought at Kagumo bookstore (1mk)
        57600             B1
            x
      2. for the cost of each carton had he bought from Kerugoya bookstore (1mk)
        57600  − 480  B1
                   x
    2. find the value of x (6mk)
      No. of cartons bought from Kerugoya bookstore = x + 4    M1
      (x + 4) ( 57600 − 480)  = 57600      M1
                   (        x         )

      57600 − 480x + 230400 − 1920 = 57600  M1
                                      x
      55680x − 480x2 + 230400 =57600x     M1
      −1920x −480x2 +230400 =0
      x2 + 4x − 480 =0                     M1
       x2 − 20x+ 24x − 480 =0
      x( x−20) + 24(x−20) =0
      x=20
      x=−24       A1
      No of cartons = 20    B1
    3. the bookseller later sold all the books he had bought each carton at ksh 720 more than he had paid for it. Determine the percentage profit he made (2mk)
      720 × 20  × 100 =25%
        57600
  2. A cylindrical metal bar of diameter 14cm and length 2m is melted and moulded into spherical balls. In the process, 5% by volume of metal is lost and what remains makes balls of radius 3.5cm.
    1. calculate the volume of metal used to make the balls. (3mk)
      22/7 × 7 ×7 × 200 =30,800cm3  M1
      Remaining = 95/100 × 30800   M1
                      = 29,260cm3    A1
    2. find to the nearest whole number the number of balls made (3mk)
      4/3 × 22/7 × 3.53 = volume of 1 ball
      No. of balls = 29260 × 21        M1
                            4×22×3.53
                      =162.86                A1
                      =162 balls             B1
    3. find the total surface area of the metal bar (2mk)
      2 × 22/7 × 72 =308
       C.S = 2× 22/7 × 7 ×200 =8800   M1
      T.S.A =8800 + 308
              =9108cm2                        A1
    4. find the total surface area of the balls made (2mk)
      162 × 4 × 22/7 × 3.52    M1
                   =24948cm2      A1
  3.  
    1. Draw the graph of y=(2x–3)(x–1) for the interval –2 ≤ x ≤ 4 (6mk)
      y=2x2 − 2x − 3x +3
      y=2x2 − 5x +3
       x   −2  −1   0   1     2   3   4 
       2x2   8   2  0  2    8  18   32 
       −5x  10   5  0  −5   −10   −15   −20 
         3   3   3  3   3   3   3    3
        y  21  10  3    0   1   6   15
      B1 B1 B1        S1  P1  C1
      graph for question 19
    2. use your graph to solve
      1. 2x2–5x+3=0 (1mk)
        y=0
        x=0
      2. 2x2=3x+2 (3mk)
           y=2x2 − 2x − 3x +3    B1
        −y=2x2 − 5x +3
          0=2x2 − 3x −2
          y =−2x +5
         x   0   −1   1 
         y   5   7  3
        x=2
        y=−0.45
  4.  
    1. Plot triangle ABC with coordinates A(1, 1), B(3, 1) and C(1,3 ) (1mk)
      Triangle ABC plotted with coordinates
    2. Plot A′B′C′ the image of ABC under an enlargement scale factor 2 centre A and write down it’s coordinates (2mk)
      A′(1,1) B′(5,1) C′(1,5)
      B1  B1   B1
    3. Plot A′′B′′C′′ the image of A′B′C′ under a reflection in the line x+y=0 and write down it’s coordinates (3mk)
      A′′(−1,−1)  B′′(−1,−5)  C′′(−5,−1)
      L1   B1  B1
    4. A′′B′′C′′ is then reflected in the line y=0 to give A′′′B′′′C′′′.Give the coordinates of A′′′B′′′C′′′ (2mk)
      A′′′(−1,−1)  B′′′(−1,5)  C′′′(−5,−1)   
      B1  B1
    5. Describe fully a rotation that maps A′′′B′′′C′′′ onto A′B′C′ (2mk)
      Rotation −90° about (0,0)
      B1  B1
  5. Three businessladies Wanjiku, Muthoni and Njoki decided to buy a lorry. The .marked price of the lorry was 2.8million shillings. The dealer agreed that the ladies could pay a deposit of 60% of the marked price and the rest to be paid within a year.The ladies raised the deposit in the ratio of 3:2:5 respectively. At the end of the year the lorry had realized 2.08million shillings which the three shared in the ratio of their contribution. However, they were required to contribute for the balance of the lorry from these earnings again in the ratio of their original contributions.
    1. calculate amount to be paid as deposit (1mk)
      60/100 × 2800000 = sh. 1,680,000   B1
    2. how much did each contribute to pay for the deposit? (3mk)
      Wanjiku = 3/10 × 1680000 =504,000   B1
      Muthoni = 2/10 × 1680000 =336,000   B1
      Njoki  = 5/10 × 1680000 =840,000      B1
    3. how much did Njoki receive at the end of the year? (1mk)
      5/10 × 2080000 = 1,040,000     B1
    4. calculate the total amount Muthoni and Njoki contributed to pay for the balance. (3mk)
      bal = 40/100 × 2800000  = 1,120,000    M1
      7/10 × 1,120,000 = 784,000    M1 A1
    5. how much money did Wanjiku remain with after paying her share of the balance? (2mk)
      Received = 3/10 × 2080000 = 624,000
      Paid 3/10 × 1120000 = 336,000   M1
                     remained  = 288,000    A1
  6.  
    1. Make a table and draw the graph of y= sinx−cosx for x in the range 0°≤x≤360°, with y values to 2dp and x intervals of 30° (6mk)
       x  0 30 60 90 120 150 180 210 240 270 300 330 360
      Sin x  0 0.5 0.87  1 0.87 0.5   0 −0.5 −0.87 −1 −0.87 −0.5  0
      Cos x 0.87 0.5   0  −0.5 −0.87 −1  −0.87 −0.5  0  0.5 0.87  1
       y −1 −0.37 0.37  1  1.37  1.37  1 0.37 −0.37 −1 −1.37 −1.37 −1

      graph representing table
      B1 B1 B1   S1 P1 C1
    2. use the graph to find the value of y when
      1. x=75° (1mk)
        0.75 ± 0.05      B1
      2. x=255° (1mk)
        0.675 ± 0.05    B1
    3. find the values of x for which y= –0.9 (2mk)
      6°, 264°    ±6°   B1 B1
  7. The figure below is a segment of a circle centre O radius r units.CM is the perpendicular bisector of AB. 
    figure showing segment of a circle
    Given that CM=1cm and AB=2√3cm,
    1. calculate the radius of the circle centre O from which the segment was cut (3mk)
      radius of centre O calculations
      r2=(r − 1)2 + (√3)2    M1
      r2 =r2 − 2r + 1 + 3     M1
      2r = 4
      r=2cm                       A1
    2. calculate the angle that chord AB subtends at the centre of the circle (2mk)
      Cos θ = ½ = 0.5
            θ =60º                   M1
      Angle at centre =120°   A1
    3. hence calculate
      1. the length of arc ACB (2mk)
        120/360 × 2 × 22/7 ×2 =4.190       M1 A1
      2. the area of the segment AMBC (3mk)
        120/360 × 22/7 × 22 − ½ ×22 sin 120   M1 M1 
        =4.190 − 1.7321 = 2.4579       A1
  8. A rectangular sheet of metal which measures 120cm by 0.8m is 1.5mm thick and is made of material whose density is 2.2 g/cm3. From each of the four corners of the rectangle, a square of side 10cm is cut off and the remaining part folded to form an open cuboid.
    1. calculate
               rectangular sheet of metal B1
      1. the capacity of the cuboid in cm3 to the nearest whole number (3mk)
        99.7 × 59.7 × 9.85   M1
        =58628cm3               A1
      2. the mass of the empty cuboid in kg to the nearest whole number (3mk)
        Outer volume = 100 × 60 × 10
                                      =  60,000
        Inner volume       =−58,628         M1
        Volume of material    1372cm3     M1
        Mass = 1372 × 2.2  =3kg             A1
                        1000
    2. the cuboid is filled with a liquid whose density is 0.75g/cm3. Calculate the mass in kg of the cuboid when full of the liquid (2mk)
      Mass of liquid
      =0.75 × 58628 = 43.971kg   M1
               1000
      Total mass = 46.971kg          A1
    3. calculate the mass of metal lost in kg (2mk)
      Volume = 4 × 10 × 10 × 0.15     M1
      =60cm3
      Mass = 60 × 2.2 =132g              A1

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