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

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Mathematics Paper 2 Form 3 End Term 2 Exams 2021 with Marking Schemes

SECTION I (50MARKS)

Answer all questions in this section

  1. Evaluate using logarithms. [4 Marks]
    mathsp2endterm2q1
  2. A rectangular card measures 5.3 cm by 2.5cm.Find 
    1. The absolute Error in the area of the card. [2 Marks]
    2. The Percentage Error in the area of the card [2 Marks]
  3. The length of a room is 4m longer than its width. Find the length of the room if its area is 32m2. [3 Marks
  4. If 20 Men can lay 36m of a pipe in 8 hours. How long would 25 Men take to lay the next 54m of the pipe? [2 Marks]
  5. Expand (2 + x)5 in ascending powers of x up to the term in x3. Hence, approximate the value of (2.03)5 to 4s.f. (4 marks)
  6. Simplify by rationalizing the denominator; [2 Marks]
            3       
    2√3 − √2
  7. A scientific calculator is marked at sh. 1560. Under hire purchase it is available for a downpayment of sh. 200 and six monthly instalments of sh. 250 each. Calculate;
    1. The Hire purchase price. [2 Marks]
    2. The extra amount paid out over the cash price. [1 Mark]
  8. Solve the equation; [3 Marks]
    log⁡ (2x − 10) − 2 log 8 = 2 + log⁡ (9 − 2x)
  9. The Equation of a circle is given by x2 + y2 − 6x + 4y − 3 = 0 . Determine the center and the radius of the circle. [3 Marks]
  10. Make x the subject of the formula in the equation. [3 marks]
    mathsp2endterm2q10
  11. In the figure below, BT is a tangent to the circle to the circle at B. AXCT and BXD are straight lines. AX=6cm, CT=8cm, BX=4.8cm and XD=5cm.
    mathsp2endterm2q11
    Find the length of;
    1. XC [2 Marks]
    2. BT [2 Marks]
  12. Find the value of x if the matrix mathsp2eneterm2q12is a singular matrix. [3 Marks]
  13. The first term of an arithmetic sequence is −7 and the common difference is 4.
    1. List the first 6 terms of the sequence [2 Marks]
    2. Determine the sum of the first 30 terms of the sequence [2 Marks]
  14. The coordinates of points A and B are (2,5) and (8, −7) respectively. Find the
    1. Coordinates of M Which Divides AB in the Ratio 1:2 [2 Marks]
    2. Magnitude of AB [2 Marks]
  15. Tap A Fills a tank in 6 hours, tap B fills it in 8 hours and tap C empties it in 10 hours.Starting with an empty tank and all the three taps are opened at the same time, how long will it take to fill the tank. [3 Marks]
  16. Grade X of Tobacco Costs Sh.81.50 per Kg and grade Y cost sh 109 per Kilogram. In what ratio must the two grades be mixed in order to make a profit of 20% when the mixture sells at sh. 112.80 per kg. [3 Marks]

SECTION II: (50MARKS)

Answer any 5 questions from this section

  1. The figure below shows triangle OAB in which M divides OA in the ratio 2: 3 and N divides OB .in the ratio 4:1 AN and BM intersect at X.
    mathsp2endterm2q17
    1. Given that OA = a and OB = b, express in terms of a and b: (4mks)
      1. AN
      2. BM
    2. If AX = sAN and BX = tBM, where s and t are constants, write two expressions for OX in terms of.a, b s and t. Find the value of s and t. Hence write OX in terms of a and b (6mks)
  2. Kamau, Njoroge and Kariuki are practicing archery. The probability for Kamau hitting the target is 2/5 , that of Njoroge hitting the target is 1/4 and that of Kariuki hitting the target is 3/7
    Find the probability that in one attempt;
    1. Only one hits the target (2mks)
    2. All three hit the target (2mks)
    3. None of them hits the target (2mks)
    4. Two hit the target (2mks)
    5. At least one hits the target (2mks)
  3.  
    1. A matrix T is given by mathsp2endterm2q19. Find T-1 [2 Marks]
    2. Wanjiku bought 20 bags of maize and 25 bags of beans at a total cost of sh. 77,000. If she had bought 30 bags of maize and 20 bags of beans, she would have spent sh. 7,000 more. 
      1. Form a matrix equation from this information. [1 Mark]
      2. Determine the cost of a bag of maize and a bag of beans. [3 Marks
    3. She sold all the maize and beans at a profit of 10% on a bag of maize and 12½ % on a bag of beans. Calculate the total percentage profit. [4 Marks]
  4. At the beginning of the year 2000, Kanyora bought two houses, one in Thika and the other in Nakuru each at 1,240,000. The value of the house in Thika appreciated at a rate of 12% p.a.
    1. Calculate the value of the house in Thika after 9 years to the nearest shilling. [2 Marks
    2. After n years, the value of the house in Thika was 2,741,245 while the value of the house in Nakuru was 2,917,231. 
      1. Find n [4 Marks]
      2. Find the annual rate of appreciation of the house in Nakuru. [4 Marks]
  5. The table below shows income tax rates. 
     Taxable income 
     K£ per month
     Rate in shs. per K£
     1 - 325  2
     326 - 650  3
     651 - 975  4
     976 - 1300  5
     1301 - 1625  6
     over 1626  7
    Mr. Wafula earns a basic salary of 30,500. He has a house allowance of sh. 6,000 per month, medical allowance of sh. 4,000 per month and transport allowance of sh. 3,000 per month. He claims a tax relief of sh. 1,056 per month.
    1. Calculate
      1. Wafula’s taxable income in k£ per month. [2 Marks]
      2. Gross tax. [3 Marks]
      3. Net Tax [2 Marks]
    2. His net income per month has the following deductions

      Health insurance fund – sh. 150 
      Loan interest – sh. 200
      Service charge – sh. 200
      Sacco loan – sh. 2,500

      Calculate his net income per month. [3 Marks]
  6.  
    1. P varies jointly as Q and the square of R. P = 18 when Q = 9 and R = 15. Find R when P=32 and Q=81. [5 Marks]
    2. A varies Directly as B and inversely as the square root of C. Find the percentage change in A When B is decreased by 10% and C increased by 21%. [5 Marks]
  7.  
    1. The first term of an arithmetic progression is 2. The sum of the first 8 terms of the AP is 240.
      1. Find the common difference of the AP. [2 Marks]
      2. Given that the sum of the first n terms of the AP is 1,560. Find n [2 Marks]
    2. The 3rd, 5th and 8th terms of another AP from the first three terms of a G.P. If the common difference of the AP is 3. Find.
      1. The first term of G. P [4 Marks]
      2. The sum of the first 9 terms of the G.P to 4 s.f. [2 Marks]

  8.  
    1. Complete the table below for the function Y=2x+ 4x − 3 [2 Marks]
       x  −4  −3  −2  −1  0  1  2
       2x2  32      1  0    8
       4x  −8  −12  −8      4  8
       −3  −3  −3  −3  −3  −3  −3  −3
       y  21    −3        
    2. On the grid provided, draw the graph of the function y=2x2 + 4x − 3for −4 ≤ x ≤2 [3 Marks]
      mathsp1endt2q24b
    3. Use your graph to solve the roots of the quadratic equations.
      1. 2x2 + x − 5 = 0 [2 Marks]
      2. 2x2+3x − 2 = 0 [2 Marks]
      3. x2 + 4x − 3 = 0 (1 mark)

 

 

 

 

 

Marking Scheme

  1.   
    mathsp2endterm2ans1
  2.  
    1.  
      mathsp2endterm2ans2
       Minimum  Actual  Maximum
       5.25  5.3  5.35
       2.45  2.5  2.55

      Max. Area = 5.35 × 2.55
                      = 13.46 cm2
      Min. Area = 5.25 × 2.45
                     = 12.8625 cm2
      Absolute Error = max − min
                                      2
                           = 13.6425 − 12.8025 = 0.78 ✓1
                                           2

    2. % error = Absolute error × 100 ✓1
                      Actual area
                   =    0.78      × 100
                      5.3 × 2.5
                   = 5.8868 % ✓1
  3.  
    mathsp2endterm2ans3
    x(x+4) = 32
    x2 + 4x − 32 = 0
    x2 + 8x − 4x − 32 = 0
    x (x+4) − 4(x+8) = 0
    (x − 4)(x+8) = 0
    x = 4 or x = -8
    ∴ x =4 since length can't be negative
    length = x + 4 = 4+4 = 8m 
  4. Men Length Hours
    20    36       8
    25    54       ?
    = 25/20 × 54/36 × 8 ✓1
    = 15 hours ✓1
  5. (2+x)5
     Co-efficients  1  5  10  10  5  1
     Expansion  25x0  24x1  23x2  22x3  21x4  20x5
     Combined  32  80x  80x2  40x3  10x4  x5
    = 32 + 80x + 80x2 + 40x3 
    (2+x)5 = 2.035
    2+x = 2.03
        x = 2.03 − 2
        x = 0.03
    ∴ 2.035 = 32 + 80 (0.03) + 80 (0.03)2 + 40 (0.03)3
                = 32 + 2.4 + 0.072 + 0.00108 
                = 34.47308 ≈ 34.47 ✓1
  6.  
    mathsp2endterm2ans6
  7.  
    1. = 200 + (2500 × 6)
      = sh. 1700
    2. 1700 − 1560
      = sh. 140 
  8. log (2x − 10) = log (100 × (9 − 2x) )
    2x − 10 = 900 − 200x
       64                1
    2x − 10 = 57600 − 12800x
    12802x = 57610
    x = 57610
          12802
        = 4.5001
        ≈ 4.5

  9. (x2 − 6x + ___) + (y2 + 4y + ___) = 3 + ___+ ___
    x2 − 6x + 6/2 + y2 + 4y + 4/2 = 3 + (6/2) + (4/2)2
    (x − 3)2 + (y+2)2 = 16
    = centre (3, −2)
    = Radius = √16
                  = 4 units
  10.  
    mathsp2endterm2ans10
  11.  
    1. 6 (XC) = 4.8 × 5
      XC = 4.8 × 5
                   6
            = 4 cm
    2. BT2 = AT.CT
            = 18 × 8 
      BT = √144
          = 12 cm
  12. For singular matrix det = 0 
    ∴ {x(x − 3)} − (4 × 1) = 0
    x2 − 3x − 4 = 0
    x2 + x − 4x − 4 = 0
    x(x+1) − 4(x+1) = 0
    (x−4) (x+1) = 0
    x = −1 or 4
  13.  
    1. 1st term = −7
      2nd term = −7 + 4 = −3
      3rd term = −3 + 4 = 1
      4th term = 1 + 4 = 5
      5th term = 5 + 4 = 9
      6th term = 9 + 4 = 13
    2. S30 = 30/2{2(−7) + (30 − 1)4}
            = 15 {−14+116}
            = 1530

  14.  
    mathsp2endterm2ans14
  15. Tap A ⇒ 6 hours
    in 1hr = 1/6 of the tank
    Tap B ⇒ 8 hours
    in 1hr = 1/8 of the tank
    in 1hr both = 1/6 + 1/8
    Tap C ⇒ 10 hours to empty
    in 1hr = 1/10
    All three in hour = 7 − 1 = 35 − 12 23 
                              24   10      120        120
    23/20 = 1 hr
    1       = ?
    1 × 1 × 120/23
    = 5.22 hours
    = 5 hrs 13 min
  16.           Grade X    Grade Y    Mixture
    B.P     sh 81.50   sh. 109
    Ratio        x     :     y
    Total     81.50x      109y     81.50x + 109y
    120/100(81.50x + 109y) = 112.80
                      x + y
    97.8x + 130.8y = 112.80x + 112.80y
    97.8x − 112.80x = 112.80y − 130.8y
    −15x = −18y
    x = 18 ⇒ x:y = 6:5
    y    15
  17.  
    1.  
      1. AN = AO + ON
             = −a + 4/5b
      2. BM = BO + OM
             = −b + 2/5a
    2. OX = OA + AX
           =  a   + sAN
           = a + s (4/5b − a
           = a − sa + 4/5sb
           = a(1 − s) + 4/5sb
      Also:
      OX = OB + BX
           = b + t BM
           = b + t(2/5− b)
           = b − tb +2/5ta
           = b (1 − t) + 2/5ta

      a(1 − s) + 4/5sb  = b (1 − t) + 2/5ta
      1 − s = 2/5t
      t = 5/2 − 5/2s ------- (i)

      4/5s = (1 − t) --------(ii)
      4/5s = 1 − (5/2  − 5/2s)
      4/5s = 1 − 5/2 + 5/2s
      4/5s =  − 3/2  + 5/2s
      3/2 = 17/10s
      s = 15/17
      t = 5/2 − 5/2s
      t = 5/2 − 5/2 × 15/17
      t = 5/17
      ∴ OX = b(1 − 5/17) + (2/5 × 5/17)a
      = 12/17b + 2/17a
  18.  
    1. (2/5 × 3/4 × 4/7) + (3/5 × 1/4 × 4/7) + (3/5 × 3/4 × 3/7)
      = 6/35 + 3/35 + 27/140
      = 9/20
    2. P(HHH) = 2/5 × 1/4 × 3/7
      = 3/70
    3. P('H'H'H) = 3/5 × 3/4 × 4/7
      = 9/35
    4. (2/5 × 1/4 × 4/7) + (2/5 × 3/4 × 3/7) + (3/5 + 1/4 + 3/7)
      = 2/35 + 9/70 + 9/140 = 1/4
    5. P(1 or 2 or 3) = 1 − (P(none hits))
                           = 1 − 9/35
                           = 26/35
  19.  
    1.  
      mathsp2endterm2ans19a
    2.  
      1. 4         5       15400
        20m + 25b = 77000
        30m + 20b = 84000
        6          4       16800
        mathsp2matrixansq19bi
      2.  
        mathsp2matrixans19bii
    3. B.P = sh. 77,000
      Maize
      S.P = 1600 × 110
                    100
            = sh. 1760
      Beans
      S.P = 1800 × 112.5
                     100
      S.P = (20 × 1760) + (25 × 2025)
           = sh. 85, 825
      Profit = sh. 85825 − 77,000
              = sh. 8,825
      % Profit = 8,825  × 100 %
                     77,000
                  = 0.115 × 100 %
                  = 11.5 %
  20.  
    1. A = P (1 + r/100)n
      = 1,240,000 (1 + 12/100)9
      = 1,240,000 × 2.773
      = sh. 3,438,618
    2.  
      1. A = P (1 + r/100)n
        2,741,245 = 1,240,000 (1 + 12/100)n
           2.211                  1 
        2,741,245
         = 1,240,000
        1,240,00      1,240,000
        2.211 = (1.12)n
        log 2.211 = log 1.12 n

        log 2.211
        = n log 1.12
         log 1.12        log 1.12
        0.3445 = n
           0.0492
        n= 7 years
      2. 2,917,231 = 1,2400,000 (1 + r/100)7
         2,917,2311,2400,000(1 + r/100)7
        1,2400,000   1,2400,000
        2.3526 = (1 + r/100)7
        7√(2.3526) = 1 + r/100
        1.13 − 1 = r/100
        100 × 0.13 = r/100 × 100
        r = 13 % p.a
  21.  
    1.  
      1. T.I = B.S + T.A
            = 30,500 + 6000 + 4000 + 3000
                                     20
           = K£ 2175
      2. Gross tax
        325 × 2 = sh. 650
        325 × 3 = sh. 975
        325 × 4 = sh. 1300
        325 × 5 = sh. 1625
        325 × 6= sh. 1950
        550 × 7 = sh. 3850

        Gross tax = Ksh. 10,350
      3. net tax = Gross tax − relief
        = 10350 − 1056
        = Ksh. 9294
    2. Total dedcutions = 9294 + 150 + (200 × 2) + 2500
                              = Shs. 12,344
      Net Income = Sh. 43500 − 12,344
                       = Shs. 31,156
  22.  
    1. P ∝ QR2
      P = kQR2
      18 = k × 9 × (15)2
      18 = 225 × 9 × k
        18   = 2025k
      2025    2025
      k = 2/225
      ∴ P = 2/225 QR2
      32 = 2/225 × 81 × R2
      225 × 32 = 162R2 × 225
          162         225       162
      444/9 = R2
      R = √444/9
         = 62/3
    2. A ∝  B/√C
      A = kB/√C
      A = k × 0.9B
            √(1.21C)
      A = 0.9kB
            1.1√C
         = 0.818kB
                    √C
      B = 100%
         = 90%
      = 90 × 13 = 0.9B
            100 
      C = 100%
         = 121%
      121 × C = 1.21C
          100
      Δ in A: 0.818kB − kB
                        √C    √C
      = k  B(0.818 − 1)
           √C
      % Δ in A =  B/√C(0.818 − 1) × 100 %
                                    kB/√C
                   = − 0.1818 × 100%
      A decreases by 18.18%
  23.  
    1.  
      1. S8 = n/2 (2a + (n−1)d)
        240 = 8/2 ((2×2) + (8 −1)d)
        240 = 4 (4+7d)
        240 = 16 + 28d
        28d = 224 
        28       28 
        d = 8
      2. 2 × 1560 = n/2 ((2×2) + (n − 1)8) × 2
        3120 = n (4+8n − 8)
        3120 = 4n + 8n2 − 8n
           4       4      4        4
        780 = n + 2n− 2n
        780 = 2n2 − n
        2n2 − n − 780 = 0
        n = −(−1) ± √(1+ 4×2×780)
                                4
        = 80/2
        n = 40 times
    2.  
      1. a+2d, a+4d, a+2d ....G.P
        a+4d = a+7d
        a+2d    a+4d
        a2 + 8ad + 16d2 = a2 + 9ad +14d2
        16d2 − 14d2 = 9ad − 8ad
        2d2 = ad
          d       d
        a = 2d
        but d=3
        ∴ a = 6 (first term A.P)
      2. C.R = 18/12 = 3/2
        Sn = a (rn − 1
                    r − 1
        S9 = 12(1.59 − 1)
                    1.5 − 1
        = 898.6
  24.  
    1.  
       x  −4  −3  −2  −1  0  1  2
       2x2  32  18  8  1  0  2  8
       4x  −8  −12  −8 −4  0   4  8
       −3  −3  −3  −3  −3  −3  −3  −3
       y  21  3  −3  −6  −3  3  13
    2.  
      mathsp2endterm2ans24
    3.  
      1.   y = 2x2 + 4x −3
        −0 = 2x2 + x −5
        y = 3x + 2
         x  0  1
         y  2  5
        x = −2 or 1.4 ±0.1
      2.   y = 2x2 + 4x −3
        −0 = 2x2 + 3x −2
           y = x  − 1
         x  0 3
         y  −1  2
        x = 0.4 or −2 ± 0.1
      3. x = 0.6 or -2.6 ± 0.1

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