Mathematics Paper 2 Questions and Answers - Kangundo Subcounty Pre Mock Exams 2021/2022

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INSTRUCTIONS TO THE CANDIDATES

  • Write your name and index number in the spaces provided above
  • This paper contains two sections; Section A and Section B
  • All workings and answers must be written on the question paper in the spaces provided below each question.
  • Marks may be given for correct working even if the answer is wrong.
  • Show all the steps in your calculations, giving your answers at each stage in the spaces below each question.
  • This paper consists of 15 printed pages.
  • Candidates should check carefully to ascertain that all the pages are printed as indicated and no questions are missing.

Section A. (50mks)
Answer all the questions in this section in the spaces provided.

  1. Find the value of x that satisfies the equation. (3mks)
    log⁡ (2x-11)- log⁡2= log⁡3 - log⁡ x 
  2. The base and the height of a right angled triangle were measured as 6.4cm and 3.5cm respectively. Determine to 1 decimal place the percentage error in calculating the area of the triangle. (3mks)
  3. The figure below shows a quadrilateral ABCD in which AB = 8cm, DC =12cm <BAD= 45º <CBD= 90º and <BCD =30º
    1
    1. The length of BD. (1mk)
    2. The size of angle ADB . (2mks)
  4. Simply the expression√48/√5+√3 leaving the answer in the form a√b+c where a, b and c are integers. (3mks)
  5.    
    1. Expand (1+x)7 up to the 4th term (1mk)
    2. Use the expansion in part (a) above to find the approximate value of (0.94)7 to 3 decimal places. (2mks) 
  6. A variable P varies directly as t3 and inversely as the square root of S. When t=2 and S = 9 P= 16. Determine the equation connecting P, t and S hence find P when S = 36 and t = 3. (3mks)
  7. Given that 2Find the values of x for which AB is a singular matrix. (4mks) 
  8. Use completely the square method to solve
    3x2+8x-6=0 Correct to 3 significant figures. (3mks)
  9. In the figure below the tangent ST meets chord Vu. Produced at T. chord SW passes through the centre O of the circle and intersect chord Vu at x. Line ST = 12cm and uT= 8cm
    3
    1. Calculate the length of chord Vu. (1mk)
    2. If wx = 3cm and Vx:xu = 2:3 . find Sx
  10. Make n the subject of the formula.
    r/PM/√n-1 (2mks)
  11. The equation of a circle is given by x2+4x+y2-2y-4=0. Determine the centre and radius of the circle. (3mks)
  12. The 5th term of an AP is 82 and the 12th term is 103.
    Find
    1. The first term and the common difference. (2mks)
    2. The sum of the first 21 terms. (2mks)
  13. Use matrices to solve the simultaneous equation. (3mks)
    2x+3y=39
    5x+2y=81
  14. Solve the following pair of simultaneous inequalities and illustrate the value on a number line. (3mks)
    3x-1 >-4
    2x+1 ≤7
  15. A triangle has sides 10cm, 7cm and 9cm. find
    1. The area. (2mks)
    2. The size of angle <BAC. (2mks)
  16. A plot of land is valued at sh 1,250,000 due to increase in demand its appreciates at the rate of 6% every six months. What will be its value after 3 ½ years. (3mks)

    Section B (50mks)
    Answer any five questions from this section on the spaces provided.
  17. In a mixed school there are 420 boys and 350 girls. The probability that a girl passes her exams in the school is 4/7 while that of a boy passing is 5/8. The probability of a girl being made a prefect is 2/11 while that of a boy is 1/8.
    1. Find the probability that a student picked at random.
      1. Is a boy and passes the exam and is not a perfect. (3mks)
      2. Is a girl, a prefect and passes the exam. (3mks)
      3. Is not as prefect and passes the exam. (4mks)
  18. The table below shows some values of the curves.
    y=2 sin⁡x and y=3 cos⁡x
    1. Complete the table for the values of y=2 sin⁡x and y=3 cos⁡x correct to 1 decimal place. (2mks)
      30º  60º  90º  120º  150º  180º  210º 240º 270º 300º 330º 360º
      y=2 sin⁡x 0 1   2   1 0 -1 -1.7       0
      y=3 cos⁡3 3   1.5 0   -2.6     -1.5       3
    2. On the grid below draw the graph of y=3 cos⁡x and y=2 sin⁡x for 0º ≤x ≤360º (5mks)
      2
    3. Use the graph to find the values of x when 3 cos⁡3-2 sin⁡x=0 (2mks)
    4. Find the difference in amplitude of y=3 cos⁡x and y=2 cos⁡x (1mk)
  19. In the figure below , OP = P ,OQ = ⏟q PQ∶ QR =1:1 and OQ: QS = 3:1
    4
    1. Determine, in terms of P and q
      1. PQ (1mk)
      2. RS (2mks)
    2. If RS:ST = 1:K and OP:PT = 1:n
      Determine.
      1. ST in terms of P, q and K. (2mks)
      2. The values of K and n. (5mks)
  20. Using a ruler and a compass only construct.
    1. Triangle PQR such that PQ = 6cm , <PQR = 60º and <RPQ = 45º (4mks)
    2. Locate the point A in the triangle when is equidistant from all the three sides of triangle PQR. (3mks)
    3. Find the distance of A from the sides of the triangles. (1mk)
    4. Drop a perpendicular height to PQ and Measure its height. (2mks)
  21. The figure below represents a cuboids EFGHJKLM in which EF= 40cm FG=9cm and GM= 30cm. N is the midpoint of LM.
    5
    Calculate correct to 3 significant figures
    1. The length of GL. (2mks)
    2. The length of FJ. (3mks)
    3. The angle between EM and the plane EFGH. (3mks)
    4. The angle between the planes EFGH and ENH. (2mks)
  22. The table below shows income tax rates for a certain year.
    Monthly income in Kshs  Tax rate in each shillings 
    1 – 9400  10% 
    9401 – 18000  10% 
    18001 – 26600 20%
    26601 – 35600  25% 
    35601 –and above  30%
    A monthly tax relief of Khs 1172 was allowed. Opunyi’s taxable income in the last band was Ksh 3,200 in month.
    1. Calculate
      1. His taxable income per month. (2mks)
      2. The amount of tax he paid in a month. (5mks)
      3. Opunyi’s salary included a medical allowance of Shs 8000. He contributed 6% of his basic salary to a sacco. Calculate his net pay. (3mks)
  23. The masses of 100 patients in a hospital were distributed as shown in the table below.
    Mass (Kg)  0-9 10-19 20-29 30-39 40-49 50-59 60-69 70-79 80-89 90-99
    Frequency 3 7 8 9 12 18 25 10 6 2
    1. State the modal class. (1mk)
    2. Calculate
      1. The mean mass of the patients. (3mks)
      2. The standard deviation of the distribution. (3mks)
    3. Find the interquartile range for the data. (3mks)
  24. OABC is a parallelogram with vertices O (0, 0) A (2, 0) B (3, 2) and C (1, 2).
    O1 A1 B1 C1 is the image of OABC under transformation matrix 6
    1. Find the co-ordinates of O1 A1 B1 C1. (2mks)
    2. On the grid provided draw OABC and O1 A1 B1 C1 (2mks)
    3. Find O11 A11 B11 C11 the image of O1 A1 BC1 under the transformation matrix 7 (2mks)
      1. On the same grid draw O11 A11 B11 C11 . (1mk)
    4. Find the single matrix that map O11 A11 B11 C11 onto OABC. (3mks)


MARKING SCHEME

  1. log⁡(2x-11)- log⁡2= log⁡3- log⁡x
    2x-11/2 = 3/x
    x(2x-11) = 6
    2x2 – 11x – 6 = 0
    2x2 – 12x + x – 6 = 0
    2x(x-6) + 1(x – 6) = 0
    2x + 1 = 0 x- 6= 0
    2x = -1 x =6
    X= - ½
    M1 for 2x-11/2 = 3/3
    M1 for attempting to solve quadritic equation
    A1 for both values of x
  2. Actual area
    = ½̸ ×6432 ×3.5 =11.2
    Limits 6.45   3.55
              6.35   3.45
    Max area ½ × 6.45 ×3.55 =11.45
    Min area ½ × 6.35 ×3.45= 10.93
    A.E. A - max⁡- min    11.45-10.95=0.25
                        2                   2 
    % Error = 0.25/1.2 ×100
    = =2.2%
    M1 for limits
    M1 for ½ ×6.35 ×3.45
    A1 for % error

  3. 8
      BD 
    =  12  
    sin⁡30  sin⁡90 
    BD = 12 × sin⁡30 = 6
                sin⁡90
         8          6    
    Sin⁡ADB   sin⁡45
    Sin⁡ADB= 8 × sin45
                          6
    = 0.942
    ADB = sin-10.942
    = 70.39
    M1 for expression to find BD
    M1 for Sin⁡ADB= 8 × sin45
                          6
    A1 for % correct Angle 
  4.   √48   × √5-√3 
    √5+√3    √5 -√3
    √240-√1444√15-12
         5-2                3
    = 4/3 √15 - 4
    M1 for multipying by conjugate
    M1 for 5-2
    A1 for % C.A
  5.    
    1. (1+ x)7
      17xº + x1 – x+ x3
         1      4     6      4
      = 1 + 4x + 6x2 + 4x2
    2. (0.94)7 = (1 + -0.06)7 x = -0.06
      1 + 4 (-0.06) + 6( -0.062) + 4 (-0.063)
      1 + 0.24 + 0.0036 =1.244
      B1 for corrrect expansion
      M1 for x= -0.06
      A1 for correct value
  6. P = Kt3
         √5
    16 = K23
            √9
    16 × 3=K   16 × -3=K
         8      8            8 
    K=6      K= -6
    P = 6t3
         √5
    P = 6 ×3= 6 × 27 = 27
           √36          6
    P = -6 ×3= -6 × 27 = -27
           √36          6
    M1 for P = 6 × 3
                       √3
    A1 correct value of P.

  7. 9
    (3+3x)(2x+2) = 6( x+7)
    6x+6+6x2+ 6x = 6x+42
    6x2+ 12x – 6x + 6 – 42 = 0
    6x2 – 6x – 36 = 0
    x2 – x – 6 = 0
    x2– 3x + 2x – 6 = 0
    x( x- 3) + 2( x-3 ) = 0
    x+ 2 = 0 x – 3 = 0
    x = -2 x = 3
    M1 for ( 3+3x)( 2x+2) =6 ( x+7)
    M1 for Attempting to solve quadratic equation
    A1 for x = -2
    A1 for x = 3
  8. 3x2+8x-6=0
    x2 + 8/8 x – 6/3=0
    x2 + 8/3 x+K=2+K
    K=(8/3 ÷2)2= (8/3 ×½ = (4/3)2= 16/9
    x28/3 x + 16/9= 2+ ( 16 )/(9 )
    ( x+ 4/3)=37/9
    x+4/3= √37/9           x= -1.944-1.333
    x=1.944-1.333     x= -3.277
    x=0.611
    M1 for x28/3 x + 16/9= 2+ ( 16 )/(9 )
    M1 for x+4/3= √37/
    A1 for both values of x
  9.    
    1. TS=8(8+x)
      144 = 64 + 8x
      80 = 8
      10 = x
      Vu = 10cm
    2. vx = 4cm (2/5 ×102)
      Xu= 10cm – 4 = 6
      4×6=3 × x       x = 4 x 62 =8cm
                                     3
      Sx = 8cm
      M1 for 144=64+8x
      A1 for C.A
      M1 for vx or xu
      A1 for 8cm
  10. r/p = m/√n-1
    r2/p2 = (m2)/(n-1 )
    r2 (n-1 )= m2p2
    r2n - r2 =m2p2
    r2 n =m2p2 +r2
     r2         r2
    n=m2p2 +r2
             r2
    M1 for squaring
    A1 for C.A
  11. x2+4x+K+y2-2y+K=4
    K= (4/2)2=4         K= (-2/2)2=1
    x2+4x+4 +y2-2y+1=4+4+1
    (x+2)2 + (y-1)2= a
    (x-a)2 +(y-b )2= r2
    a= -2   b=+1  r=3
    Mfor competing the square
    A1 for centre
    A1 for radius
  12.      
    1. 5th = a+4d=82
      12th= a+11d=103
      -7=-21 
      d=3
      a+4d=82
      a+12=82
      a=82-12
      = 70
    2. = n/2( 2a+(n-1)d
      = 2½ (2×70+20 ×3)
      = 2½ ((140+60)
      = 2100
      M1 for -7d= -21 (or equivalent)
      A1 for term 1
      M1 for substitution
      A1 for C.A

  13. 10
    x=-2/11×39+ 3/11×81=15
    y= 5/11×39 + -2/11×81=3
    M1 for determinant
    A1 for correct values of x and y
  14. 3x-1 > -4     2x+1≤7
    3x >-4+1        2x≤6
    3x > -3          x ≤3
    x> -1
    - 1 <x ≤3
    Mfor 3x > -3
    A1 for x ≤3 and x >-1
    B1 for number line with correct arrow
  15. Area =P=10+7+9
    = 26
    S= 13
    A = √(13(13-10)( 13-7)(13-9))
    =30.5
    11
    ½ ×10 ×7 ×sinBAC=30.5
    = sin⁡BAC= 30.5/35
    = 0.8714
    BAC=60.6
    M1 for for substitution of a, b c and S in the formula
    ACorrect Area
    M1 for sin⁡θ= 30.5/35
    A1 for correct angle
  16. A= P( 1+ r/100 )n
    = 1250000 (1+ 6/100)7
    =1250000 × 1.067
    =1879537.82
    Mfor substitution
    M1 for (1+ 6/100)7
    A1 for Amount
  17.    
    1. B and passes and not prefect
      6/11 × 5/× 7/= 105/352 
    2. G and prefect and pass
      5/11 × 2/11 × 4/7= 40/847
    3. B or G
      B NP passes or G NP passes
      6/11 × 5/× 7/85/11 × 9/11 × 4/7
      = 105/352 + 180/847
      = 0.5108
      M1 for 7/(8 )
      M1 for 6/11 × 5/× 7/8
      A1 for C.A
      M1 for 5/11 × 2/11 × 4/7
      A1 for C.A
      M1 for 6/11 × 5/× 7/85/11 × 9/11 × 4/7
      A1 for 105/352 
      A1 for 180/847
      A1 for 0.5108
  18.    
    15
  19.    
    1.      
      1. PQ=PO+OQ
        = - P+q
      2. RS = RQ+Q S
        = -PQ+ 1/3 OQ
        = P- q+ 1/3 q
        P - 2/3 q
    2.    
      1. ST/RS = K/T
        ST=KRS
        K=(P- 2/3q)
        =KP - 2/3 qK
      2. Expressing RT in two ways
        RT = RS + ST
        = P- 2/3 q+KP- 2/3Kq
        =(1+K)P +(-2/32/3 K)q
        RT=RP+PT
        = 2PQ + nOP
        = -2(q- p) + nP
        = (2+ n)p – 2q
        (1 +K) P + (-2/32/3K)=(2+n )p-2q
        -2/2/3 K= -2        1 + K = 2+ n
        K=2                       1+ 2 = 2+ n
        1 = n
        B1 for PQ
        M1 for P- q + 1/3q
        A1 for P- 2/3q
        M1 for K(P-2/3q)
        A1 for correct vector
        M1 for P- 2/3 q+KP- 2/3 Kq
        M1 for (1 +K) P + (-2/32/3K)q
        M1 for equating the two expression
        A1 for K
        A1 for n
  20.      
    12
  21.    
    1. GL2= 92 + 302
      GL=31.3
    2. FJ=FH2+ HJ2
      FH= 40+ 92
      FH =HI
      FJ2= 41+302
      FJ2= 50.8
    3. EM and EFGH
      EM projection = EG
      <GEM
      16
      Tan θ = 30/41=0.75
      Q = 36.87
      M1 for 92 )+ 302
      A1 for correct lenght
      M1 for FH2=40+ 92
      M1 for FJ2= 41+302
      A1 for 50.8
      M1 for Tan θ= 30/41
      M1 for projution
      A1 for Q
      M1 for Tan θ
      Afor angle
  22. Taxable income
    = 35600 + 3200
    = 38800
    1. 1st 9400 ×10/100=940
      Next 8600× 15 /100×8600=1290
      Next 8600× 20/100×8600=1720
      Next 9000 × 25/100×9000=2250
      Next 3200 × 30/100 ×3200=960
      (Total tax = & 71 60
      less relief 1172 )
      Shs 5988
    2. Basic salary = taxable income – allowances
      = 38800 – 8000
      Sacco = 6/100 ×30800 = 1848
      Net salary = T.1 – (PAYE + Sacco
      = 38800 – (5988 + 1848)
      = 38800 – 7836
      = Ksh 30964
      M1 for addition
      A1 for 38800
      M1 for first slab
      M1 for 2nd and 3rd slabs
      M1 for last 2 slabs
      A1 for gross tax
      A1 for Net tax
      M1 for 1848
      M1 for substration
      A1 for C.A
  23.    
    Mass  X F Fx dx-x d2 Fd2 x2 Fx2
    0-9 4.5 3 13.5       20.25 60.73
    10-19 14.5 7 101.5        210.25 1471.75
    20-29 24.5. 8 196       600.25 4802
    30-39 34.5 9 310.5       1190.25 10712.25
    40-49 44.5 12 534       1980.25 23769
    50-59 54.5  18 981       2970.25 53464.5
    60-69 64.5 25 1612.5       4160.25 104006.25
    70-79 74.5 10 745       5550.25 55502.5
    80-89 84.5 6 507       7140.25 42841.5
    90-99 94.5 2 189       8930.25 17860.5
        100 5190          
    1. Modal class
      = 60 – 69
    2.    
      1. (Mean =ƸFx/ƸF = 5190/100 = 51.9
      2. standard deviation
        √ƸFx2/ƸF- (ƸFx/ƸF)2
        = 314485/100- 51.92
        = 3144.86 – 2696.61
        =√448.24
        =21.17± 0.1
        B1 for Modal class
        M1 for ƸFx
        M1 for ƸF
        A1 for Mean
        M1 for 314485/100- 51.92
        M1 for Square root
        A1 for C.A

  24. 13
    M1 for finding cordinates
    A1 for cordinates
    B1 for OABC drawn
    B1 for O1 A1 B1 C1 drawn
    M1 for Attempting to find cordinates
    of O11 A11 B11 C11
    A1 coordinates of O11 A11 B11 C11
    B1 for O11 A11 B11 C11 drawn
    M1 for Attempting to find matrix
    A1 for a= -½
    A1 for matrix
    14

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