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

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. 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 Find 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
   
   
   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/_P = ^M/_√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)
       

       xº	0º	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)
       
    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
    
    
    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.
    
    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 
    
    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 B¹ C¹ under the transformation matrix  (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. 
   
     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-√144 = 4√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 – x² + 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= 8 K   16 × -3=K
        8      8            8 
   K=6      K= -6
   P = 6t³
        √5
   P = 6 ×3³ = 6 × 27 = 27
          √36          6
   P = -6 ×3³ = -6 × 27 = -27
          √36          6
   M₁ for P = 6 × 3³ 
                      √3
   A₁ correct value of P.
7. 
   
   (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
   x²+ ⁸/₃ x + ¹⁶/₉= 2+ ( 16 )/(9 )
   ( x+ ⁴/₃)² =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 x²+ ⁸/₃ x + ¹⁶/₉= 2+ ( 16 )/(9 )
   M₁ for x+⁴/₃= √3⁷/₉ 
   A₁ for both values of x
9.    
   1. TS² =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
    M₁ for 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. 
    
    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
    M₁ for 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
    
    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
    A₁ Correct 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
    M₁ for substitution
    M₁ for (1+ ⁶/₁₀₀)⁷
    A₁ for Amount
17.    
    1. B and passes and not prefect
       ⁶/₁₁ × ⁵/₈ × ⁷/₈ = 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 × ⁵/₈ × ⁷/₈ + ⁵/₁₁ × ⁹/₁₁ × ⁴/₇
       = ¹⁰⁵/₃₅₂ + ¹⁸⁰/₈₄₇
       = 0.5108
       M₁ for 7/(8 )
       M₁ for ⁶/₁₁ × ⁵/₈ × ⁷/₈
       A₁ for C.A
       M₁ for ⁵/₁₁ × ²/₁₁ × ⁴/₇
       A₁ for C.A
       M₁ for ⁶/₁₁ × ⁵/₈ × ⁷/₈ + ⁵/₁₁ × ⁹/₁₁ × ⁴/₇
       A₁ for ¹⁰⁵/₃₅₂ 
       A₁ for ¹⁸⁰/₈₄₇
       A₁ for 0.5108
18.    
    
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 +(^-2/₃- ²/₃ K)^q
          RT=RP+PT
          = 2PQ + nOP
          = -2(q- p) + nP
          = (2+ n)p – 2q
          (1 +K) P + (-²/₃- ²/₃K)^q =(2+n )p-2q
          ^-2/₃ - 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 + (-²/₃- ²/₃K)^q
          M₁ for equating the two expression
          A₁ for K
          A₁ for n
20.      
    
21.    
    1. GL²= 9² + 30²
       GL=31.3
    2. FJ² =FH²+ HJ²
       FH² = 40+ 9²
       FH =HI
       FJ²= 41² +30²
       FJ²= 50.8
    3. EM and EFGH
       EM projection = EG
       <GEM
       
       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= 41² +30²
       A₁ for 50.8
       M₁ for Tan θ= ³⁰/₄₁
       M₁ for projution
       A₁ for Q
       M₁ for Tan θ
       A₁ for 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. 
    
    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