Mathematics Paper 2 Questions and Answers - Asumbi Girls Pre Mock Examinations 2023

INSTRUCTIONS TO THE CANDIDATES

• This paper contains two sections; Section A and Section B
• All workings and answers must be written on the question paper
• 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.

FOR EXAMINER’S USE ONLY
Section A

Section B

QUESTIONS

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  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.
    (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	15%
    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	4 0 – 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).
    O¹A¹ B¹ C¹ is the image of OABC under transformation matrix 
    1. Find the co-ordinates of O¹A¹ B¹ C¹. (2mks)
    2. On the grid provided draw OABC and O¹A¹ B¹ C¹ (2mks)
       
    3. Find O¹¹A¹¹ B¹¹ C¹¹ the image of O¹A¹ B¹ C¹ under the transformation matrix  (2mks)
       1. On the same grid draw O¹¹A¹¹ B¹¹ C¹¹ . (1mk)
    4. Find the single matrix that map O¹¹A¹¹ B¹¹ C¹¹ onto OABC. (3mks)

MARKING SCHEME

1. log (2x - 11) - log 2  = log 3  - log x
   2x - 11 = 3
        2        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 = -1/2
2. Actual area 
   =1/2 x 64 x 3.5 = 11.2
   limits 6.45 / 6.35    3.55 / 3.45
   Max area 1/2 x 6.45 x 3.55 = 11.45
   Min area 1/2 x 6.35 x 3.45  = 10.93
   A.E A = max - min    11.45 - 10.95 = 0.25
                       2                    2
   % error = 0.25 x 100
                   1.2
   = 2.2%
3.    BD    =  12    
   sin 30     sin 90
   BD = 12 x sin 30 = 6
              sin 90
          8      =     6      
   sin ADB      sin 45
   sin ADB = 8 x sin 45
                        6
   = 0.942
   ADB = sin^-10.942
   = 70.39
4. 
5.  
   1. (1 + x)
      1⁷x⁰ + 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(-0.006²) + 4(-0.06³)
      1 + 0.24 + 0.0036 = 1.244
6. 
7. x = -2   x = 3
8. 3x² + 8x - 6 = 0
   x² + 8/8x - 6/3 = 0
   x² + 8/3x + K = 2 + K
   K = (8/3 ÷ 2)² = 8/3 x 1/2 = (4/3)2 = 16/9
   x² + 8/3x + 16/9 = 2 + 16/9
   (x + 4/3)² = 3⁷/₉
   x + 4/3 = √3⁷/₉
   x = 1.944 - 1.333
   x = 0.611
9.                         
   1. TS² = 8(8 + x)
      144 = 64 + 8x
      80 = 8
      10 = x 
      Vu = 10 cm
   2. Vx = 4 cm (2/5 x 10)
      Xu = 10 cm  - 4 = 6
      4 x 6 = 3 x x    x = 4 x 6 = 8
                                       3
      Sx = 8 cm
10. 
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 = s
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
       = 2¹/₂ (2 x 70 + 20 x 3)
       = 2¹/₂ (140 + 60)
       = 2100
13. 
14. 3x - 1 > -4     2x + 1 ≤ 7
    3x > -4 + 1    2x ≤ 6
    3x > -3       x ≤ 3
    x > -1
    -1 < x ≤ 3
15. Area = P  =10 + 7 + 9
    = 26
    S = 13
    
16. A = P (1 + r/100)ⁿ
    = 1250000 (1 + 6/100)⁷
    = 1250000 x 1.06⁷
    = 1879537.82
17.             
    1. B and passes and not perfect
       6/11 x 5/8 x 7/8 = 105/352
    2. G and perfect and pass
       5/11 x 2/11 x 4/7 = 40/847
    3. B or G
       B NP passes or G NP passes
       =6/11 x 5/8 x 7/8 + 5/11 x 9/11 x 4/7
       = 105/352 + 180/947
       = 0.5108
18. 
19.               
    1.            
       1. PQ = PO + OQ
          = -P + q
       2. RS = RQ + QS
          =-PQ + 1/3OQ
          =P - q + 1/3q
          P - 2/3q
    2.              
       1. Kp ^-2/₃ qK
       2. 1 = n
          
          k = 2
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
22. Taxable income
    = 35600 + 3200
    = 38800
    1. 9400 x 10/100 = 940
       Next 8600 x 15/100 x 8600 = 1290
       Next 8600 x 20/100 x 8600 = 1720
       Next 9000 x 25/100 x 9000 = 2250
       Next 3200 x 30/100 x 3200 = 960
       Total tax = 7160
       less relief 1172
       sh 5988
    2. Basic salary  = taxable income - allowances
       = 38800 - 8000
       sacco = 6/100 x 30800 = 1848
       net salary  = T.I - (PAYE + sacco)
       = 38800 - (5988 + 1848)
       = 38800 - 7836
       = Ksh 30964
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	23763
    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 = Σfx/Σf = 5190/100 = 51.9
       2. standard deviation
          
          = 314485/100 - (51.9)2
          = 3144.86 - 2696.61
          = √448.24
          = 21.17 ± 0.1
    3.      
24.             
    1.                
    2.            
    3.               
    4.