Mathematics Questions and Answers - Form 3 Mid Term 2 2022

INSTRUCTIONS TO CANDIDATES.

• This paper consistsof twosection I and II.
• Answer ALL questions in sectionIand only five questions from section II.
• Marks may be given for correct working even if the answer is wrong
• Non-programmable electronic calculators may be used.

Questions

SECTION I (50 MARKS)
Answer ALL Questions from this section in the spaces provided

1. Use logarithms to evaluate (4mks)
   5.246 x log 0.2349
        0.06364^½
2. Make 4 the subject of the formula. (3mks)
   t = ^2m/_n √(^L-A/_3k)
3. Express the recurring decimal below as a fraction:  leaving your answer in the form of ^a/_b where a and b are integers. (2mks)
4. Translation Q is represented by the column vector  and another translation R by the column vector . A point S is mapped onto a point T by Q and a point T is mapped into a point U by R. If point U is (8, - 4), determine the co-ordinates of point S. (3mks)
5. Find the values of a and b. Given: (4mks)
       3       +      3√5   = a + b√5
   3 + √5         3 - √5
6. The figure below shows a circle centre O. Chord AB subtends 300 at the centre. If the area of the minor segment is 5.25cm2, find the radius of the circle. (3mks)
   
7. The position vectors for points A and B are 2i + 5j + 3k and 4i – 7j + 3k respectively. Express vectors AB in terms of unit vectors i, j and k. Hence find the length of AB leaving your answer in simplified surd form 3marks
8. Given that the matrix has no inverse, find x. (2mks)
9. In the figure below ABC is a tangent to the circle at point B.Given that BE =6.9cm, FE=7.8cm, GE=4.1cm, DC=11.2cm and ED = xcm.Determine the length BC; give your answer in four significant figures. (4mks)
   
10. Given that matrix , Find matrix B such that: A + B (3mks)
11. A quantity P varies partly as t and partly as the square of t.When t = 20, p = 45 , and when t = 24 , p = 60.
    1. Express p in terms of t. (2mks)
    2. Find p when t = 32. (2mks)
12. Achang’a deposited sh. 20 000 in a saving account. Find the interest after two years. If the interest was paid at 16% per annum compound semi-annually. (3mks)
13. Using a ruler and a pair of compasses only construct triangle ABC such that BC=6cm, <ABC=75o and BCA=45o. Drop a perpendicular to BC from A to meet BC at O hence find the area of triangle ABC (4 mks)
14. Five men working 8 hours daily complete a piece of work in 3 days. How long will it take 12men working 5hours a day to complete the same work. (2mks)
15. Find the integral values of x which satisfy 6 < 2x + 1 and 5x – 29 < - 4 . (3mks)
16. In a fund-raising committee of 45 people, the ratio of men to women is 7: 2.Find the number of women required to join the existing committee so that the ratio of men to women changes to 5:4. (3mks)

SECTION II (50 MARKS )
Attempt any five questions from this section

17. James’ earning are as follows:- Basic salary 38,000 p.m, House allowance Sh. 14, 000p.m Travelling allowance Sh. 8,500p.m. Medical allowance sh. 3,300
    The table for the taxable income is as shown below
    

    Income tax in k£ p.a	Tax in Sh. Per pound
    1 – 6000 6001 – 12000 12001 – 18000 1001 – 24 000 24001 - 30 000 30001 - 36000 36001 – 42 000 42001 – 48 000 Over 48 000	2 3 4 5 6 7 8 9 10

    
    
    1. Calculate Jame’s taxable income p.a (2mks)
    2. Calculate Jame’s P.A.Y.E if he is entitled to a tax relief of Sh. 18 000 p.a(4mks)
    3. James is also deducted the following per month:-
       NHIF Sh. 320
       Pension scheme Sh. 1000
       Co-operative shares Sh. 2000
       Loan repayment Sh. 5000
       Interest on loan Sh. 500
       1. Calculate James’ total deduction per month in Ksh. (2mks)
       2. Calculate his net salary per month. (2mks)
18. In the figure below, ABCD is a square.Points P,Q,R and S are the midpoints of AB,BC,CD and DA respectively.
    
    
    1. Describe fully:
       1. A reflection that maps triangle QCE onto triangle SDE. (1mk)
       2. An enlargement that maps triangle QCE onto triangle SAE. (2 mks)
       3. A rotation that maps triangle QCE onto triangle PEB. (3 mks)
    2. The triangle ERC is reflected on the line BD. The image of ERC under the reflection is rotated clockwise through an angle of 900 about P.
       Determine the images of R and C:
       1. Under the reflection ( 2mks)
       2. After the two successive transformations ( 2mks)
19. The figure below shows a pulley system where a conveyor belt is tied round the two wheels. The radius of the large wheel is 180cm and the distance between the centres of the wheel is 300cm and XOY = 1400
    
    Determine
    1. Length XV (2mks)
    2. Length VBW (3mks)
    3. Length XAY(2mks)
    4. The total length of the conveyor belt
20. From Jane’s home, her school is 150m on a bearing of N60oE. The shopping centre is on a bearing of 150o from the school and 110o from Jane’s home. The church she attends is 180 m from her home on a bearing of 320o.. Using a scale of 1cm to represent 30m
    1. Show the relative positions of the four points (4mks)
    2. Use the drawing to determine
       1. The distance of the school from the shopping centre (1mks)
       2. The distance of her home from the shopping centre (1mks)
       3. The distance and bearing of the church from the shopping centre (2mks)
       4. The distance and bearing of the school from the church (2mks)
21. P varies directly as V and inversely as the square root of R. Given that P = 180, R = 25 when V = 9.
    1. Find P when V = 6 and R = 26. (3 marks)
    2. Find the value of V when P = 360 and R = 0.64. (3 mks)
    3. If V increased by 16% and R decreases by 25%, find the percentage change in P. (4mks)
22.     
    1. Complete the table below. (2mks)
    2. Draw the graph of y = sin (x+30) and y=cos(x-15) for -30≤X≤2700on the same grid. Take 1cm to represent 30o on x-axis and 1cm to represent 0.2units on y-axis. (5 mks)
    3. Using your graph drawn (b) above
       1. Find the values of x for which cos (x-15) –sin (x+30) = 0 (2mks)
       2. Estimate the angle corresponding to cos(x-15) = 0.6. (1 mk)
23. A tailor bought a number of suits at a cost of sh 57600 from Jimco clothing. Had he bought the same suits from Gikondi clothing, it would have cost him sh 480 less per suit. This would have enabled him to buy 4 extra suits for the same amount of money.
    If x represents the number of suits bought;
    1. Write an expression of the cost per suit bought from
       1. Jimco clothing (1mk)
       2. Gikondi clothing (1mk)
    2. Form an equation in x and determine the number of suits the tailor bought (4mks)
    3. The tailor later sold each suit for sh 720 more than he had paid for it. Determine the percentage profit. (4mks)
24.             
    1. The first term of an arithmetic progression (AP) is 2. The sum of the first 8 terms of the AP is 256.
       1. Find the common difference of the AP (2mks)
       2. Given that the sum of the first n terms of the AP 416, find n (2mks)
    2. The 3rd, 5th, and 8th terms of another AP forms the first three terms of a geometric progression (GP).If the common difference of the AP is 3.
       Find
       1. The first term of GP (4mks)
       2. The sum of the first 9 terms of the GP to 4 s.f (2mks)

Marking Scheme 

1. 5.246 x log 0.2349
       0.06364 ½
   = 5.246 x 1.3708
           0.6364 ½
   =-(5.246 x 0.6292)
         0.063664 ½
   No             Log
   5.246       0.7198
   0.6292    1.7988
                   0.5186 0.    5186 -
   0.06364   x   2.8038   1.4019
                                       1.1167
   10^-1 x 1.309
   = -0.1309
2.  t² = ^4m²/_n² (4m²/3k)
   t² = 4m²l - 4m²A
              3n²k
   3n²t²k = 4m²l - 4m²A
   4m²A = 4m²l - 3n²t²k
   A= 4m²t - 3n²t²k
              4m² 
3. let x = 4.37272
   10x = 43.72
   100x = 437.2
   1000x = 4372.72
   1000x - 10x = 4372.72 - 43.72
   990x = 4329
   x 4329 =   481  
       990      110   
4.    
    
5. 3 (3 - √5) + 3√5(3 + √5)
                9 -5
   = 9-3√5 + 9√5 + 15
                4
   24 + 6√5
          4
   = 6 + ³/₂√5
   ∴ a = 6 and b = 2/3
6. ³⁰/₃₆₀ x ²²/₇ x r² - ¹/₂r² Sin 30° = 5.25
   ¹/₄₂r² - ¹/₄r² = 5.25
   ¹/₈₄r² = 441
   r = ± 21cm
   r = 21cm
7. 
   
   AB =
   AB = 2i – 12j + Ok
   =|AB| = √(2² + (-12)² + 0²)
   = √4 + 144
   = 148
   = 12.17
8. 4(5 - x)- 6x = 0
   20 - 4x - 6x = 0
   10x = 20
   x = 2
9.  FE x ED =  GE x EB
   7.8x = 4.1 x 6.9
   7.8x = 28.29
   x = 3.627
   = 253.4224cm
   BC² = CD x CF
   y² = 11.2 x 22.627
   = 253.4224
   y = √253.4224 = 15.92cm
10.   
    
11.            
    1. p ± x t and p x t2
       ∴ p = mt + nt2
       45 = 20m + 400n
       60 = 24m + 400n
       60 = 24m + 576n
       9 = 4m + 80n .....(i) x 2
       5 = 2m + 48n ....(ii) x 4
       18 = 8m + 160n
       20= 8m + 192n 
       -2 = -32n
       n = 2/32 = 1/16
       and m = 5 - (48 x 1/16)1/2 = 7/2
       hence p = 7/2t + 1/16t2
    2. p = (7/2 x 32) + (1/16 x 32 x 32) = 112 + 64 = 176
12. A = P(1 + r/100)n
    A=20000(1 + 8/100)4
    = 27210 
    I = 27210 - 20000
    = Sh 7, 210
13.    
         
14. Men hours days
        5      8      3
        12     5     x
    x = 5 x 8 x 3
            12 x 5
    = 2 days
15. Area Scale Factor = Determinant of the matrix
    ⁶⁰/₁₀ = x (x + 3)(x + 3) - 12
    6= x² + 3x - 12
    (x² - 3x) + (6x - 18) = 0
    (x + 6)(x - 3) = 0
    X= -6 or 3
    X=3
16. men = 7/9 x 45 = 35
    Women = 2/9 x 45 = 10
    let the additional number of women be x
    ³⁵/_10+x = ⁵/₄
    50 + 5x = 140
    5x = 90
    x = 18 women          
17.        
    1. IF = 38000 + 14000 + 8500 + 3300 = Ksh 63, 800pm
       IF = (63800 x 12) = 38280
                    20        
    2. 1^st  6000x2=12000
       2^nd 6000x3 =18000
       3^rd  6000x4 =24000
       4^th  6000x5 =30000
       5^th  6000x6 =36000
       6^th  6000x7 =42000
       7^th  2280x8 =1842
       Tax due =Ksh.180,240pa
       Less relief     18,000
       Tax payable   162,240
        PAYE = 13,520
    3. Total deduction
       =13520+320+1000+2000+5000+500
       =Ksh.22,340
    4. 38280-22340 =Ksh.15,940
18.    
    1.        
       1. Reflection in the line PR or ER 
       2. Enlargement centre E 
          Scale factor -1 
       3. Rotation about point R 
          Through 900 
          clockwise 
    2.     
       1. R - S B1 
          C - A B1
       2. R’ - Q B1
          C’ - E B1 10
19.            
    1.     
        
    2.    
       
       •        
         1.   length XV = 300 cm 70°
            = 281.9cm A1
         2. Arc length VBW =
            140/360 x 2πr = 140/360 x 2x22/7 x 180
            = 439.88cm
         3. Length XAY
            C = θ/360 2πr = 220/360 x 2x22/7 x 180 = 691.24
         4. Length of the conveyor belt
            =691.2+439+(2x281.9)
            =1694.88
20.             
    1.    
        
    2.           
       1. 7 x 30 = 210m ,
       2.  9.2 x 30 = 276m
       3. 13.7 x 30 = 411m
          bearing= 310° ± 1°
       4. 7.5 x 30 = 225m
          Bearing =110° ± 1°
21.              
    1. Pα ^V/_√R 
       P = K ^V/_√R
       180 = K = ⁹/_√25 
       K= 180 x 5
                 9
       K = 100
       180 - ^100V/_√R 
       P = 100 x 6 = 117.67
                  √26
    2. P = 100V   360 =    100V  
                  √R             √R0.64
       360 x 0.8 = V
          100
       V= 2.88
    3. V = %116 R = 75%
       P = 100 x 1.16
                √ 0.75
       P = 133.945 - After change
       133.945 - 100 = 33.945%
22.            
    1.  
       

       x	-30	0	30	60	90	120	150	180	210	240	270
       Sin (x - 30)			0.87			0.5	0.00		0.87	-1.00
       Cos (x - 15)		0.97		0.71		-0.26	0.71

    2.     
       
       1. cos (x – 15)- sin (x – 30) = 0
          cos (x – 15 ) = sin(x + 30)
          x = 36 ±2 or x = 216±2
       2. co-ordinates of the turning point of y = cos (x – 15) on the negative section (198;-1)
       3. Cos (x – 15 ) = 0.6
       4. X-15 = 66°
23.            
    1.  
       1. 57600
             x
       2. 57600
            x + 4
    2. 57600 - 57600 = 480
          x           x+ 4    
       X²+4x=480=0
       X²+24x-20x-480=0
       X(x+24)-20(x+24)=0
       (x+24)(x-20)=0
       X=20
    3.  Price per suit
       57600
          20
       =sh 2880
       % profit       720     x 100 =25%
                         2880
24.      
    1. S_n = ^A/₂ {2a + (n-1)d}
       a = 2, n = 8 and S_n = 156d
       8/2 {2 x 2 + (8-1)d}= 156
       4(4 + 7d)= 156
       16 + 28d = 156
       ⇒28d = 156 - 16
       28d = 140 
       d=5
    2. ⁿ/₂ {2 x 2 + (n-1)5}= 416
       ⁿ/₂ (4 + 5n - 5)= 416
       4n + 5n² - 5n = 416 x 2 ⇒ 4n - 5n + 5n² = 832
       5n² - n - 832 = 0
       n = n 1 ± √(1) - 4 x 5 - 832 =   1 ± 129
                                10                     129
       n = 1 ± 129 = 130 = 13
                10          10
       n = 1 - 129 = -128 = n = 12.8
                  10        10
       ignore n = 12.8
       ∴ n = 13
    3. ∴ a + 2d, a + 4d , a + 7d
          a + 4d   =    a + 7d  
             a + 2d       a + 4d
       (a + 4d)(a + 4d) = (a + 2d)(a + 7d)
       a² + 4ad + 4ad + bd² = a² + 2ad + 14d²
       8ad + 16d² = 9ad
       16d² = 9ad - 8ad
         16d² = 1d
         9d
       a = 16d
       but d = 3
       ⇒a = 16 x3
       = 48
       r =   a + 4d =   48 x 4 x 3    =   60   =   10   
              a + 2d     48 + (2 x 3)    54         9  
       S_n = a(rⁿ - 1)
                  r -1
       48 (¹⁰/₉) - 1
          ¹⁰/₉ - 1
       =683.067