Mathematics Paper 1 Questions and Answers - Alliance Mock Examinations 2022

QUESTIONS

SECTION 1 (50 marks)
Answer ALL the questions in the spaces provided.

1. Evaluate correct to 4 significant figures;
   4 x 6 + 25 + 0.05 + ¹/₅
   (-3)+(-6)+(23)-6 of 3
2. Given that Tan 650=3+ √5, determine without using mathematical tables nor calculator, Tan 25°, leaving your answer in the form a+b√c, where a, b, and c are rational numbers. (3mks)
3. Use mathematical tables to find y, correct to four significant figures.
   ¹/_y = ¹/_24.3 + ¹/_13.1 (3mks)
4. The three sides of a right angled are (x-1), (2x+8) and the hypotenuse (3x+1), find the area of the triangle. (3mks)
5. The figure below shows the graph of log P against log Q.
   
   Given that P = aQ", find the values of a and n. (3mks)
6. A salesman is paid a salary of Sh. 10,000 per month. He is also paid a commission on sales above Sh. 100,000. If in one month he sold goods worth Sh. 500,000 and his total earnings amounted to Sh. 56,000. Calculate the rate of commission. (3mks)
7. Solve the following equation, giving your answer correct to 4 decimal places. (3mks)
   8^x +5+2^3x = 35
8. Express in surd form and simplify by rationalizing the denominator; (3mks)
   1 + cos 30°
   1 - sin 60º
9. Solve; 4 ≤ 3x - 2 <9+ x
   Hence list all the integral values that satisfy the inequality. (3mks)
10. Simplify the expression;
    6b - 3ab-2a + a²
           a² - 9b²
11. A construction company employs technicians and artisans. On a certain day 3 technicians and 2 artisans were hired and paid a total of ksh 9000. On another day the firm hired 4 technicians and 1 artisan and paid a total of ksh 9,500. Calculate the cost of hiring 2 technicians and 5 artisans in a day. (3mks)
12. On the line provided below, by construction, locate a point such that the ratio AB: BQ = 5: -2
    
13. Given that log 7 = 0.8451 and log 6 = 0.7782, find log 25.2
14. A triangle Twith vertices A (2.4), B (6,2) and C(4,8) is mapped onto a triangle Twith vertices A' (10,0), B' (8,-4) and C' (14,-2) by a rotation. (2mks)
    1. On the grid provided draw triangle T and its image T¹ (2mks)
    2. Determine the centre and angle of rotation that maps Tonto T¹.
15. Velocity of a particle moving on a straight line is given by V = (2t + 10)ms^-1, where t is the time taken in seconds. Find the distance covered in the 3rd second.  (3mks)
16. A point R (0, 2) has its image R' (4, 6) under an enlargement with scale factor 3. Find, without drawing, the centre of enlargement. (3mks)

SECTION II (50 marks)
Answer only five questions from this section in the spaces provided.

17. The diagram below shows two intersecting circles of radii 20cm and 15cm such that their centres A and B are 30cm apart.
    
    Calculate to 2 decimal places; a) The area of the sector ACD (3mks)
    1. The area of sector BCD (3mks)
    2. The length of the chord CD (2mks)
    3. The area of the quadrilateral ABCD (1mk)
    4. The shaded area (1mk)
18. A solid S is made up of a frustrum of a cone whose upper part is replaced with cylindrical part.
    The height of the solid is 4.5m, the common radius of the cylindrical part and the conical part is 0.9m, and the height of the conical part is 1.5m.
    Taking π as ²²/₇
    1. Calculate, correct to 1 decimal place, 
       1. The volume of solid S. (4mks)
          
       2. The total surface area of solid S (4mks)
    2. A square based pillar of side 1.6m has the same volume as solid S. Determine the height of the pillar, correct to 1 decimal place. (2mks)
19. A bus left Mombasa and travelled towards Machakos at an average speed of 60km/h. After 2½ hours, a car left Mombasa and travelled along the same road at an average speed of 100km/h. If the distance between Mombasa and Machakos is 500km, determine:
    1.    
       1. The distance of the bus from Machakos when the car took off.
       2. The distance the car travelled to catch up with the bus.
    2. Immediately the car caught up with bus, the car stopped for 25min. find the new average speed at which the car travelled in order to reach Machakos at the same time as the bus. (4mks)
20.    
    1. Given that a = (⁴₃), c= (^-2_-5) and 3a – 2b + 4c = (¹⁰_-19) find b. (3mks)
    2. In the figure below, OAB is a triangle. A is the point (2,8) and B the point (10,2). C, D, and E are the mic-points of OA, OB, and AB respectively, while F is on AB such that AF = ²/₃AB
       
       
       1. Find the position vectors of point C (2mks)
       2. Find the length of vector AB (3mks)
       3. If vector OA = a and vector OB = b, write DF in terms of vectors a and b (2mks)
21.    
    1. Using a ruler and a pair of compasses construct a parallelogram ABCD, where AB - 8cm, BC = 6cm and angle ABC = 120°. (3mks)
    2. In the diagram draw the diagonal BD and construct the circumcircle to triangle ABD. (3mks)
    3. Construct a perpendicular from C to meet AB produced at X. Measure CX. (2mks)
    4. Calculate the area of triangle ACD. (2mks)
22. Mr. Pesa bought two types of items from a wholesaler. He bought 5 of type A at Ksh. 1,250 each and 20 of type B at Ksh. 650 each. He is to sell these items at a retail price of Ksh, 1,400 for each item of type A and Ksh. 700 for each of type B.
    1. Find the total expenditure of Mr. Pesa. (2mks)
    2. Assuming that Mr. Pesa sold all the items, calculate his percentage profit, to 1 decimal place.(3mks)
    3. Mr. Pesa learned of a new variety of the same items of type A and type B and decided to return the remaining stock in exchange for the new variety. The remaining stock consisted of one item of type A and 10 of type B. The prices of the new variety were Ksh. 1,500 and Ksh. 800 of type A and B respectively. If Mr. Pesa bought the same number of items as before;
       1. What was the value of the returned goods if a depreciation of 10% is allowed? (3mks)
       2. Find how much money he was to add in order to get the new variety? (2mks)
23. Triangle ABC has vertices A(-2,0), B(-5,3), and C(1,3).
    1. Find the coordinates of a point that is equidistant from points A, B, and C (5mks)
    2. If the triangle is circumscribed, find the equation of the circumcircle in the form ax² + by² + cx + dy + k = 0, where a, b, c, d, and k are constants. (2mks)
    3. Determine the equation of the tangent to the circle at point C, in the form of y=mx+C (3mks)
24.    
    1. Complete the table below for the functions y = 2Cos x and y = Sin 2x, for (-180º < x < 180°) (2mks)
       

       x	-180	-150	-120	-90	-60	-30	0	30	60	90	120	150	180
       2x	-360	-300	-240	-180	-120	-60	0	60	120	180	240	300	360
       2cosx	-2	-1.73		0		1.73	2	1.73	1		-1		-2
       sin2x		0.87	0.87	0	-0.87		0	0.87	0.87		-0.87		0

    2. On the grid provided, draw on the same axis the graphs of y = 2Cos x and y = Sin 2x, for(-180° SXS 180°) (4mks)
    3. Use the graphs in (b) above to determine;
       1. The amplitude and period of the graph y = 2Cos x (2mks)
       2. The values of such that; 2Cosx - Sin 2x = 0 (1mk)
       3. The difference in the values of y when x = -45° (1mk)

MARKING SCHEME

1. 24 + 0.8 + ¹/₅ = ²⁵/_5.5 = 4.545
   0.5 + 23 - 18
2. Tan25 =   1   
                3+√5
   = 3 -√5 = 3 -√5 = ¾ - ¼ - √C
       9 - 5       4    
3. 24.3 = 0.04115
   13.1 = 0.07634
   ¹/_y = 0.04115 + 0.07634
   = 0.11749
   y =     1      
        0.11749
   rec 0.11749
   y = 8.509
4. (3x + 1)² = (x - 1)² + (2x + 8)²
   9x² + 6x + 1 = x² - 2x + 1 + 4x² + 32x + 64
   4x² - 24x - 64 = 0
   x = -2 or 8
   Area = ½ x 7 x 24 = 84sq units
5. logP = log a + nlog a
   log a = 1
   10¹ = a
   gradient = n
   21 - 1 = 4
    5 - 0
6. commission = 56000 - 10000
   = 46000
   46000 = 400000 x r/100
   r = 11.5%
7. 2^3x + 2^3x = 30
   let 2^2x = y
   2y = 30
   y = 15
   2^2x = 15
   log2^2x = log 15
   2x = log 15
            log 2
   x = 1.953
8. 1+^√3/₂ (1+^√3/₂) = 1 + √3 + ¾
   1-^√3/₂  (1+^√3/₂)        1 - ¾
   = ⁷/₄ + √3 = 7 + 4√3
          ¼
9. 4 ≤ 3x - 2
   6 ≤ 3x 
   2 ≤ x 
   3x - 2 < 9 + x
   2x < 11
   x < 5.5
   2 ≤ x < 5.5
   x = 2, 3, 4
10. num = 36(2 - a) - a(2 - a)
    = (36 - a)(2 - a)
    den = (a - 3b)(a + 3b)
    -1(-3b + a)(2 - a)
     (a - 3b)(a + 3b)
    =  a - 2  
       a + 3b
11. 3t + 2a = 9000
    4t + 1a = 9500
    3t + 2(9500 - 4t) = 9000
    -5t = -10000
    t = 2000
    a = 1500
    2(2000) + 5(1500)
    = 11,500shs
12.    
    
13. log 6² x 7
            10
    = 2log6 + log 7 - log 10
    2(0.7782) + 0.8451 - 1 = 1.4015
14.      
    1.    
       
    2. (4, -2) , -90º
15. S = ∫2t + 10 dt
    t² + 100 + c
    t = 0 s = 0, therefore c = 0
    s = t² + 10t
    3² + 10(3) = 39
    S = ³₂∫2t + 10 dt
    ²[t² + 10t + c]³₂
    (3² + 10(3) + c) - (2² + 10(2) + c)
    (39 + c) - (24 + c)
    15
16. 
    
    6 - y = 3
    2 - y
    6 - y = 6 - 3y
    y = 0
    4 - x = 3
    0 - x
    4 - x = -3x
    4 = -2x
    -2-x
    (-2, 0)
17.         
    1. 15² = 30² + 20² - 2(30)(20) cos θ
       θ = 26.38
       CAD = 52.77
       52.77 x ²²/₇ x 20² = 184.28
        360
    2. 20² = 15² + 30² - 2(15)(30)cos θ
       θ = 36.34
       CBD = 72.68
       72.68 x ²²/₇ x 150 = 142.76
        360
    3. sin 52.77 = ^x/₂₀
       x = 15.92
       15.92 x 2 = 31.48
    4. (½ x 20 x 20 sin 52.77) + (½ x 15 x 15 sin 72.68)
       = 266.64
    5. 266.64 - (184.28 + 142.76 - 266.64)
       = 206.24
18.         
    1.    
       1. ^4.5/₃ = ^r/_0.9
           r = 135
          (¹/₃ x ²²/₇ x 1.35² x 4.5 - 1/3 x ²²/₇ x 0.9² x 3)
          = 6.046
          ²²/₇ x 0.9² x 3 = 7.637 ≈ 7.6
       2. ²²/₇ x 1.352 = 5.728
          ²²/₇ x 0.92 = 2.546
          ²²/₇ x 2 x 0.9 x 3 = 16.97
          L = √1.35² + 4.5² = 4.698
          L = √0.9² + 3² = 3.132
          ²²/₇ x 1.35 x 4.698 - ²²/₇ x 0.9 x 3.13
          = 11.074
          total surface area 5.728 + 2.546 + 16.97 + 11.074
          = 36.318 ≈ 36.3
    2. 1.6 x 1.6 x h = 7.6
       h = 2.968 ≈ 3.0
19.      
    1.      
       1. 500 - (60 x 2.5)
          = 350km
       2. t = ¹⁵⁰/₄₀ = 3¾hrs
          d = 100 x 3¾ = 375km
    2. t = ¹²⁵/₆₀ = 2hrs 5 mins
       2hr 5 mins - 25mins = 1hr 40mins = 1²/₃hrs
       S = 125/1²/₃ = 75km/h
20.        
    1. 3(⁴₃) - 2(^x_y) + 4(^-2_-5) = (¹⁰_-19) 
       (¹²₉) - (^2x_2y) - (⁸₂₀) = (¹⁰_-19)
       (^2x_2y) = (^-6₈)
       b = (^-3₄)
    2.    
       1. OC = ½(²₈) = (¼)
       2. AB = (¹⁰₂) - (²₈) = (⁸_-6)
          IABI = √8² + (-6)² = 10
       3. DF = DB + BF 
          ½b + ¹/₃(a - b)
          ½b + ¹/₃a - ¹/₃b
          ¹/₆b +¹/₃h
21.      
    1.  
       
    2. 5.2 ± 0.1
    3. ½ x 5.2 x 8 = 20.8cm²
22.      
    1. (1250 x 5) + (650 x 20) = 19250
    2. S.P = (1400 x 5) + (7000 x 20) = 21000
        1750   x 100 = 9.09
       19250
       ≈9.1%
    3.    
       1. Returned (1 x 1250) + (10 x 650) = 9750
          9750(1 - ¹⁰/₁₀₀) = 6975
       2. (1500 x 5) + (800 x 20) = 23500 - 6975 = 16525
23.      
    1. grad AB
         3- 0   = ³/_-3 = -1
       -5 + 2
       (-2+-5 , 0+3)
           2        2
       (-3.5, 1.5)
       y-1.5 = 1
       x+3.5
       y - 1.5 = x + 3.5
       y = x + 5
       x = -2
       y = -2 + 5
       = 3
       0 = (-2,3)
    2. Centre (-2,3)
       (1 + 2)² + (3 - 3)² = r²
       r² = 9
       (x + 2)² + (y - 3)² = 9
       x² + 2x + 4 + y² - 6y + 9 = 9
       x² + y² 2x - 6y + 4 = 0
    3. equation of tangent x = 1
24.      
    1.      
       

       x	-180	-150	-120	-90	-60	-30	0	30	60	90	120	150	180
       2x	-360	-300	-240	-180	-120	-60	0	60	120	180	240	300	360
       2cosx	-2	-1.73	-1	0	1	1.73	2	1.73	1	0	-1	-1.72	-2
       sin2x	0	0.87	0.87	0	-0.87	-0.87	0	0.87	0.87	0	-0.87	-0.87	0

    2.   
       
    3.       
       1. Amplitude = 2
          period = 360
       2. x = -90 or 90
       3. 2.4