Mathematics Paper 1 Questions and Answers - Cekenas Mock Exams 2022

QUESTIONS

SECTION I: (50MKS)
ATTEMPT ALL THE QUESTIONS FROM THIS SECTION

1. Without using a calculator or a mathematical table evaluate (3mks)
   
2. In a disco hall, four lights signals are programmed at intervals of 8 seconds, 10 seconds, 12 seconds and 15 seconds. What is the earliest time they will give out light signals simultaneously if the last time they did this 1915 Hrs. (3mks)
3. Simplify  (3mks)
     12x² - 16x  
   20 - 11x - 3x²
4. The number of sides of two regular polygons differ by one. If the sum of the interior angles of these polygons is in the ratio 3:2 calculate the number of sides of each polygon and name them. (4mks)
5. Under an enlargement scale factor -4 the image of point A(4,3) is A¹(-4,-5). Calculate the coordinates of the centre of enlargement. (3mks)
6. Given that A+(²₀ ⁰₃) = (²₀ ³₁) find AT the transpose of matrix A. (2 mks)
7. If   , find h +k (3mks)
8. The figure below shows a circle inscribed in a square. The length of the arc AB is ⁴⁴/₇√7cm. Calculate the area of the shaded region. π = ²²/₇ (4mks)
   
9. Two students Paul and Omondi standing 10m apart on the same side of a tall building on a horizontal ground. Paul who is closer to the building sees the roof top at an angle of 70º, while Omondi at an angle of 46.8º, if the building, Paul and Omondi lies on a straight line, calculate the height of the building correct to 3 s.f. (4 mks)
10. A trapezium was created by joining two similar triangles. What is the perimeter of the quadrilateral? (3mks)
    
11. Mr Wajakoya purchased animal feeds for his 10 cows which would last for 7 days. After feeding his cows for 4 days, he sold 4 cows. How many more days will the feeds last the remaining cows. (3mks)
12. Use the tables of logarithms to evaluate. (4mks)
    (6.79 x 0.391)^-3/₄
            log5
13. A pentagonal prism has each of its sides as 5cm and length 10cm. Find the volume of the prism.(3 mks)
14. Solve the following inequalities and state the integral solutions. (3mks)
    ½(24 - 4x) > 6(3x - ⁴/₃) ≥ -²/₃(42 + 3x)
15. When a shirt is sold at sh 126 a loss of x% is made. If the same shirt is sold at sh 154, a profit of x% is realized. Find the buying price of the shirt. (3mks)
16. Without using calculator or mathematical tables, find the value of x in sin60º = cos (3x-45)º given that 0º  ≤ x ≤ ^πc/₂  (2mks)

SECTION II : ANSWER ANY FIVE QUESTIONS

17. The figure below shows a line that passes through points P, Q and R. Given that its equation is ^-7y/₃₀ + ^x/₁₀ = 1 and that point Q is equidistant from x and y-axis
    
    
    1. Calculate the co-ordinates of points Q. (4mks)
    2. Calculate the size of angle formed by the line and y-axis. (3mks)
    3. Find the equation of the perpendicular to the line through Q in the form of ^x/_a + ^y/_b = 1 (3 mks)
18. The figure below shows a histogram of a certain data.
    
    
    1. Given that the frequency of the last class is 30, prepare frequency distribution table for the data. (3mks)
    2.        
       1. State the modal class. (1 mk)
       2. Estimate the mean height (3mks)
    3. Using a vertical line show where the median lies and state the median. (3mks)
19. Towns A and B are 420km apart. Two Lorries departed from A at the same time travelling towards B. Lorry X travelled at an average speed of 15km/h less than Y and reached 1 hour and 24 minutes later.
    1. Calculate the average speed of lorry Y. (5mks)
    2. How far was X from A when Y reached B. (2mks)
    3. A van left town B heading towards A at the time lorries X and Y left A. If the van travelled at an average speed of 90km/h, how far from A did it meet Lorry Y. (3mks)
20. The vertices of a triangle PQR are P(0,0), Q(6,0) and R(2,4).
    1. Draw triangle PQR on the grid provided. (1mk)
    2. Triangle P¹Q¹R¹ is the image of triangle PQR under an enlargement scale factor   and centre (2,2), write down the co-ordinates of P¹Q¹R¹ and plot on the same grid. (3mks)
    3. Draw triangle P¹¹Q¹¹R¹¹ the image of triangle P¹Q¹R¹ under a positive quarter turn about point (1,1). State the co-ordinates of P¹¹Q¹¹R¹¹. (3mks)
    4. Draw triangle P¹¹¹Q¹¹¹R¹¹¹ the image of P¹¹Q¹¹R¹¹ under a reflection in the line y=1. (2mks)
    5. State the type of congruence between triangle P1Q1R1 and P¹¹¹Q¹¹¹R¹¹¹. (1mk)
21. The diagram below represents a solid consisting of a hemispherical bottom and a conical frustum at the top. O₁O₂ =4cm, O₂B=R=4.9cm and O₁A=r=2.1cm.
    
    Calculate to 1 d.p,
    1. The height of the chopped off cone and hence the height of the bigger cone. (2mks)
    2. The surface area of the solid. (4mks)
    3. Calculate the volume of the solid. (4mks)
22. In the triangle below P and Q are points on OA and OB respectively such that OP:PA=3:2 and OQ:QB =1:2 AQ and PQ intersect at T. Given that OA= a and OB= b
     
    1. Express AQ and PQ in terms of a and b (2mks)
    2. Taking BT=kBP and AT=hAQ where h and k are real numbers.
       1. Find two expressions for OT in terms of a and b (2mks)
       2. Use the expressions in b(i) above to find the values of h and k. (3mks)
    3. Show that B, T and P are collinear. (3mks)
23. A ship leaves port M and sails on a bearing of 0500 heading towards island L. Two navy destroyers ships sail from a naval base N to intercept the ship. Destroyer A sails such that it covers the shortest distance possible. Destroyer B sails on a bearing of 010º to L. The bearing of N from M is 100º and distance NM=300km. Using a scale of 1cm represents 50km, determine; 
    1. The position of M, N and L. (5 mks)
    2. The distance travelled by destroyer A. (2mks)
    3. The distance travelled by destroyer B. (2mks) 
    4. The bearing of N from L. (1mks)
24.    
    1. Determine the x and y-intercepts of the curve y=(x-1) (x+3)2 (3mks)
    2. Find the stationary points of the curve above and state their nature. (4mks)
    3. Sketch the curve. (3mks)

MARKING SCHEME

1. Num
   ⁶/₇ x ¹⁴/₃ = 4 ÷ 8 x ^-2/₃
   = ½ x ^-2/₃ = ^-1/₃
   Den
   -6 + 2 x -3
   -6 + -6
   = -12
   ^-1/₃ ÷ -12 = ^-1/₃ x ^-1/₁₂ = ¹/₃₆
2. L.C.M = 23 × 3 × 5 =120
   ¹²⁰/₆₀ = 2mins
   Next time = 1915
    +  2 
   1917 Hrs 
3.   4x(3x - 4)  
   (5+x)(4-3x)
   = -4x 
      x+5
4. Let the polygons have n& n-1 sides
   (n-2) 180: (n-1-2)180=3:2
   180n - 360 = ³/₂
   180n - 540 
   900 -180n
   n=5
   Therefore n=5 → pentagon
   -1=4 → Quadrilateral
5. -4[⁴₃ ^-x_-y] = [^-4_-5 ^-x_-y] 
   -16 +4x=-4-x
   -12+4y = -5 - y
   x=2.4     y=1.4
   centre (2.4, 1.4)
6.    
   
7. 
   
   3 + h = 3
   h = 0
   = ¹/₃ - k = -4
   k = 4¹/₃
   h + k = 0 + 4¹/₃ = 4¹/₃
8. ⁴⁴/₇√7 = ⁹⁰/₃₆₀ x ²²/₇ x r
   r = ⁴⁴/₇√7 x ⁴/₁ x ⁷/₂₂ = 8√7
   Area of square – Area of Circle
   (16√7 x 16√7) - ²²/₇ x (8√7)²
   = 1792 - 1408 = 384
   Area of the shaded part
   = ¾ x 384
   = 288cm² 
9. 
   
   Tan70 = ^h/_x ; h = xtan70
   tan46.8 =    h     ; h = (x + 10)tan46.8
                  x + 10
   xtan70 = (x + 10)tan46.8
   2.7475x = 1.06589x + 10.6589
   1.6826x = 10.6589
   x = 6.3347
   h = 6.3347tan70
   = 17.4m
10. BC/12 = DC/6 = 12/9
    DC =12 x 6 = 8
                9
    BC = 12 x 12 = 16
                  9
    Perimeter = 16 + 8 + 9 + 6 = 39cm
11. 1 COW -70 DAYS 
    1cow takes ¹/₇₀ of feeds
    ¹/₇₀ x 10 = ¹/₇
    ¹/₇ x 4 = ⁴/₇
    Remaining cows 10 - 4 = 6 ⇒ ¹/₇₀ x 6 = ³/₃₅ of feeds per day
    1 day - ³/₃₅
    ? - ³/₇
    1 x ³/₇ x ³⁵/₃ = 5days
12. (     Log5      )^¾
    6.79 x 0.391
    

    No	Log
    0.6990 6.79 0.391 3.675 x 10-1 0.3675	1.8445 0.8319 1.5922 0.4241 1.4241 x ¾ = 2.2612                           4 4 + 2.2612       4 1.5653

13. C.A = 5 x ¼ x 5 x 5tan54
    C.A = Vol = 5 x ¼ x 5 x 5tan54 x 10
    = 430.1cm³
14. 12-12x>18x-8 -28-2x
    12-2x>18x-8
    20>20x
    1>x
    18x-8 -28-2x
    20x -20
    x -1
    -1 x<1
    integral values -1, 0
15. Let B.P be T
    %loss = (T - 126) x 100 = x%
                       T
    %profit = (154 - T) x 100 = x%
                         T
    (T - 126)100 = (154 - T)100
         T                       T
    T - 126 = 154 - T
    2T = 280
    T = sh140
16. 60+3x-45=90
    3x=75
    x=25º
    360º - 2π^c
    25º= ?
    x = 5πº
          36
17.        
    1. y - intercept = -30/7
       P(0, -30/7)
       x - intercept = 10
       R(10,0)
       
       (5,-¹⁵/₇₎
    2. Tanℑ = ^Δx/_Δy
       tanℑ = ¹⁰/_30/7
       ℑ = tan-1(⁷⁰/₃₀) = 66.80º
    3. grade of line PQR = ³/₇
       Grade of ⊥r line = ^-7/₃
       y - ^-15/₇ = -7
         x - 5       3
       y + ¹⁵/₇ = ^-7/₃x + ³⁵/₃
       y = ^-7/₃x + ²⁰⁰/₂₁
       ²¹/₂₀₀(y + ⁷/₃ x) = ²⁰⁰/₂₁ x ²¹/₂₀₀
       ^21y/₂₀₀+ ⁴⁹/₂₀₀x = 1
18.      
    1.    
       

       class	mid-point	frequency	fx
       4-9 10-19 20-39 40-49	6.5 14.5 29.5 44.5	12 36 24 30 ∑f = 102	78 522 708 1335 ∑fx = 2643

       5 = ³⁰/₁₀n ; n = ⁵/₃
       2 = ^f/₂₀ x ⁵/₃ ⇒ f = 24
       f = 6 x10 x ³/₅ = 36
       f = 4 x 5 x ³/₅ = 12
       Modal class 10-9 
    2. x = ∑fx = 2643 = 25.91
             ∑f       102
    3. Area of the bars
       (4 x 5) + (10 x 6) + (20 x 2) + (10 x 5) = 170
       Media = ¹⁷⁰/₂ = 85
       (4 x 5) + (10 x 6) + (2 x y) = 85
       2y = 85 - 80
        y = 2.5
       Verticallineat 19.5 + 2.5 = 22
19.        
    1. Time by Lorry X= ⁴²⁰/_x hrs
       Time by Lorry Y = ⁴²⁰/_x+15
       ⁴²⁰/_x - ⁴²⁰/_x+15 = 1hr 24mins
       420(x + 15)-420x = ⁷/₅
             x(x + 15)
       420x + 6300 - 420x = ⁷/₅(x² + 15x)
       7x² +105x-31,500=0
       x²+15x-4500=0
       x² +75x-60x-4500=0
       x(x+75) - 60(x+75)=0
       (x-60) (x+75)=0
       x = 60  or  x = -75 ignore
    2. Therefore speed of lorry Y = 60+15 = 75km/h
       Time by Y = ⁴²⁰/₇₅ = 5.6hrs = 5hrs 36mins
    3. Distance by X from A = 60×5.6=336km
       R.S = 90+75 =165
       time they took to meet = ⁴²⁰/₁₆₅ = 2.54hrs
       Distance from A = 75 ×  2.54= 190.9km
20.      
    1.  
       
    2. P¹(1,1)  Q¹ (4,1)  R1 (2,3)
    3. P¹¹ (1,1)  Q¹¹(1,4)  R¹¹ (-2,2)
    4. Indirect congruent 
21.        
    1. L.S.F = ^4.9/_2.1 = ⁷/₃
       ^4th/_h = ⁷/₃
       12 +3h = 7h
       4h=12
       h=3cm
       H=7cm
    2. L = √7² + 4.9² = 8.5446
       l = √3² + 2.1²= 3.6620
       Top = ²²/₇ x 2.12 = 13.86cm2
       C.S.A = ²²/₇ x 4.9 x 8.5446 - ²²/₇ x 2.1 x 3.6620
       = 131.59 - 24.17 = 107.42cm²
       TSA = 13.86 + 107.42 + 150.92 = 272.2cm²
       Total Vol= 246.5+12.21
    3. ²/₃ x ²²/₇ x 4.9³ = 246.5cm^{3
       1}/₃ x ²²/₇ x 4.9² x 7 - ¹/₃ x ²²/₇ x 2.1³ x 3 =408.71cm³ 
       = 176.07 - 13.86 = 162.21
       Total vol = 246.5 + 12.21
       = 408.71cm³
22.      
    1. AQ = AO + OQ
       = a + ¹/₃b
       PQ = PO + OQ
       = ^-3/₅a + ¹/₃b
    2.      
       1. OT = OA + AT
          a + h (-a + ¹/₃b) = a - ha + ³/₅hb
          OT = OB + BT
          = b + kBP = b + k(³/₅a - b)
          -b - kb + ³/₅ka
       2. 1-h = ³/₅k    ³/₅h=1-k
          5 - 5h = 3k    h = 3 - 3k
          5 - 5(3 - 3k) = 3k
          -10 + 15k = 3k
          k = ⁵/₆; h = 3 - 3(⁵/₆) = ½
    3. BT = kBP
       BT = ⁵/₆BP
       SO BT//BP and point B common so B, T &P are collinear.
23.      
    1.    
       
    2. 5.3 × 50
       = 265km
    3. 10.8×50 
       =540 km
    4. 190º
24.      
    1. x-intercept, y=0
       0=(x-1) (x+3)²
       x=1, 
       x=-3
       Y-intercept, x=0
       y= (-1) (9)
       =-9
    2. at stationary point. ^dy/_dx = 0
       (x-1) (x² + 6x+9)
       =x³ + 5x² + 3x-9
       ^dy/_dx = 3x²+ 10x+3
       3x²+10x+3=0
       3x+9x+x+3=0
       (3x+1) (x+3) =0
       x= -¹/₃  or x=-3
       (-¹/₃ , -9¹³/₂₇)min, (-3,0)max
    3.