Mathematics Paper 1 Questions and Answers - Wahundura Boys Mock Examination 2023

INSTRUCTIONS

• The paper contains two sections: Section I and II.
• Show all the steps in your calculations, giving your answers at each stage in the spaces below each question.
• Marks may be given for correct working even if the answer is wrong.
• Non-programmable silent electronic calculators and KNEC Mathematical tables may be used, except where stated otherwise.

QUESTIONS

SECTION 1 (ANSWER ALL QUESTIONS IN THIS SECTION)

1. The difference between two numbers is 4 and their product is 165. What is the sum of their reciprocals (4 marks)
2. Solve for x and y in the equation:
   4^x (27)^y =72   (2 marks)
3. Simplify completely: (4 marks)
   
4. Equal squares as large as possible are ruled off on a rectangular board 54cm by 78cm. find the number of those squares (3 marks)
5. A regular polygon has an interior angle of 150° and a side of length 10cm.
   1. find the number of sides the polygon has (1 mark)
   2. Find the area of the polygon (3 marks)
6. The line passing through the points P (5, b) and Q (2, 3) is perpendicular to the line 4y + 3x = 1. Determine the value of b (3 marks)
7. Solve the simultaneous equations below by matrix method (4 marks)
   Log₂(2x + 5y) = 4
   Log₁₇ (3x + 4y) = 1
8. The cost of a pair of shoes was Ksh. 800. A dealer raised the price by 10% and later reduced it by 20%. What is the current price of the pair of shoes (2 marks)
9. Given that sin∝= √5⁄3, where ∝ is an acute angle, find without using mathematical tables:
   1. tan2 ∝ - sin2∝ (2 marks)
   2. sin (90 - ∝)° (1 mark)
10. Given that position vector OT = and the column vector TN = , find the position vector ON (2 marks)
11. A pupil stands 20m away from a flag post. The angle of elevation of the top of the post is 30°. Find how far he has to walk towards the post for the angle of elevation to be 82°. (Assume the ground between him and the post is flat. Give the answer to one decimal place (3 marks)
12. A cone 80mm high has mass 7Kg. A frustrum of this cone is formed by cutting off the top 30mm of the cone as shown below. Find the mass of the frustrum (4 marks)
    
13. 10 million shillings was shared by three brothers. Salat got ¼ . Kipgwen got ½ . While Sang’ got 1⁄5. The remainder was donated to a children home. How much was donated to children home (3 marks)
14. The figure below is a velocity – time graph for a car.
    
    Find the total distance travelled by the car (2 marks)
15. Find the integral values of x which satisfy the compound inequality (3 marks)
    3x - 2≤8 + 5x>12 + 6x
16. Using a ruler and a pair of compasses only,
    1. Construct triangle ABC in which BC = 8cm, angle ABC= 112.5° and angle BAC = 45° (2 marks)
    2. Drop a perpendicular from A to meet CB produced at P. Measure length AP and find the area of the triangle ABC (2 marks)

SECTION II (50 marks)
Answer only five questions in this Section

17.  A swimming pool is 20m by 12m and it slopes gently from a depth of 1m at the shallow end to a depth of 3m at the deep end.
    1. Calculate the volume of water in the swimming pool (in m³) when it is full. (3 marks)
    2. If the swimming pool is to be drained by a pump which removes water at the rate of 2.5m³ per minute, how long will it take this pump to drain the swimming pool it was full? Give your answer in hours (3 marks)
    3. If the sides of the swimming pool and the floor are to be covered with square tiles of side 20cm, find to the nearest 100, the number of tiles required (4 marks)
18. The position vectors of A, B and C are A = 3i -2j, B = - 6i+4j and c = -9i -3j.
    1. State the column vectors
       1. AB (2 marks)
       2. CB (2 marks)
    2. Find the distance from A to C (2 marks)
    3. Find the coordinates of the midpoint of AC (2 marks)
    4. If C’ is the image of C under translation vector  find the coordinates of C’ (2 marks)
19. The points A’B’C’ are the images of A(4,1) B(O,2) and C(-2,4) respectively under a transformation represented by the matrix m = 
    1. Write down the coordinates of A’B’C’ (3 marks)
    2. A”B”C” are the images of A’B’C’ under another transformation whose matrix
       N =  Write down the co-ordinates of A”B”C” (3 marks)
    3. Transformation M followed by N can be replaced by a single transformation P. Determine the matrix for P (2 marks)
    4. Determine the inverse of matrix P (2 marks)
20.  
    1. Find the equation of a straight line which is perpendicular to 3y – 5x = 5 and passes through the point of intersection of the lines 2x + y = 5 and x – 2y = 0 (5 marks)
    2. A straight line y = 2⁄3 x- 2⁄3 meets the x axis at point T.
       1. Determine the coordinates of T (2 marks)
       2. Determine the acute angle between the line y = 2⁄3 x- 2⁄3 and x – axis (2 marks)
       3. Find the y –intercept of line y = 2⁄3 x- 2⁄3 (1 mark)
21. Bujumba secondary school intends to buy a certain number of chairs for Ksh. 16200. The supplier agreed to offer a discount of Ksh. 60 per chair which will enable the school to get 3 chairs more. Taking y as the originally intended number of chairs.
    1. Write an expression in terms of y for:
       1. Original price per chair (1 mark)
       2. Price per chair after discount (1 mark)
    2. Determine:
       1. The number of chairs the school originally intended to buy (4 marks)
       2. Price per chair after discount (2 marks)
       3. The amount of money the school would have saved per chair if it got the intended number of chairs at a discount of 15%. (2 marks)
22. The figure ABCD is a quadrilateral in which AB=7cm, angle DAB = angle ABC = 90°. Angle APD = 76°. Point P lies on AB such that AP = 2cm and PB is 5cm. length BC=1cm. AB produced meets DC produced at Q.
    
    Calculate to one decimal place
    1. The length AD (1 mark)
    2. The size of angle CPD (2 marks)
    3. The area of quadrilateral ABCD (2 marks)
    4. The length BQ (2 marks)
    5. The area of triangle CDP (3 marks)
23. Andai recorded data on observation of time spent by a local university’s first year bachelor of commerce students in the library as follows:
    

    Time spent in minutes	11-20	21-30	31-40	41-50	51-60
    Cumulative frequency	70	170	370	470	500

    
    Calculate:
    1. The mean (4 marks)
    2. The median (3 marks)
    3. On the grid provided, draw a frequency polygon from this data (3 marks)
       
24. The displacement S metres of a moving particle after t seconds is given by S = 2t³ – 5t² + 4t + 3.
    Determine:
    1. The velocity of the particle when t = 4 seconds. (3 marks)
    2. The value of t when the particle is momentarily at rest (3 marks)
    3. The displacement when the particle is momentarily at rest (2 marks)
    4. The acceleration of the particle when t = 10 seconds (2 marks)

MARKING SCHEME

SECTION 1

1. x - y = 4        ; xy = 165
   x = 4 + y
   y (4 + y ) = 165
   y2 + 4y - 165 = 0
   (y - 11) (y + 15) = 0
   when y= 11       when y= -15
             x= 15                x= -11
   1⁄11 + 1⁄15= - 26⁄165
   or
   - 1⁄11 + - 1⁄15= - 26⁄165
   
   
2. 2 ^2x×3^3y= 2³×3²
   2x=3       3y=2
   x=1 1⁄2      y=2⁄3
   
   
3. 2x² - 9x-5 = (2x+1)(x-5)
   1 ⁄ x-5 - x+6 ⁄ (2x+1)(x-5) = 2x+1-x-6 ⁄ (2x+1)(x-5)
   =x-5 ⁄ (2x+1)(x-5)
   =1 ⁄ 2x+1
   
   
4. G.C.D of 78 and 54
   
   GCD = 2×3 = 6cm
   78⁄6 × 54⁄6 = 117 squares
   
   
5.  
   1. Exterior angle = 30°
      No.of.sides = 36030 = 12 sides
   2. 
      
      Tan 15° = ^s⁄_h
      h = ⁵ ⁄_tan15 = 18.66 cm
      Area of polygon = 1⁄2×1018.66×12
      =1119.6cm²
      
      
6. y = -3⁄4 x + 1⁄4
   M₂ = 4⁄3
   ^b-3 ⁄ _5-2 = 4⁄3
   3b = 21
   b=7
   
   
7. 2x+5y =16
   3x+4y =17
   
   de=8-15=-7
   
   x=3  y=2
   
   
8. 800 × 110⁄100 × 80⁄100
   sh=704
   
   
9.  
   1. 
      
      3²-(√5)² = 4
      √4 = 2
      Tan∝ = √5⁄2
      Tan²∝ - Sin² ∝= 5⁄4-5⁄9=25⁄36
   2. Sin (90-∝) =cos∝
      =2⁄3
      
      
10. ON = OT+TN
    
    
    
11.  
    
    Tan 30° =^h⁄20       h=11.547m
    Tan82 =^11.547 ⁄20-d
    20-d =¹¹⁵⁴⁷ ⁄_Tan82  =11.6228
    d=20-1.6228=18.377
    d=18.4m
    
    
12. mass of frustrum=mass of whole-part cut off
    Lsf=30:80=3:8
    vsf=(3:8)³ =29:512
    mass cut off =27⁄512 ×7kg =0.3691kg
    mass of frustrum =7-0.3691
    =6.6308
    
    
13.  Remainder = 1-1⁄5-1⁄2-1⁄4
    =1⁄20
    donated amount =120 ×10000000
    =sh500000
    
    
14. Distance=1⁄2[30+15]×50
    =1⁄2× 45×50
    =1125m
    
    
15. 3x - 2 ≤ 8 + 5x
    -2x ≤ 10
    x ≥ -5
    8 + 5x >12 + 6x
    -x > 4
    x < -4
    integral value= -5
    
    
16. 
    
    AP=4.2 ± 0.1
    Area= 1⁄2 × 8 × 4.2
    =16.8cm²
    
    construction oa angle ABC 1mk
    complee triangle                  1mk
    length of AP                         1mk
    Area                                      1mk
    
    

SECTION 2

17. 
     
    1. cross sectional area = 1⁄2 × 12 ×  4
       =24m²
       volume = 24 × 20
       =480m³
    2. 1min = 2.5m³
       48cm³ = ^{480 × 1}  ⁄ _2.5
       =192minutes
       =13hrs 12min
    3. 
       
          =12.1655m
       Area of floor = 12.655 × 20
       =243.31
       Total area = 243.31+24+24+20+60
       371.31 0.20.²
       =9282.75
       =9300(nearest 100)
       
       
18.   
    
19.    
    
20.  
    1. 2x + y = 5            
       2(x - 2y = 0)           
       
       2x + y = 5
       -                         
       2x - 4y = 0                          
       0 + 5y = 5                        
          y = 1
          x = 2  
       
       3y - 5x = 5
        3y = 5x + 5
       y = ^5x⁄3  + 5⁄3
       
       gradient of perpendicular line = 3⁄5
       ^{y - 1} ⁄ _{y - 2} = -3⁄5
       5y - 5 = -3x + 6
       5y = -3x -11
    2.  
       1. 2⁄3x - 2⁄3 = 0
          2⁄3x = 2⁄3
          x = 1=T(1,0)
       2. Tanθ = 2⁄3
           θ =33.69°
       3. y = -2⁄3
21.  
    1.  
       1. ¹⁶²⁰⁰/ _y
       2. ¹⁶²⁰⁰/ _y - 60 or ¹⁶²⁰⁰/ _{y + 3}
    2.  
       1. ¹⁶²⁰⁰/ _y - ¹⁶²⁰⁰/ _y+3 = 60
          16200 (y + 3) - 1600y = 60y (y + 3)
          16200y + 48600 - 16200y = 60y² + 180y
          60y² + 180y - 48600 = 0
          y² + 3y - 810 = 0
          y² 30y - 27y - 810 = 0
          y(y + 30) - 27(y + 30) = 0
          (y - 27)(y + 30) = 0
          y = 27 or y = -30
          chairs = 27
       2. ¹⁶²⁰⁰/ ₃₀ = ksh540
       3. ¹⁵/ ₁₀₀ ×¹⁶²⁰⁰/ _2T
22.  
    1. Tan 76° = ^AD/₂
       AD = 2 tan76 = 8.02156
       =8.0(1dp)
    2. Tan (CPB) = 1⁄5
       angle CPB = 11.31
       angle CPD = 180 - 11.31 - 76
       = 92.69
       =92.7 (1dp)
    3. 1⁄2 (8 + 1) × 7 = 31.5
    4. 1⁄8 = ^x/ _{x + 7}
       x + 7 = 8x
       7x = 7
       x = 1
       BQ = 1cm
    5.   
       
23.   
    1.  
       

       Time	F	X	FX	CF
       11 - 20	70	15.5	1085	70
       21 - 30	100	25.5	2550	170
       31 - 40	200	35.5	7100	370
       41 - 50	100	45.5	4550	470
       51 - 60	30	55.5	1665	500
       sum	500		16950

       
       mean = ^∑ƒ/ _{∑ƒ
       = 16950/ 500}
       =33.9
    2. median position = ⁵⁰⁰/ ₂ = 250
       median class = 31 - 40
       median = 30.5 + ^{250 - 170}/ ₂₀₀ × 10
       =30.5 + ⁸⁰/ ₂₀₀ ×10
       =30.5 + 4
       =34.5
    3.     
       
24.   
    1. 5 = 2t³ - 5t² + 4t + 3
       V= ^ds/ _dt = 6t² - 10t + 4
       at t = 4
       V= 6(4²) - 10(4) + 4
       =60m/s
    2. 6t² - 10t + 4 = 0
       3t² - 5t + 2 = 0
       3t² - 3t - 2t + 2 = 0
       3t(t-1) -2(t-1)=0
       (3t-2)(t-1) = 0
       t = 2⁄3 s or t = 1sec
    3. s= 2 (2⁄3)³ - 5(2⁄3)² + 4(2⁄3) + 3
       = 4 1⁄27m or 4.037m
       s= 2 (1³) - 5(1)² + 4(1) + 3
       =4m
    4. a = dv/ dt = 12t - 10
       =12(10) - 10
       =110m/s²