Mathematics Paper 1 Questions and Answers - Lugari Constituency Joint Pre Mock Exams 2023

1. Evaluate −4−4+−15÷5+−3−4÷2             (3 marks)
                        84÷−7+3−−5 
   Numerator = −4(−7 −3 −2)
                     = −4 x −12 = 48
   Denominator = −12 + 3 + 5 = −4
                    =  48  
                        −4     = −12                                         
2. Simplify completely the expression: 6x²y²−20xy+16        (3 marks)
                                                                  2x2y2−8
   Numerator =  6x²y²−20xy+16                                          =     (6xy − 8)(xy − 2)
                     =  6x²y²−12xy−8xy+16                                         2(xy + 2)(xy − 2)
                     = 6xy(xy − 2)−8(xy−2)                                    =  3xy − 4
                     = (6xy − 8)(xy − 2)                                                xy + 2
   Denominator =  2x2y2−8
                         =  2(xy + 2)(xy −2)                   
3. Given that sin (x + 60)⁰ = cos (2x), find tan (x + 60)⁰       (3 marks)
   x + 60 + 2x = 90
                 3x = 30
                   x = 10
              Tan(10 + 60) = 2.7475
4.  A triangle whose vertices are P(1, 1), Q(2, 1) and R(1.5, 2) is first rotated about (0, 0) through 1800 followed by an enlargement scale factor 3 with the centre at the origin. Find the co-ordinates of the vertices of the final image.          (4 marks) 
              P¹(−1, −1),  Q¹(2, −1)  R¹(−1.5, −2) B₁B₁
              P¹¹(−3, −3), Q¹¹(−6, −3) R¹¹(−4.5, −6) B₁B₁                               
5. List all the integral values of x that satisfy the inequalities;          (3 marks)
                           x−³/₂≤2x+1<5
                       −³/₂ ≤ x + 1
                       ^_5/₂ ≤ x
                         2x < 4
                           x < 2
                        −2½ ≤ x < 2
                                             = −2, −1, 0, 1 all intergral values                      
6. A bus travelling at an average speed of x km/h left station at 8.15 am. A car, travelling at an average speed of 80km/h left the same station at 9.00 am and caught up with the bus at 10.45 am. Find the value of x.   (3 marks)
                 Distance covered by gas at 9.00am  ⁴⁵/₆₀ X x = ¾x km   
                  Relative speed = (80 − x)km/h
                  Catch up time = 10.45
                                             9.00 
                                            1¾hrs   
                        Distance = s x t
                         ¾x = ⁷/₄ ( 80 − x)
                               ¹⁰/₄x = 140
                            x  =  56                                                      
7. Find the size of angle RPQ of a triangle PQR in which PQ = 9 cm, QR = 12 cm and RP = 6 cm  (3 marks)
                            12² = 9² + 6² − 2 x 9 x 6 Cos P
                              144 = 117 − 108Cos P
                             < RPQ = Cos ^−1 27/₋₁₀₈
                                                     = 104.48°
8. Use squares, square roots and reciprocals tables to evaluate, to 4 significant figures, the expression:  (3 marks)                             
                                                 1     +    3     
                                            √27.56    (0.071)² 
                √27.56 = 5.2497
                 (0.071)² = (7.1 x 10⁻²)²  = 50.41 x 10⁻⁴   
                  = ¹/_5.2497 + ³/_{5.041 x 10−3} 
                  = 0.1908 + 3 x 0.1984 x 10
                  = 595.2               
9. From a point 20m away on a level ground the angle of elevation to the bottom of the window is 27° and the angle of elevation of the top of the window is 32°. Calculate the height of the window.      (3 marks)
   Height of the window = 20 tan 32 − 20 tan 27
                                     =  2.30689
10. Solve for x in the equation: 3^2x+1+4×3^2x+1− 45 = 0    (3 marks)
                                     3.3^2x + 4.3.3^2x = 45
                                     15.3^2x = 45
                                      15         15
                                      3^2x = 3
                                      2x = 1           
                                                        x = ½                           
11. Three spacecrafts in different orbits go around the earth at intervals of 3, 6 and 7 hours respectively. An engineer at an observatory on earth first observes the three crafts cruising above one another at 6.35 a.m. At what time in a similar configuration if the all revolve around the earth from east to west?          (3 marks)
                                                      
         L.C.M = 2 x 2 x 7
                   = 42hrs
          6.35
         42.00
          48.35
        −24    
          24.35           = 12.35am on the third day
12. Use the mid - ordinates rule to estimate the area enclosed by the curve y = x2, x-axis and lines x=2 and x=5 using 3 strips      (3 marks)
                            h = 5 − 1 = 1
                                      3 
    

    x	2.5	3.5	4.5
    y	6.25	12.25	20.25

    
    A = 1(6.25 + 12.25 + 20.25)
       = 38.75sq units
13. A piece of wire 18 cm long is to be bent to form a rectangle. If its length is x cm, obtain an expression for its area. Hence calculate the dimensions of the rectangle with maximum area from the expression      (4 marks)
                                                                 Width = 18 − 2x 
                                                                                    2
                                                                           = 9 − x
                                                                       A  = x(9 − x)
                                                                           = 9x − x^{2 
                                                                       dy}/_dx = 9 − 2x = 0
                                                                            x = 4.5cm
                                                                     width = 9 − 4.5 = 4.5cm                                                         
14. The cost of providing a commodity consists of transport, labour and raw materials in the ratio 8:4:12 respectively. If the transport cost increases by 12%, labour cost by 18% and raw materials by 40%, find the percentage increase of producing the new commodity.                     (3 marks)
               Let the total cost be x 
               new cost = ⁸/₂₄x X ¹¹²/₁₀₀ + ⁴/₂₄x X ¹¹⁸/₁₀₀ + ¹²/₂₄x X¹⁴⁰/100
                                       =  127x
                                   100 
                % Increase = ¹²⁷/₁₀₀x − x x 100%  = 27% Increase
                                             x       
15. Vector OP = 6i + j and OQ = − 2i +5j. A point N divides PQ in the ratio 3:−1. Find ON in terms of i and j.         (3 marks) 
                     
16. In the figure below ABCDE is a cross-section of a solid ABCDEPQRST. The solid has a uniform cross-section. Given that AP is an edge of the solid, complete the sketch showing the hidden edges with a broken lines.     (3 marks)
                                                                                                                                                                                  

                                                                                    SECTION II (50 Marks)

                                                 Answer only five questions from this section in the spaces provided. 

17. A carpenter constructed a closed wooden box with internal measurements 1.5m long, 0.8m wide and 0.4m high. The wood used in constructing the box was 1.0cm thick and had a density of 0.6g/cm³.
    1. Determine:-
       1. Volume of the wood used in constructing the box in cm³. (4 marks)
          Internal volume = 150 x 80 x 40 = 480,000cm³
          External volume = 152 x 82 x 42 = 523,488cm³
          Volume of the wood = 523,488 − 480,000
                                          = 43,488cm³
       2. Mass of the box in kilograms. Give answer to one decimal place.       (2 marks)
          M = V X D
             = 43,488 X 0.6 = 26092.8g
             = 26.1kg
    2. Identical cylindrical tins of diameter 10cm, height 20cm with a mass of 120g each, were packed into the box.
       Calculate:-
       1. The maximum number of the tins that can be packed.  (2 marks)
          ¹⁵⁰/₁₀ x ⁸⁰/₁₀ x ⁴⁰/₂₀
          = 480 tins
       2. The total mass of box and the tins in kg.   (2 marks)
          = 120 x 480 + 26.0928
                   1000
             = 83.6928kg
18. Two factories A and B produce both chocolate bars and eclairs. In factory A, it costs Kshs x and Kshs y to produce 1 kg of chocolate bars and 1 kg of eclares respectively. The cost of producing 1 kg of chocolate bars and 1 kg of eclairs in factory B increases by the ratio 6:5 and reduce by the ratio 4:5 respectively.
    1. Given that it costs Kshs 460 000 to produce 1 tonne of chocolate bars and 800kg of eclares in factory A and Kshs 534 000 to produce the same quantities in factory B, form two simplified simultaneous equations representing this information.    (3 marks) 
       1000x + 800y = 460,000
             5x + 4y = 2300
       In factor B
       Cost of 1kg of chocolate = ⁶/₅x 
       Cost of 1kg of eclairs = ⁴/₅y 
       1000 x ⁶/₅x + 800 x ⁴/₅y = 534000
                  15x + 8y = 6675
    2. Use matrix method to find the cost of producing 1 kg of chocolate bars and 1 kg of eclaires in factory A.       (5 marks)
                          
    3. Find the cost of producing 100 kg of chocolate bars and 50 kg of eclaires in factory B.  (2 marks)
       ⁶/₅ x 415 x 100 + ⁴/₅ x 50 x 56.25
                    = ksh 52,050
19. Two friends Jane and Bob live 40 km apart. One day Jane left her house at 9.00 a.m. and cycled towards Bob’s house at an average speed of 15 km/h. Bob left his house at 10.30 a.m. on the same day and cycled towards Jane’s at an average speed of 25 km/h.
    1. Determine :
       1. The distance from Jane’s house to where the two friends met.         (4 marks)
          Distance covered by Jane at 10.30 am = 15 x 1.5 = 22.5hk
          Distance between them at 10.30 am = 40 − 22.5 = 17.5km
          Relative speed = 15 + 25 = 40km/hr
          Time taken before the met = ^17.5/₄₀ = ⁷/₁₆ hrs
          Distance from Jane's house = 22.5 + ⁷/₁₆ x 15 = 29.0625km
       2. The time they met.       (2 marks)
          = 10.30
                  26.15
            10.56.15 am
       3. How far from Bob's house when they met.        (2 marks)
          40 − 29.062 = 10.938
    2. The two friends took 10 minutes at the meeting point and then cycled to Bob’s house at an average speed of 12 km/h. Find the time they arrived at Bob’s house.               (2 marks )
       T = ^D/_S = ^10.938/₁₂ = 54min 41 seconds
                   = 10.56.15
                           10
                           54.41
                       12.00.56 am 
20. The masses to the nearest kilogram of some students were recorded in table below.
     
    

    Mass(kg)	41-50	51-55	56-65	66-70	71-85
    Frequency	8	12	16	10	6
    	0.4	1.2	0.8	1.0	0.2

    
    
    1. Complete the table above to 1 decimal place.      (2 marks)
    2. On the grid provided below, draw a histogram to represent the above information. (3 marks)
       
    3. Use the histogram to:
       1. State the class in which the median mark lies.                   (1 mark)
          56 - 65
       2. Estimate the median mark                     (2 marks)
          55.5  +    6    = 59.25
                     2 x 0.8
       3. The percentage number of students with masses of at least 74kg.                (2 marks)
          ⁵/₂₆ x 100% = 8.929%
21.  
    1. straight line L₁ whose equation is 9y−6x = −6 meets the x-axis at Z. Determine the coordinates of Z      (2 marks)
                                     − 6x = − 6
                                          x = 1
                                        Z(1, 0) 
    2. A second line L₂ is perpendicular to L₁ at Z. Find the equation of L2 in the form ax + by = c, where ,b and c are integers.    (3marks)
                                       L1 y = ²/₃x − 6
                                      L2 y − 0 = −3
                                           x − 1      2
                                         y = ⁻³/₂x + ³/₂
                                          3x + 2y = 3
    3. A third line L₃ passes through the point (2,5) and is parallel to L1. Find:
       1. The equation of L3 in the form ax+by=c, where a, b and c are integers.           (2 marks)
                                y − 5 = 2
                                x − 2    3
                                y − 5 = ²/₃x − ⁴/₃
                                        2x − 3y = −11
       2. The coordinate of point R at which L2 intersects L3.           (3 marks)
                                   3x + 2y = 3
                                   2x − 3y = −11
          
                                   6x + 4y = 6
                                   6x − 9y = −33
                                          13y = 39
                                             y = 3
                                             x = −1
                                       R(−1, 3)
22.  
    1. Complete the table below for the equation y = 2x² + 3x − 11 (2 Marks)
       

       x	−5	−3	−2	−1	0	1	2	3
       2x2	32	18	8	2	0	2	8	18
       3x	−12	−9	−6	−3	0	3	6	9
       − 11	−11	−11	−11	−11	−11	−11	−11	−11
       y	−9	−2	−9	−12	− 11	− 6	3	16

    2. On the grid paper provided draw the graph of y = 2x² + 3x − 11 (3 Marks)
       
    3. On the same axes draw the graph of y = 2x + 1 (1 Marks)
    4. Use your graph to solve the quadratic
       1. 2x² + 3x −11 = 0 (2 Mark)
          X = −3.2 OR X =1.7
       2. 2x² + x − 12 = 0 (2 Marks)
          Y = 2x² + 3x − 11
           0 = 2x² + x − 12 
              y = 2x + 1               x = −2.7 or 2.3
23. Five points, P, Q, R, V and T lie on the same plane. Point Q is 53km on the bearing of 055° of P. Point R lies 162° of Q at a distance of 58km. Given that point T is west of P and 114km from R and V is directly south of P and S40°E from T.
    1. Using a scale of 1:1,000,000, show the above information in a scale drawing.        (3 marks)
       
    2. From the scale drawing determine:
       1. The distance in km of point V from R.                 (2 marks)
          6.4 x 10 = 64km
       2. The bearing of V from Q.           (2 marks)
          029° + 180 = 209°
       3. Calculate the area enclosed by the points PQRVT in squares kilometers.     (3 marks)
          = ½ x 41 x 49 + ½ x 53 x 49Sin125 + ½ x 64 x 58Sin96
          = 3914 km
24. The displacement S of a particle after t seconds is given by S = 4t³ − ⁵/₂t² − 3t + 3.
    Determine the:
    1. velocity of the particle when t = 3 (3 marks)
       V = 12t2 − st − 3
            at t = 3
       V = 12(3)² − 5(3) − 3
          = 90 m/s
    2. value of t when the particle is instantaneously at rest (3 marks)
       12t² − st − 3 = 0
       12t² − 9t + 4t − 3 = 0
       3t(4t − 3) + (4t − 3) = 0
       (3t + 1)(4t − 3) = 0
       3t + 1 = 0
                 t = − ¹/₃ or
       4t − 3 = 0
                t = ¾ seconds
    3. displacement when the particle is instantaneously at rest. (2 marks)
       s = 4(¾)³ − ⁵/₂(¾)² − 3 x (¾) + 3
          = 1¹/₃
    4. acceleration of the particle when t=2 seconds (2 marks)
       a = 24t − 5
          = 24(2) − 5
          = 43m/s²