Mathematics Paper 1 Questions and Answers - Mokasa II Joint Mock Exams 2023

• The paper contains two sections A and B.
• Answer ALL questions in section A and anyfive from section B in the spaces provided below each question.
• Marks may be given for correct working even if the answer is wrong.
• Non-programmable silent electronic calculator and KNEC mathematical tables may be used, except where stated or otherwise.

SECTION A (50 MARKS)

Answer all questions in this section in the spaces provided

1. Translation T is represented by the column vector  and another translation U by column vector .  A point P is mapped to a point Q by T and then Q is mapped to a point R by U.  If R has coordinates (7,-4) determine the coordinates of P.      (3 marks)
2. A fruit dealer blends the fruit juice in a common container to the brim before choosing the quantities in which to distribute them. She can pack them in either 20 litres, 24 litres or 36 litres can before selling them. If she chooses 20 litres cans she remains with 13 liters while when she uses 24 litres 17 litres remain in a container and 29 litres remain when distributed in 36 litres cans. Determine the leastcapacity of her container in litres.    (3 marks)
3. A family spent ²/₅ of their income on food, ¹/₃ of the remainder on water bill and saved the rest.  If sh. 1200 more is spent on food than water bill.  Find how much they  saved?     (3 marks)
4. Evaluate without using tables or calculator.   2 marks)
   
5. Simplify:    (3 marks)
     x+2    −     3      
    x²−1       2(x−1)
6. A hollow cylindrical alloy of length 40mm has a mass of 352g.  If its internal radius is 3cm and its 0.01m thick; calculate the density of the metal used to make the alloy in kg/m³. Take pi as 22/7. (3 marks)
7. In the figure below, O is the centre of the circle. AOis parallel to BC and angle BDC=130°.
   
   A polygon whose exterior angle ∠AOB above is formed. Find;
   1. Sum of interior angles of the polygon. (3 marks)
   2. Name of the polygon. (1 mark)
8. A salesman dealing in iphones earns a basic salary and commission as follows.  
   

   Sales	Commission
   For sales up to ksh.  150,000	0%
   For sales above ksh. 150,000
   -       First ksh. 85,000 3%	3%
   -       First ksh. 85,000 3%	4%
   -        Any amount above	5%

   In the month of December 2022, the salesman earned a basic salary of  ksh. 25,000 and he sold 100 iphones at a discount of 10%.  Calculate;
   1. His total sales in the month of December if the iphones were marked at  Ksh. 5000.        (1 mark)
   2. His total earnings that month.       (3 marks)
9. Using a ruler and pair of compasses only.  Draw parallelogram ABCD in which AB = 8 cm, BC = 5cm and <ABC is 105°.  Construct circle touching edges AB, AD and DC and measure its radius.       (4 marks)
10. Solve for x in the equation;     (3 marks)
    25^x −  5^(2x−1) = 500     
11. A line L1 whose equation is 5x − 3y − 15 = 0 passes through the points A and B.  Point A is on the x-axis while point B is equidistant from x and y-axes and ties in the fourth quadrant of a Cartesian plane.  Calculate the exact coordinates of points  A and B. (3 marks)
12. Matrix R =  transforms point P to a point P1 lying in a straight line.  Find value of x.      (3 marks)
13. A cone is made from a sector whose dimensions are given below.  Calculate the volume of the cone. (Take π= ²²/₇).  OA = OB = 7cm radius of the sector and <AOB=120°.Give your answer to 3 significant figures.
    (4 marks)
    
14. Given that 6 cos⁡ (x+15)° − 6 sin⁡(2x−30)° = 0, find tan⁡(2x − 10°) leaving your answer in surd form.    
    (3 marks)
15. Solve the inequality 3x−10≤ 6x−26 <4x−6 hence express the solution as a combined inequality.  (2 marks)
16. Two trains T₁ and T₂ travelling in same direction, on parallel tracks, are just12m apart.  Train T₁ is 72m long and travelling at 108km/h.  T₂ is x mlong and travelling at 72km/h.  Find the length of T₂ in metres if they pass each other completely after 16.2s. (3 marks)

SECTION B (50 MARKS)

Answer any FIVE questions from this section in the space provided

17. The frequency distribution table below shows the mass in kilograms (kg) of parcels delivered by a courier company to their office in Lubao.
    

    Mass (kg)	50 − 99	100 − 199	200 − 299	250 − 349
    Number of parcels	380	230	160	1700
    Frequency density

    1. Complete the table above to 1dp    (2 marks)
    2. On the grid provided below, draw a histogram to represent the information shown in the table above.
    3. Draw a vertical line in the histogram where the median lies. Show your calculations using the graph.
       (3 marks)
    4. Use the histogram to determine the number of parcels with a mass of 300kg or more delivered by the courier company.     ( 2 marks)
18. A solid which comprises of a cylindrical solid and a base which is a frustum of a pyramid.  The cylindrical part which is open at the top has a portion of the cylinder cut off along a dotted line as shown.  Radius of the cylinder is 9cm and height 28cm.  
    
    Given that the original pyramid had slant edge of 32.5cm and HG = 24cm, GF = 7cm, BC = 4.9cm and AB = 16.8cm; Find to 1d.p;
    1. the surface are of the solid. (6 marks)
    2. the volume of the solid. (4 marks)
19. Point B is 110km on a bearing of S60°E from point A.  Point C is 90km on a bearing of 065 from B.  Point D is 075 from A and 336° from C. 
    1. Using a scale of 1cm to represent 20km, draw a diagram to show the relative positions of A, B, C and D.         (5 marks)
    2. Use your diagram to determine;
       1. the distance and the bearing of B from D. (2 marks)
       2. the distance and compass bearing of A from C. (3 marks)
20. In the figure below, ABCD is a quadrilateral in which AB=17 cm, AD=16 cm and angle ABC=150°. ABis parallel to DC and AB=BD.
    
    Calculate correct to 2 decimal places;
    1. The length of BC. (3 marks)
    2. The length of AC. (2 marks)
    3. The size of angle ACD. (3 marks)
    4. Area of the quadrilateral. (2 marks)
21. Odete and Akala entered into a business partnership in which they contributed  ksh. 120,000 and ksh. 150,000 every year respectively. After one year Chelang’a joined the business and contributed ksh. 90,000.
    1. Calculate the ratio of their investment after 3 years of business. (3 marks)
    2. It was agreed that 30% of the profits after 3 years be used to cater for the cost of running the business, while the remaining would be shared proportionally.  Calculate each person’s share, if the profit made after three years has ksh. 187,000.   (  4   marks)
    3. If each of them invested their shares back in the business, find their new individual investments at the beginning of the fourth year.  (  3 marks)
22. A stone is through vertically upwards such that its height is s meters above the ground after t, seconds is given by s = 40t² −10t³.  Find;
    1. Exact maximum height reached by the stone. (4 marks)
    2. The instant at which the stone attains maximum velocity. (3 marks)
    3. Its velocity after 1 second. (3 marks)
23. The velocity,V m/s, of a particle at any time is , t sec., is given by the equation                   
    V = t² − 4t + 5.
    1. Complete the table below;              (2 marks)
       

       t	1	1.5	2	2.5	3	3.5	4	4.5	5	5.5	6	6.5	7	7.5	8
       v	2				2		5				17

    2. Use mid-ordinate rule with seven strips to estimate the area enclosed by the curve V = t² − 4t + 5, the x- axis and the lines t = 1 and t = 8.   (2 marks)
       Find the exact area of the region in (b) above.          (3 marks)
    3. Calculate percentage error introduced by using mid- ordinate rule in (b) above to 2d.p. (3marks)
24.  
    1. On the grid provided, draw a square (s) with vertices (6,3), (7,0), (9,4) and (10,1). Draw also two straight lines AB and AC where the coordinates of A ,B and C are (1,−2), (3,2) and (0,1) respectively
       S¹ is the image of S under reflection in the line AB and S¹¹ is the image of S¹ under reflection in the line AC. Draw S¹ and S¹¹.
    2. Describe the transformation which maps S onto S¹¹ if the transformation is
       1. Translation     (1 mark)
       2. Rotation       (2 marks)

MARKING SCHEME

1.   
   
   
   
2. 20−13 = 24−17 = 36−29 = 7
   Lx−1/.C.M of 20, 24 & 36 = 360
   360 − 7 = 353litres
3. food = ²/₅x
   water = ¹/₅x
   saved = ²/₅x
   ²/₅x − ¹/₅x = ¹/₅x
   ¹/₅x = 1200
   x = 6000
   ²/₅ × 6000
   = Sh. 2,400
4.  
   
5. 2(x+2) − 3(x+1) 
       2(x² −1)
         1−x       
   2(x−1) (x+1)
      −1(x−1)     
   2(x−1)(x+1)
      −1    
   2(x+1)
6.  ²²/₇(4² − 3²)4 = 88cm³
   density = ³⁵²/₈₈ = 4g/cm³
   4 × 1000 = 4000 kg/m³
7.  
   1. ³⁶⁰/₄₀ = 9
      180(9 − 2)
       = 1260
   2. Nonagon
8.  
   1. 0.9 × 5000
      Sh. 4500
      4500 × 100
       =Ksh. 450,000
   2. 85,000 ×  0.03  = 2550
      85,000 × 0.04   = 3400
      130,000 × 0.05 = 6500   
                               = 12,450
      12450 + 25000 = Ksh 37,450
9.  
   
10. 5^2x − 5^(2x−1) = 500
    m − ^m/₅ = 500
    5m − m = 2500
    4m = 2500
      m = 625
    625 = 5^2x
    5⁴ = 5^2x
    4 = 2x
    x = 2
11. 5x − 15 = 0
     5x = 15
     x = 3
    A(3,0)
    5x − 3y − 15 = 0 ......(i)
                y = −x .........(ii)
        8x = 15
        x = ¹⁵/₈
    B( ¹⁵/₈, −¹⁵/₈ ) or B(1.875 , −1.875)
12.  x(x−2) − 24 = 0
    x² − 2x − 24 = 0
    x = −4 or 6
13. A.L = ²⁴⁰/₃₆₀ × ²²/₇ × 2 × 7 = ⁸⁸/₃ 
    ⁸⁸/₃ = ²²/₇ × 2 × 2
       R = ¹⁴/₃
    h = √(7² − (¹⁴/₃)² = 5.217
    v = ¹/₃ × ²²/₇ ×(¹⁴/₃)² × 5.217
       = 119cm³ 
14. (x+15) + (2x−30) = 90
    3x − 15 = 90
            3x = 105
              x = 35
    tan(2 × 35 − 10)
    tan 60 = √3
15. 3x −10 ≤ 6x − 26
    16 ≤ 3x
    ¹⁶/₃ ≤ x
    6x − 26 < 4x − 6
    2x < 20
    x < 10
    ¹⁶/₃ ≤ x ≤ 10
16. x+72+12 = 16.2
         10
    x + 84 = 162
    x = 162 − 84
    x = 78m
17. 1. 
       

       Mass (kg)	50 − 99	100 − 199	200 − 299	250 − 349
       Number of parcels	380	230	160	1700
       Frequency density	7.6	2.3	3.2	1.3

    2.  
       
    3. 50 × 7.6 + 2.3 × 100 +50 × 3.2  + 1.3 × 100 = 900
       450 = 7.6 × 50 + 2.3y
       2.3y = 70
            y = 30.43
       30.43 + 99.5
         = 129.9347826 ≅ 129.9
    4.  349.5 − 300 = 49.5
       49.5 × 1.3 = 64.35
       = 64 parcels
18.  
    1. (24+16.8)1.7434 = 71.16
       (7+4.9)3.86 = 45.93
       24 × 7         = 168
       (16.8 × 4.9) = 82.32 − (²²/₇ × 1.52) = 75.25
       (2 × ²²/₇ × 1.5 × 28) − (¹/₂ × 4 × ²²/₇ × 1.5²)
       264 − 14.14 
         = 249.86
       Total Surface Area
       71.16 + 45.93 + 168 + 75.25 + 249.86 
       = 610.2cm²
    2.  (¹/₃ × 24 × 7 × 30) = 1680cm³
       (¹/₃ × 16.8 × 4.9 × 27.12) = 744.1728
       Vol of frustum = 1680 − 744.1728
       = 935.8272
       (²²/₇ × 1.5 × 28) − (²/₃ × ²²/₇ × 1.5³)
       198 − 7.07143 = 190.93
       Vol of solid 
       935.8272 + 190.93
       = 1126.8cm³   
19.  
    1.  
       
    2.  
       1. 5.5 ± 0.1 
          (110 ± 2)km
          210° ± 2°
       2. 8.8km ± 0.1
          176km ± 2
          N86°W ± 2°   
20.  
    1. 16² = 17² + 17² − 2 × 17 × 17 CosB
       −322 = −578 Cos B
       B = 56.14°
          17    =        BC       
       Sin 30      Sin 56.14
       BC = 28.23cm
    2. AC² = 17² + 28.23² − 2 × 17 × 28.23 Cos 150
        = 43.79cm
    3.   16    =       43.79     
       Sin C      Sin 118.07
       C = Sin⁻¹(Sin 118.07 × 16)
                              43.79
          = 18.81°
    4. ¹/₂ × 17² × Sin 56.14 + ¹/₂ × 43.79 × 17 Sin 93.86
       119.993 + 371.3706
       = 491.36cm²  
21.   
    1. 120,000 × 3 = 360,000
       150,000 × 3 = 450,000
       90,000 × 2 = 180,000
       4 : 5 : 2
    2. 0.7 × 187,000 = Sh. 130,900
       ⁴/₁₁ × 130,900 = 47,600
       ⁵/₁₁ × 130,900 = 59.500
       ²/₁₁ × 130,900 = 23,800
    3. Odete = 120,000 × 4 + 47,600 = Ksh. 527,600
       Akola = 150,000 × 4 + 55.500 = Ksh. 655,500
       Chelang'a = 90,000 × 3 + 23,800 = Ksh. 293,800
22.  
    1. 80t − 30t² = 0
       t(80 − 30t) = 0
       t = ⁸⁰/₃₀ = ⁸/₃
       t = 
       S = 40 × (⁸/₃)²  − 10 × (⁸/₃)³
          = 284⁴/₉ − 5120
                              27
        = 94²²/₂₇ 
    2. 80 − 60t = 0
        80 = 60t
       t = ⁴/₃s
       ₌ 1¹/₃s
    3. v = 80t − 30t² 
         = 80 × 1 − 30 ×1²
        = 50m/s
23.  
    1.  
       

       t	1	1.5	2	2.5	3	3.5	4	4.5	5	5.5	6	6.5	7	7.5	8
       v	2	1.25	1	1.25	2	3.25	5	7.25	10	13.25	17	21.25	26	31.25	37

    2.  A = 1(1.25 + 1.25 + 3.25 + 7.25 + 13.25 + 21.25 + 31.25)
       = 78.75 sq units
    3. A = ⁸∫(t² − 4t + 5)
       ¹[¹/₃t³ − 2t² + 5t]⁸
       (⁵¹²/₃ − 2 × 64 + 40) − (¹/₃ − 2 + 5)
          = 79¹/₃ sq units
    4. % error = 79¹/₃ − 78.75
                               79¹/₃ 
         = 0.74%
24.  
    1. S¹(2,5)  (1,8)  (−1,4)  (−2,7)
       S¹¹( −4,3)  (−5,6)  (−2,7)  (−5,6)
       
    2.   
       
       1. Translation vector 
       2. Positive quarter - turn about (1, −2) or + 90 turn about (1, −2)