Mathematics Paper 1 Questions and Answers - Maranda High Pre Mock Exams 2023

Instructions to Candidates

• This paper consists of two sections; Section I and Section II.
• Answer all the questions in Section I and only five questions from Section II
• Show all the steps in your calculations, giving your answers at each stage
• 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.

SECTION I (50 marks)

Answer all questions in this section in the spaces provided

1. Evaluate          (2 marks)
2. Given that log₁₀2 = 0.3010 and log₁₀3 = 0.4771 without using tables or calculator find log 0.036 correct to 4 significant figures. (3 marks)
3. Simplify                   (3 marks)
   ax − 2a − 5x + 10
           8 − 2x²
4. Solve the equation sin (150 ÷ x)° = cos (12x)° for which x is acute. (3 marks)
5. Under an enlargement, the image of the points A (3,1) and B (1,2) are A' (3,7 )' and B' (7,5). Find the centre and scale factor of enlargement. (4 marks)
6. A shopkeeper bought a bag of sugar. He intends to repack the sugar in 40g , 250g and 750g .Determine the least mass in grams of sugar that was in the bag. (3 marks)
7. Solve for x and y in 3^3y ÷ 3^2x = 6561 and 2^2x × 2^3y = 1.      (4 marks)
8. The figure below shows a regular tetrahedron PQRS.
   
   Draw its net and measure the length of the straight path of PS through the midpoint T over the edge QR. (3 marks)
9. The figure below shows three circles touching each other externally.
   
   If the centres of the circle form a triangle with sides of length 9 cm, 8 cm and 7 cm, calculate the radii of each the circles. (3 marks)
10. A Kenyan company received M Us dollars. The money was converted into Kenya Shillings in a bank which buys and sells foreign currencies as shown below.
    

    	Buying (Kshs)	Selling (Kshs)
    1 Sterling pound	125.78	126.64
    1 US Dollar	75.66	75.86

    1. If the company received Kshs.15, 132, 000, calculate the amount, M Us Dollars. (2 marks)
    2. The company exchanged the above Kenyan shillings into Sterling pounds to buy a car in Britain. Calculate the cost of the car to the nearest Sterling Pound. (2 marks)
11. 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)
12. The poistion vector of points A and B are −10i −6j + 9k and −5i + k respectively. Calculate leaving your answer in surd form. (3 marks)
13. Solve the inequality 5 − 2x < ½x < x+2 and represent the solution on a number line (3 marks)
                                                              3
14. Use tables of reciprocals and square roots to evaluate.    3     + √0.4036    (3 marks)
                                                                                            0.521          
15. Two farmers Salat and Ahmed share a grazing field. Salat puts 120 cows for 18days and Ahmed puts 150 cows for 16 days. If the rented land was charged Kshs.8 550, how much should Ahmed pay if they share proportionally? (3 marks)
16. Construct triangle XYZ in which XY = 4.5 cm, YZ = 6 cm and XZ = 5.5 cm. Construct a circle that touch ZY, XY and XZ produced and also opposite to ∠YXZ. (3 marks)

SECTION II (50 marks)

Answer only five questions from this section in the spaces provided

17. A salesman dealing in mattresses earns a basic salary and commission as follows:
    

    Commission

    For sales up to Ksh 150 000           0% For sales above Ksh 150 000 First Ksh 85 000                              3% Next Ksh 85 000                              4% Any amount above Ksh 320 000     5%

    1. In the month of January 2023, the salesman earned a basic salary of Ksh 45 000 and he sold 110 mattresses at Ksh 5 000 each. Calculate:
       1. His total sales in the month of January 2023. (1 mark)
       2. His total earnings that month. (3 marks)
    2. In the month that followed, his basic salary was decreased by 10%. If he received a total earning of Ksh 47 450 in that month, calculate:
       1. Total sales that month. (4 marks)
       2. The number of mattresses sold in that month. (2 marks)
18. In the figure below, PQRSTU is a regular hexagon.
    
    1. Describe fully:
       1. a reflection that maps SCR onto SCT. (1 mark)
       2. an enlargement that maps SCR on PCU. (2 marks)
       3. a rotation that maps SCR to UCT. (3 marks)
    2. The ΔPQC is reflected on the line RU. The image of ΔPQC under the reflection is then rotated through an angle −120° about point C. Determine the images of P and Q:
       1. Under the reflection. (2 marks)
       2. After the two successive transformations. (2 marks)
19. A cylindrical tin of radius 7 cm contains water to a height of 19 cm. When a conical solid of radius 14 cm is fitted into the cylindrical tin to a depth of 18 cm as shown in the figure below, the water completely fills the space between the cylindrical tin and part of the cone inside the tin.
    
    1. Taking π to be ²²/₇, calculate correct to one decimal place:
       1. The volume of part of the cone that is not in contact with water. (3 marks)
       2. The surface area of part of the cone that is not in contact with water. (4 marks)
    2. Calculate the height of the cylindrical tin. (3 marks)
20. The figure below shows a trapezium on a Cartesian plane with the co-ordinates A (p ,3) , B(2,5) , C (x, y) and D(5, 2) . Line AD is parallel to BC.
    
    Given that the line AB makes an angle of 45° with the positive x-axis
    1. Determine the equation of line AB in the form y = mx + c     (2 marks)
    2. State the value of p in the coordinates of point A (2 marks)
    3. Find the equation of BC in the form ax + by = c. (2 marks)
    4. Given that the gradient of CD = −5. Find its equation hence coordinates of point C.
       (4 marks)
21. Two aeroplanes T1 and T2 leave airport A for airports B and D respectively. Aeroplane T1 flies on a bearing 050° at an average speed of 300 km/hr for 40 minutes. Aeroplane T2 flies on a bearing 120° at an average speed of 360 km/hr for 45 minutes. An airport C is directly south of B and east of A.
    1. Calculate the distance between:
       1. Airports A and B. (1 mark)
       2. Airports A and D. (1 mark)
    2. Calculate the distance correct to four significant figures between:
       1. Airports A and C. (2 marks)
       2. Airports C and D. (3 marks)
    3. Calculate the true bearing to the nearest degree of airport D from C. (3 marks)
22. The figure below shows a frequency polygon representing the scores of Form 4 students in a Mathematics Test.
    
    1. Generate the Frequency Distribution of the data under the columns given below in the table below.    (5 marks)
       

       Marks	Frequency (f)	Mid points (x)	fx	c.f.

    2. State the modal class. (1 mark)
    3. Estimate
       1. the mean mark (2 marks)
       2. the median mark (2 marks)
23. In the diagram below, the coordinates of points O, P and Q are (0, 0), (2, 8) and (12, 8) respectively.
                             →                   →       →                                                              →      →
    A is a point on OQ such that 2OA = OQ. Line OP is produced to R is such as OR = 3OP.
    
    1.                    →
       Find vector RA . (3 marks)
    2.                                        →                  →   →                             →
       Given that point L is on PQ such that PL : LQ : 2:3, find vector RL . (4 marks)
    3. Show that R, L and A are collinear. (3 marks)
24. In the figure below, ABCD is a trapezium in which AB = 17 cm, AD = 16 cm and angle ABC = 150°. AB is parallel to DC and AB = BD.
    
    1. Calculate the area of triangle ABD (2 marks)
    2. Calculate correct to two decimal places:
       1. The length BC (2 marks)
       2. The length AC (3 marks)
       3. The size of angle ACD (3 marks)

MARKING SCHEME

1.       r = 4.4141
   100r = 441.4141
     99r = 437
         r = ⁴³⁷/₉₉
         x = 2.2121
   100x = 221.2121
     99x = 219
   ⁴³⁷/₉₉ − ²¹⁹/₉₉
     = ²¹⁸/₉₉
   = 2²⁰/₉₉ 
2. log 0.036 = log (³⁶/₁₀₀₀)
                  = log(2² × 3²)
                              10³
                  = 2 log 2 + 2 log 3 − 3 log 10
                  = 2(0.3010) + 2(0.4771) − 3
                  = − 1.4438
3. a(x−2) − 5 (x−2) =  (a−5) (x−2)    
           2(4−x²)          2(2−x) (2+x)
                 ₋₁
    (a−5)(x−2)
   2(2−x)(2+x)
   −1(a−5)
     4+2x
    5−a  
   4+2x
4. ¹⁵⁰/_x + 12x = 90°
   12x² − 90x + 150 = 0
   x = 90 ±√(8100−4(12×150))
                          24
   x = 90 ± 30
             24
     = 5° or 2.5°
5. Let centre (x, y)
   3−x = 7−x
   3−x    1−x
   (3−x)(1−x) = (3−x)(7−x)
   3−3x−x +x² = 21−3x−7x+x²
   −18 = −6x
        x = 3
   5−y = 7−y
   2−3    1−y
   5−5y−y +y² = 14−2y −7y+y²
      y = 3
   Centre (3, 3)
   Scale factor = 7−3
                          1−3
                        = −2
6.  
   
   L.CM = 10 × 4 ×  25 × 3
             = 3000g
7. 3^3y ÷ 3^2x = 3⁸ 
   2^2x × 2^3y = 2⁰
     3y − 2x = 8
     3y + 2x = 0 −
           −4x = 8
               x = −2
               y = 1¹/₃ 
8.  
   
9.  
   
                                   −r₃ + r₁ = 1 .......(i)
                                     r₃ + r₁ = 7........(ii)
                                          2r₁ = 8
   r₁ = 4cm
   r₂ = 5cm 
   r₃ = 3cm
10.  
    1. 75.66M = 15132000
                M = 15132000
                            75.66  
       = 200,000Us Dollars
    2. 15132000 = 119488.3133
          126.64 
       = £ 119,448
11. Distance by bus in 45 min
      = x × ⁴⁵/₆₀ = 0.75x
    Time to catch up is 1¾hrs
    0.75x = ⁷/₄
    80−x 
    3x = 560 − 7x
      x = 56km/hr
12.  
    
13. 5 − 2x < ½x
    5 < ⁵/₂x
    10 < 5x
     2 < x
    ½x < x+2
               3
    3x < 2x + 4
     x < 4
    2<x<4
    
14. 3(      1         ) + (40.36 × 10⁻²)^½ 
       5.21 × 10⁻¹ 
    3(0.1919 × 10) + 6.353 × 10⁻¹ 
    5.757 + 0.6353
       = 6.3923
15. Amount per day = 8550 = 475
                                    18
    Ahmed = 475 × 150  ×  16
                              270
                = Ksh.4222²/₉ 
16.  
    
17.  
    1.  
       1. 110 × 5000
          Ksh 55,000
       2. 45000 + ³/₁₀₀ × 85000 + ⁴/₁₀₀ × 85000 + ⁵/₁₀₀ × 230000
          45000 + 2550 '+ 3400 + 11500
          = Ksh 62450
    2.  
       1.  B. salary = 90/100 × 45000 = 40500
          Total commission = 47450 − 40500
                                       = Ksh 6950
          First 85000 = Sh. 2550 commission
          Next 85000 = Sh. 3400 commission
          Amount above 320000 = Ksh. 1000 commission
          ⁵/₁₀₀ of (x − 320000) = 1000
          Total sales (x) = 340,000/=   
       2. 340000 = 68 mattresses
             5000
18.  
    1.  
       1. Rotation through an angle of +60° about c.
       2. An enlargement centre C, scale factor −1
       3. Rotation about C through an angle of + 120°
    2.  
       1. T and S respectively
       2. R anf Q respectively
19.  
    1.  
       1. ¹⁴/₇ = ^H/₁₈
          H = 36cm
          V = ¹/₃ × ²²/₇ (14² × 36 − 7² × 18)
             = 6468cm³ 
       2. π(RL − rL) + πR²
          L = √(14² + 36²) = √1492 = 38.63cm
          L = √(7² + 18²) = √373 = 19.31cm
          Surface Area = ²²/₇ (1492 × 14 − 7√373) + ²²/₇ × 14²
                                = 1274.67 + 616
                                = 1890.7cm²
    2. Volume of water = ²²/₇ × 7² × 19 = 2926cm³
       Volume of the cone in contact with water = ¹/₃ × ²²/₇ × 7² × 18 = 924cm³
       Volume if the cylinder = 2926 + 924 = 3850
                         Height =    3850     = 25cm
                                        ²²/₇ × 7² 
20.  
    
    
    1. Gradient of AB = tan 45° = 1
       y − 5 = 1
       x − 2
       y = x + 3
    2. 3 − 5 = 1
       p − 2
       p = 0
    3. y − 5 = − 1
       x − 2
       x + y = 7
    4. Eqn of Cd
       y + 2 = −5 ⇒ 5x + y = 23
       x − 5
       5x + y = 23 ............(i) 
         x + y =   7 ............(ii)−
             4x = 16
       x = 4, y = 3 
       C (4, 3)
21.  
    1.  
        
       1. 300 × ⁴⁰/₆₀
           = 200km
       2. 360 ×  ⁴⁵/₆₀
           = 270km
    2.  
       1.  
          
          AC = 200 Cos 40° 
                = 153.2km
       2.  
          
          CD² = 270² + 153.2² − 2 ×  270 ×  153.2 Cos 30°
                 = 24725.6904
          CD = 157.2km
    3.  
       
       157.2   =     270      
       Sin 30°   Sin ∠ACD
       ∠ACD = Sin⁻¹(270Sin30°)
                                    157.2
                 = Sin⁻¹ 0.8585
                 = 59.15°
       ∠ACD = 180 − 59.15
                  = 120.85°
       True bearing = 270°− 120.85°
                            = 149.15°
22.  
    1.  
       

       Marks	Frequency (f)	Mid points (x)	fx	c.f.
       30-40  40-50  50-60  60-70  70-80	6  18  30  34  12	35  45  55  65  75	210  810  1650  2210  900  Σfx = 5780	6  24  54  88  100

    2. 60-70
    3.  
       1. Σfx = 5780
           Σf      100
                = 57.80
       2. 50 + (50−24) 10
                       30
          = 58.67
23.   
    1.  
       
    2.  
       
    3.  
       
24.   
    
    
    1. √(25(25−16)925−17)(25−17))
       √14400
       = 120cm²
    2.  
       1. ∠DAB = ∠ADB = Cos^{−1 8}/₁₇ = 61.93°
             17     =      BC      
          Sin 30°    Sin 56.14
          BC = 17 Sin 56.14
                        Sin 30°
              = 28.23cm
       2. AC² = 17² + 28.23² − 2 × 17 × 28.23 Cos 150°
          AC = √(1917.1614)
               = 43.79cm
       3.     43.79     =        16      
          Sin 118.07     Sin∠ACD
          ∠ACD = Sin⁻¹(16 Sin 118.07)
                                          43.79
                    = Sin⁻¹ 0.3224
          ∠ACD = 18.81°