Mathematics Paper 1 Questions and Answers - Kapsabet Mocks Exams 2023

Instructions to candidates

• The paper contains TWO Sections: Section I and Section II.
• Answer ALL the questions in Section I and Only five questions from Section II. 
• 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.
• Candidates should check the question paper to ascertain that all the pages are printed as indicated and that no questions are missing.
• Answer all the questions in English.

                                                                                         SECTION I (50 marks)

                                                           Answer all questions in this section in the spaces provided

1. All odd numbers from 1 -10 are arranged in descending order to form a number.
   1.  
      1. Write the number (1 mark)
      2. Write the total value of the second digit of the number formed in (a) (i) (1 mark)
      3. Express the value of the number in (a) (ii) as a product of its prime factors in power form. (2 marks)
2. A shopkeeper bought a bag of sugar. He intends to repack the sugar in 40 g , 250 gand 750 g . Determine the least mass in grams of sugar that was in the bag. (3 marks)
3. Given that log₁₀ 2 = 0.3010 and log₁₀ 3 = 0.4771 without using tables or calculator find log0.036 correct to 4 significant figures. (3 marks)
4. Evaluate                                            (3 marks)
                  ½ of ³/₂ + 1½(2½ − ²/₃)
                            ¾ of 2½ ÷ ½
5. Using the grid provided below, solve the simultaneous equation (3 marks)
   3x − 4y = 10
   5x + 7y = 3
                      
6.  Given that a chord of length 10 cm subtends an angle of 1.2°at the circumference of the circle.
   Calculate the radius of the circle. (3 marks)
7. When a shopkeeper sells articles at Ksh 24.05, he makes a 30%profit on the cost price. During a sale, he reduced the price of each article to Ksh 22.95. Calculate the percentage profit on an article sold at the sale price. (3 marks)
8. The size of one interior angle of an irregular polygon is 80°. Each of the other interior angles is 128°. Find the number of sides of the polygon. (3 marks)
9. Simplify  (2 marks) 
             
10. Given the inequalities x − 6 ≤ − 3x + 2 < − 2x + 9
    1. Solve the inequality (3 marks)
    2. Represent on a number line (1 mark)
11. The diagram below represents a right rectangular based pyramid of 5 cm by 4 cm. The slant edge of the pyramid is 6 cm. Draw and label the net of the pyramid. (3 marks)
                                                    
12. Vectors OA = 4i + 3j, OB = −2i − j and OC = − 5i − 3j. Show that points A, B and C are collinear.  (3 marks)
13. Find the period, amplitude and phase angle of the function 2y = 3 sin(½x − 60°)    (3 marks)
14. Simplify     20 − 11x − 3x²     (3 marks)
                         16x − 12x²
15. Write the following ratios in ascending order 2:3, 15:16 , 7 : 6 , 13:15   (3 marks)
16. 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)
    
                                                                               SECTION II (50 marks)
                                                  Answer only five questions in this section in spaces provided
17. A straight line passes through P (−1,1) and Q (3, 4) .
    1. Find the length of line PQ (2 marks)
    2. Find the equation of the perpendicular bisector of line PQ, leaving the equation in the form y =  mx + c  (4 marks)
    3. Determine the equation of line parallel to line PQ and passes through point (2,3), leaving your answer in double intercept form. Hence state the y intercept. (4 marks)
18. The marks scored by 30 students in test were recorded as follows
                        41 43 34 28 19 22
                        32 38 22 18 25 33
                        30 41 36 31 28 37
                        35 34 19 22 29 23
                        29 44 26 27 29 36
    1. Starting with the class 18 - 22  , make a frequency distribution table for the data. (2 marks)
    2. Using the frequency distribution in (a) above calculate :
       1. the mean (2 marks)
       2. the median (3 marks)
    3. Draw a frequency polygon to represent the data. (3 marks)
       
19. The solid below is made up of hemispherical part and a frustum of cone. The top and bottom radius of the frustum are 5 cm and 15 cm respectively. The vertical height of the frustum is 24 cm.
                                                                    
    1. Determine the vertical height of the cone from which the frustum was cut. (2 marks)
    2. Calculate
       1. The volume of the solid correct to 2 decimal places (3 marks)
       2. The surface area of the solid correct to 2 decimal places (5 marks)
20.  
    1.  
       1. Draw the graph of the function y = 2x² − 3x − 5 for − 2 ≤ x ≤ 3 (5 marks)
          
       2.  
          1. Use the graph to solve the equation 2x² − 3x − 5 = 0       (1 mark)
          2. Use the graph to solve the simultaneous equation y = 2x² − 3x − 5 and y = −2x − 2 (3 marks)
          3. Write down the quadratic equation which the line y = −2x − 2 is solving. (1 marks)
21. The diagram below shows the speed time graph for a bus travelling between two stations, the bus starts from rest and accelerates uniformly for 75 seconds. It then travels at constant speed for 150 seconds and finally decelerates uniformly for 100 seconds.
                                               
    1. Given that the distance between the two stations is 5225 m. Calculate
       1. maximum speed in km/hattained by the bus. (3 marks)
       2. the acceleration of the bus (2 marks)
    2. A van left Nairobi at 8.30 a.m and travelled towards Mombasa at an average speed of 80 km/h. At 8.30 am a car left Nairobi and travelled along the same road at an average speed of 120 km/h .
       1. Calculate the distance covered by the car to catch up with the van. (4 marks)
       2. Find the time of the day when the car caught up with van. (1 mark)
22. On the Cartesian plane below, triangle PQR has vertices P(2, 3), Q(1, 2) and R(4, 1) while triangle P’’Q’’R’’ has vertices P’’(-2, 3), Q’’(-1, 2) and R’’(-4, 1).
                                               
    1. Describe fully the transformation which maps triangle PQR onto triangle P’’Q’’R’’. (1 mark)
    2. On the same plane, draw triangle P’Q’R’, the image of triangle PQR under a reflection y x = − x in the line (2 marks)
    3. Describe fully a single transformation which maps triangle P’Q’R’ onto triangle P’’Q’’R’’ (2 marks)
    4. Draw triangle P’’’Q’’’R’’’ such that it can be mapped onto triangle PQR by a positive quarter turn about (0, 0) (3 marks)
    5. State a pair of triangles that is
       1. oppositely congruent (1 mark)
       2. directly congruent (1 mark)
23. The equation of the curve is y = x³ − 2x² − 1
    1. Determine
       1. the stationary points (4 marks)
       2. the nature of the stationary points in (a) (i) above (2 marks)
    2. Determine
       1. the equation of the tangent to the curve at x =1 (2 marks)
       2. the equation of the normal to the curve at x =1 (2 marks)
24. The boundaries of ranch AB, BC , CD and DA are straight lines such that B is 075° from A and a distance of 50 km. C is due east of B and a bearing of N80°E from A. D is due south of C and a distance of 70 km.
    1. Using a scale of 1 cm to represent 10 km. show the relative positions of ABCD. (3 marks)
    2. From the scale drawing, determine
       1. the distance in kilometres between B and C (2 marks)
       2. the bearing of Afrom D (2 marks)
       3. the shortest distance from Ato border CD (1 mark)
    3. Calculate the area of the ranch in square kilometer. (2 marks)

                                                                               MARKING SCHEME

	Marks and comments
1.  97531 7000 23 x 53 x 7	B1  B1  B2
	4
2.    40 = 23 x 5       250 = 2 x 53         750 = 2 x 3 x 53        L.C.M = 23 X 3 X 53                      = 3000g	B1  B1  B1
	3
3.         = 2log2 + 2log3 − 3log10        = 2(0.3010) + 2(0.4771) − 3        = − 1.4438        = − 1.444	M1  M1  A1
	3
4. Num   3/4 + 1½ x 11/6 = 3/4 + 11/4 = 14/4        Den 3/4 x 5/2 ÷ 1/2 = 15/4      = 14/4 x 4/15 = 14/15	B1  B1  B1
	3
5.            point of intersection 2, 1 ( − )	L1    L1    B1
	3
6.     angle at the centre 1.2° x 2 = 2.4°          2.4 x 180 = 137.51°          137.51° ÷ 2 = 68.75°          sin 68.75° = 5/r           r =  5.365 cm	M1  M1   A1
	3
7.   cost price = 100 x 24.05 = 18.5                            130         = 22.95 − 18.5 = 4.45         = 4.45 x 100%            18.5         = 24.05%	M1  M1  A1
	3
8.                           80 + 128(n –1) = (n – 2)180                        80 + 128n − 128 = 180n – 360                          − 48 + 360 = 180n − 128n                                      312 = 52n                                        n = 6	M1  M1   A1
	3
9.                                        (34)¾ − 5 − 1                          = 27 − 5− 1 = 21	M1   A1
	2
10. (a)                            x − 6 ≤ − 3x + 2                            4x ≤ 8                              x ≤ 2                            − 3x + 2 < 2x + 9                            − 7 < x                            − 7 < x ≤ 2       (b)	B1  B1  B1  B1
	4
11.	B3 correct drawn   and labeled
	3
12.                                                    A, shared point hence point A, B and C are collinear	B1     B1         B1
	3
13.                            amplitude = 1.5               period 360 ÷ 0.5 = 72°               phase angle − 60°	B1   B1   B1
	3
14.                                Num 3x2 + 11x − 20                             3x2 + 15x − 4x − 20                             3x(x + 5) −4(x + 5)                             (3x − 4)(x + 5)                    Den 4x(4 − 3x)                    = (3x − 4)(x + 5)                         4x(4 − 3x)                    = −x −5                           4x	M1     M1     A1
	3
15.                                                    2 x 80 = 160                                   3 x 80    240                                   15 x 15 =  225                                   16 x 15     240                                    7 x 40  = 280                                    6 x 40     240                                    13 x 16 = 208                                    15 x 16    240                                           7 : 6, 15:16, 13:15, 2 :3	B1      B1    B1
	3
16.                                                                                                                                        −2(3 − x) = 3 − x                                                       −6 + 2x = 3 − x                                                           3x = 9                                                           x = 3                                                        −2(1 − y) = 7 − y                                                        −2 + 2y = 7 − y                                                             3y = 9                                                               y = 3                                                              (3,3)	M1            A1        M1        A1
	4
17. (a)                                                                         = √32 + 42                                                 = 5        (b)                                                                                           Gradient of PQ = 4 − 1 = 3                                                                          3 + 1    4                                              m2 = − 4/3                                                             y − 2.5 = − 4                                                x − 1         3                                               3y = − 4x + 11.5                                                y = − 4/3x + 35/6           (c)                                                           y − 3 = 3                                               x − 2    4                                               4y − 12 = 3x − 6                                               4y − 3x = 6                                                6     6      6                                                − x + y = 1                                                   2   1.5                                               y intercept = 1.5	M1     A1      B1        B1    M1         A1    M1    M1     A1     B1
	10
18. (a)                                 Marks  Freqency   midpoint (x)   fx   cf   18  – 22                  7                       20           140   7  23  – 27                  3                       25             75  10  28  – 32                  8                       30  240  18  33  – 37                  6                       35  210  24  38  − 42                  4                       40  160  28  43  – 47                  2                       45   90  30       (b) (i)                           ∑fx = 915                           ∑f       30                                                    =  30.5           (ii)                                = 28.5 + ( 15 − 10)5                                              8                                                    =  31.625        (c)	B1 all classes    B1 correct frequency 1    B1 correct cf         M1     A1     M1     A1                       B1 correct midpoints       B1 correct plotting    B1 frequency polygon 1
	10
19. (a)                                                x       =  5                                             x + 24     15                                             15x = 5x + 120                                              10x = 120                                                x = 12cm        (b) (i)                         Volume = (1/3 x  22/7 x 152 x 36 − 1/3 x 22/7 x 52 x 12) + 2/3 x 22/7 x 153                                      = 98,845.71 − 314.29) + 7071.43                                      = 15242.86            (ii)	M1      A1    M1    M1    A1    B1    B1    M2    A1
	10
20. (a) (i)                          (ii) x = − 1 and x = 2.5      (b)                       (c)                   2x2 − 3x − 5 = −2x − 2                       2x2 − x − 3 = 0	S1 correct scale   P1 correct plotting   C1 correct curve  award P1 and C1 with  correct turning point (0.75, −6)        B1 both correct    L1 correct line and    B2 all correct values of x and y    B1
	10
21.    ½(150 + 325)h = 5225 h = 22 m/s    = 79.2 km/h 22 − 0 = 0.2933 m/s     75   time difference 8.30 am − 8.00 am = 0.5hr Distance covered by van 0.5 × 80 = k40;km/hr R. S = 120 − 80 = 40 km/h time taken by van to catch up = 40/40 = 1hr distance covered by car 1 × 120 = 120 km/h time of the day = 8.30 am + 1hr = 9.30am	M1  A1  B1  M1A1  B1  B1  B1  B1  B1
22.                           (a)  Reflection on in line x = 0        (c)  Rotation through -90° about (0,0)        (e)  P' Q' R' and PQR , P'' Q'' R'' and P''' Q''' R''',              PQR and P'' Q'' R'' , P' Q' R' and P''' Q''' R'''	B2 ∠P1Q1R1   B3 ∠P111 Q111 R111                        B1      B2      B1      B1
	10
23. (a) (i)                                dy = 3x2 − 4x                                dy                                     3x2 − 4x = 0                                  x(3x − 4) = 0                                  x = 0                                  x = 11/3                                           y = (0)3 −2(0)2 − 1 = −1                                 (0, −1)                                 y = (4/3)3 − 2(4/3)2 − 1 = −25/27                                            ( 11/3, −25/27)            (ii)                          (0, −1) maximum                          ( 11/3, −25/27) minimum      (b) (i)                          when x = 1, gradient = 1                          when x = 1, y = −2                           y + 2 = − 1                           x − 1                           y + 2 = −x + 1                            y = −x − 1           (ii)                           gradient  = 1                            y + 2 = 1                            x − 1                            y + 2 = x − 1                            y = x − 3	M1        A1    B1    B1           B1      B1         M1    A1    M1     A1
	10
24. (a)                              (b) (i)    2.7 ± 0.1 x 10 = 27km ± 1          (ii)                   306° ± 1°                   9.6 ± 0.1 x 10 = 96 km ± 1          (iii)   7.7 ± 0.1 x 10 = 77 km ± 1      (c)              (½ x 56 x 77) + ½(77 + 27)14 = 2884km2	B3 locating B,C and D                          B2   B1   B1   B1   M1 A1
	10