Mathematics Paper 1 Questions and Answers - Form 4 End Term 2 Exams 2023

• This paper consists of two sections: Section I and Section II.
• Answer all questions in section I and any five questions in Section 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.
• KNEC Mathematical tables may be used.

SECTION I (50 Marks)

Answer all the questions in this section

1. Evaluate without using mathematical tables or a calculator;                                        (3 marks)
   0.0084×1.23×3.5
        2.87×0.056       
2. Use the prime factors of 7056 and 74088 to evaluate                                                (4 marks)
   ∛74088
     √7056
3. A bus left Nairobi and travelled towards Busia at an average speed of 90 km/h. After 2 2/9 hours, a car left Nairobi and travelled along the same road at an average speed of 170 km/h. If the distance between Nairobi and Busia is 800 km, Determine the distance the car travelled to catch up with the bus.           (3 marks)
4. Factorize 16a+12b − 8a² − 6ab                                                                                  (2 marks)
5. Find the area of the triangle below correct to 4 .s.f                                                         (2 marks)
   
6. Without using mathematical tables or a calculator, solve for x in:                                        (3 marks)
     8^x   = 3√3
   9^x+1
7. B is on a bearing of N30°E from A and C is due East of B. The distance from A to B is 600km and the distance from B to C is 400km. Calculate the distance from A to C.         (3 marks)
8. The length of three wires were 30m, 36m and 84m .pieces of wire of equal length were cut from the three wires. Calculate the least number of pieces obtained                                                (3 marks)
9. Given that sin x = ⁴/₅ where x is an acute angle, find  the value of cos⁡(180-x)°                     (3 marks)
10. A Kenyan bank buys and sells foreign currencies as shown below 
    

    	Buying (Ksh)	Selling (Ksh)
    1 Hong Kong dollar	9.74	9.77
    1 South African rand	12.03	12.11

    A tourists arrived in Kenya with 105 000 Hong Kong dollars and changed the whole amount to Kenyan shillings. While in Kenya, she Spent Ksh. 403 897 and changed the balance to South African rand before leaving for South Africa. Calculate the amount, in South African rand that she received.                        (3 marks)
11. Given the column vectors  and that  p ̌ = 2b̌ − 4b̌ + 3c ̌
    1. Express  P as a column vector                                                                           (2 marks)
    2. Hence find its magnitude                                                                                        (2 marks)
12.  
    1. Using a ruler and a pair of compass only construct triangle PQR such that PQ = 4 cm, QR = 6 cm and ∠PQR =60°     (2 marks)
    2. Hence construct an escribed circle to triangle PQR touching line PR.                         (2 marks)
13. Solve the inequality sand state the integral values of x.          (3 marks) 
    4 − 3x < x + 12 ≤ − 3x + 29 
                                        2
14. Determine the point of intersection of the lines 2x − 3y = 8 and x + 2y = −3 using matrix method.     (3 marks)
15. Use graphical method to solve the pair of simultaneous equations                                       (3 marks)
    2x+y=3
    5x+3y=7
16. Njoroge boarded a bus that took off at 9.45pm on a Sunday to visit his sister in Nairobi. The bus took 8 hours 20 minutes to arrive in Nairobi. After 10 minutes he took a taxi that took 42 minutes to arrive at his sister’s home. Find the day and time in 24hr system when Njoroge arrived at his sister’s home.                (3 marks)

SECTION II (50 Marks)

17. The figure below ABCDEFGH is a frustum of a right pyramid .The altitude of the frustum is 2cm. 
    
    Calculate
    1. The altitude of the pyramid                                                                                (5 marks)
    2. The volume of the frustum                                                                                 (5 marks)
18. A line L₁ with then equation −mx + 2y = 3 passes through the point (−3,1).
    Calculate;
    1. The value of m.                                                                                                           (2 marks)
    2. The angle the line makes with x – axis correct to 4 significant figures.                     (2 marks)
    3. The equation of line L₂ parallel to L₁ and passes through (2,6). Leave your answer in the form               
        ^x/_a + ^y/_b = 1.                                                                                                                    (3 marks)
    4. The equation of the line L₃ perpendicular to L₁ at (−3,1). Give your answer in the form                     
       ax + by = c                                                                                                          (3 marks)
19. The vertices of quadrilateral OPQR are O (0,0),P(2,0),Q(4,2) and R(0,3). The vertices of its image under a rotation are O’(1,−1),P'(1,−3) Q'(3,−5) and R'(4,−1).
    1.  
       1. On the grid provided, draw OPQR and its image O'P'Q'R'                             (2 marks)
       2. By construction, determine the centre and angle of rotation.                          (3 marks)
    2. On the same grid as (a) (i) above, draw O''P''Q''R'', the image of O'P'Q'R'  under a reflection in the line y = x               (3 marks)
    3. From the quadrilaterals drawn, state the pairs that are:
       1. Directly congruent;                                                                    (1 mark)
       2. Oppositely congruent                                                                 (1 mark)
20. Below is a cylinder of diameter 21m in which a hole in the shape of a cuboid has been drilled through the centre as shown below.
    
    Given that the cylinder has a height of 40m
    1. Determine the total surface area of the cylinder                                       (7 marks)
    2. Volume of the solid.                                                                                  (3 marks)
21. The table below shows the marks obtained by 40 students in an examination.
    

    Marks	5 - 14	15 - 29	30 - 34	35 - 44	45 - 49
    Frequency f	2	2	7	15	x

    1. Find the value of x.                                                                                       (2 marks)
    2. On a grid, draw a histogram to represent the data.                                       (5 marks)
    3. By drawing a straight line on the graph, determine the median mark.          (3 marks)
22. The figure below shows a triangular garden XZY.XZ=6cm, XY=4cm and angle XZY =40°. 
    
    A point N lies on the line ZY such that ZN = 4cm. Find correct to 2 decimal places
    1. ∠ZNX                                                                                                             (3 marks)
    2. Length of ZY                                                                                                 (3 marks)
    3. Length of NY                                                                                                 (2 marks)
    4. Area of the garden                                                                                          (2 marks)
23.  
    1. Draw the graph of the function y = 10 + 3x − x² for −2 ≤ x ≤ 5            (4 marks)
       

       x	−2	−1	0	1	2	3	4	5
       y

    2. Use trapezoidal rule with 5 strips to estimate the area of the curve from x = −1 to x = 4      (2 marks)
    3. Use integration to find the actual area of the curve from x = −1 to x = 4.    (2 marks)
    4. Find the percentage error of the estimated area to the actual area of the region correct to 2 decimal places                (2 marks)
24. The displacement of a particle after t seconds is given by s = 40t³ − t² − 3t+3.
    Find
    1. Displacement of the particle when t = 2 (2 marks)
    2. The values of t when the particle is momentarily at rest (3 marks)
    3. The velocity when the particle is momentarily at rest (2 marks)
    4. The acceleration when the particle is momentarily at rest (3 marks)

MARKING SCHEME

No	Workings	Remarks
1.	0.0084 × 1.23 × 3.5 × 107        2.87 × 0.056 × 107      84 × 123 × 35        287 × 56 × 100 9/40	M1 M1 A1
2.	74088 = 23 × 33 × 73 7056 =   24 × 32 × 72 ∛(74 088) =    2 × 3 × 7   √(7 056)       22 × 3 × 7 = 1/2	B1 B1 M1 A1
3.	Distance between them after 2 2/9 hrs = 20/9 × 90 = 200 km Relative speed = 170 − 90 = 80 km/h Time taken = 200/80 = 2 hrs 30 mins Distance covered = 2.5 × 170=425 km	M1 M1 A1
4.	16a − 8a2 + 12b − 6ab  8a(2−a) + 6b(2−a) (8a + 6b)(2−4)	M1 A1
5.	A = 1/2 × a × b × sin⁡θ = 1/2 × 6 × 8 × sin⁡70° = 22.55 cm2	M1 A1
6.	81x ÷ 9x+1 = 31.5 34x ÷ 32x+2 = 31.5 4x − 2x − 2 = 1.5  2x = 3.5  x = 1.75	M1 M1 A1
7.	AC2 =6002 + 4002 − (2×600×400×cos⁡ 120° ) AC=√(760 000)   =2683.3 km	M1 M1 A1
8.	GCD of 30, 36 and84 = 2 × 3 = 6 Longest piece = 6 m No of pieces = 30/6 + 36/6 + 84/6 = 5 + 6 + 14 = 25 pieces	B1 M1 A1
9.	x = sin−1⁡ (4/5) = 53.1 cos⁡ x = 3/5 = 0.6 180 − x = 180 − 53.13 = 126.87° Cosine is negative in 2nd quadrant cos ⁡126.87 = −3/5	M1 M1 A1
10.	HK dollars → Ksh 105 000 × 9.74 = Ksh.1 022 700  Balance = Ksh 1 022 700 − 403 897 = Ksh 618 803 Ksh→ rand 618 803 = 51 098 rands   12.11	M1 M1 A1
11.	|p|=√(42 + 192 )       =19.42 units	M1  A1 M1 A1
12.		B2 for triangle PQR  B1 for angle bisectors  B1 for circle
13.	4 − 3x < x+12 x > −2 2x + 24 ≤ −3x + 29 5x ≤ 5 x ≤ 1 Integral values are −1,0 and 1	B1 B1 A1
14.	Coordinates of point of intersection are(1,-2)	M1 M1 A1
15.	Point of intersection = ( 2, −1) x = 2, y = −1	B2 for both lines B1 for values of x and y
16.	Total time taken = 8hr 20 min⁡ + 10 mins + 42 mins = 9hr 12 mins 9:45 p.m + 9 hr 12 mins=7:07 a.m 0707h on  monday	B1 M1 A1
17.	Taking vertex of the pyramid as V, mid-point of AC as N and mid-point of FH as K NC = ½ (√(64+25)) = 4.72 cm KH = ½ (√(16+6.25)) = 2.36 cm VN = VK + KN 4.72 = 2+x 2.36      x 4.72x = 4.72 + 2.36x x = 2 cm VN = 2+2 = 4cm Volume of frustum = vol of bigger pyramid − vol of small pyramid =(1/3 × 8 × 5 × 4) − (1/3 × 4 × 2.5 × 2)                                =53.33 − 6.677                                =46.66 cm3	B1 B1 M1 A1 B1 M1 A1 M1 A1
18	−m(−3) + 2(1) = 3 3m + 2 = 3 m = 1/3 tan⁡ θ = 1/3       θ = tan−1 (1/3) = 18.43° m2 = 1/3 1/3 = (y−6)           (x−2) 3y − x = 16 3y/16 − x/16 = 1 m3 = −3 −3 = y−1  1     x+3 −3(x + 3) = y − 1 3x + y = −8	M1 A1 M1 A1  M1  M1  A1  M1 M1 A1
19.
20.	Curved surface area of cylinder = 2πrh                  = 2 × 22/7 × 10.5 × 40 = 2640 m2 Area of top and bottom circular  parts  2{(22/7 × 10.5 × 10.5) − (7×7)} = 595 m2 Surface area of internal hollow parts = 7 × 40 × 4 = 1120 m2 Total surface area =2640+595+1120= 4355 m2 Volume of solid = Volume of cylinder -volume of hollow  part                                    = (22/7 × 10.5 × 1.05 × 40) − (7 × 7 × 40)                                   =13860 − 1960 = 11900 m3	M1A1 M1 A1 M1 M1 A1 M1M1 A1
21.	x = 40 − (2+2+7+15)    =14 Histogram  Marks  Mid points   Frequency   Freq. density   5 - 14  9.5  2  0.2  15 - 29  22  2  0.1  30 - 34  32  7  1.4  35 - 44  39.5  15  1.5  45 - 49  47  14  2.8 Total area of bars = (10×0.2)+(15×0.1)+(5×1.4)+(10×1.5) +(5×2.8)                             = 2 + 2 + 7 + 15 + 14 = 40 median=  40/2 = 20 2 + 2 + 7 + 1.5x = 20 1.5x = 9      x = 6 Median = 34.5 + 6 = 40.5	M1  A1  B2 for table  B1 ... scale  B1 ... Axis  B1 ...graph M1 M1 A1
22.	t2 = 36 + 16 − (2 × 6 × 4 × cos⁡ 40°) t = √(52 − 36.77) = 3.90 cm ∠ZNX = sin−1 (6 × sin⁡40) = 81.46°                             (3.9)   3.9    =       4         sin⁡Y     sin⁡98.54 ∠XYZ = sin−1⁡ (3.9 × sin⁡98.54) =74.62°                                  4 ∠YXZ=180 − (40 + 74.62) = 65.38° ZY = (42 + 62 − 48 cos⁡65.38 )½ = 5.66 cm NY = 5.66 − 4 = 1.66 cm Area of triangle = 1/2 × 6 × 4 × sin⁡ 65.38° = 10.91 cm2	M1  M1A1  M1 M1 A1 M1 A1 M1 A1
23.	x  −2  −1  0  1  2  3  4  5  y  0  6   10   12   12   10   6   0  A = 1/2 × 1{(6+6) + 2(10+12+12+10)}     = 0.5 × (12+88)         = 50 sq units A=  =(40 + 3(8) − 64/3) − (−10 + 1.5 + 1/3) = 505/6 sq units % error =(505/6 − 50) × 100 = 1.64%                      505/6	B2 for table  B2 for graph  M1  A1  M1  A1  M1 A1
24.	s = 40(2)3 − (2)2 − 3(2)+3    = 313 m At rest, V = 0 V = dS/dt =120t2 − 2t − 3  120t2 − 2t − 3 = 0 t = −3/20 or t = 1/6 t = 1/6 sec   V=120(1/6)2 − 2(1/6) − 3    =0 m/s a = dV/dt = 240t − 2 a = 240(1/6) −2    =38m/s2	M1 A1 M1 M1 A1.. only real value M1 A1 M1 M1 A1