Mathematics Paper 1 Questions and Answers - Form 4 Term 3 Opener 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 questions in the spaces provided in this section

1. When a certain number x is divided by 30, 45 or 54, there is always a remainder of 21. Find the least value of the number x.               (3 marks)
2. Without using a calculator, evaluate                                                                            (3 marks)
    (2³/₇ − 1⁵/₆) ÷ ⁵/₆    
   (²/₃  of 2¹/₄ − 1¹/₇)
3. Find by construction the image of the triangle ABC when it is reflected in the mirror line M.        (3 marks)
   
4. Solve the following inequality and state the integral values       (3 marks)
   2x − 1 < 7 + x < 3x + 2
5. The sum of interior angles of a triangle is given by (10x−2y)° while that of a hexagon is given by (30x+24y)°. Calculate the values of x and y                                            (3 marks)
6. A bus left Mombasa at 7.00 a.m. travelling at an average speed of 80 km/hr towards Nairobi. After two hours, a car started and travelled in the same direction at 120 km/hr. Calculate how far from Mombasa the car caught up with the bus.        (3 marks)
7. Simplify the expression:                    (3 marks)
   3x² − 4xy + y²
       9x² − y² 
8. Electricity bill comprises of a meter rent charge of Ksh. 250, a fixed charge of Ksh. 130, a charge of Ksh. 1.50 per unit and a fuel cost charge of 5% of the total charges. Melton, a domestic consumer of KPLC consumed 898 units in a certain month. Find the bill for that month.                                                                                                                           (3 marks)
9. A translation maps triangle ABC onto A'B'C'  where A(1,−1), B(2,2), C(3,1) and C'(−1,3). Find:
   1. Translation vector.                                                                                          (1 mark)
   2. Coordinates of A' and B'.                                                                               (2 marks)
10. Using elimination method, solve the simultaneous equations.                                  (3 marks)
    4x + 3y = 10
    2x + 3y = 8
11. Given that x is an acute angle and that sin⁡(x − 20°) − cos⁡(3x − 50°) = 0, find the value of x in degrees.       (2 marks)
12.  
    1. Using a ruler and pair of compass only, construct a parallelogram ABCD such that AB=6 cm, BC=4.8 cm and angle ABC = 150°    (3 marks)
    2. Drop a perpendicular from D to meet AB at M. Measure DM.                               (1 mark)
13. Manzu and Halima started a journey to Nairobi from Kisumu at 8.30 a.m , after 1 hour Manzu stopped on the way for 35 minutes .If Halima arrived in Nairobi 6 hours earlier than Manzu, at what time did Manzu arrive in Nairobi.            (3 marks)
14. The figure below shows a right pyramid with sides AB = 13 cm, BC = 5 cm and CD = 7 cm. 
    
    Determine the volume of the solid.                                                                            (3 marks)
15. Find the possible values of x in the equation 9x² = 27^2x+12                                   (4 marks)
16. Find the equation of the tangent to the curve y = 4x³  – 2x²  – 3x + 5 at the point (2, 1)            (4 marks)

SECTION II 50 Marks

Attempt any five questions ONLY this section

17. In the following diagram, the gradient of the lines L₁ and L₂ area 1.2 and 2 respectively, and x-coordinates of points A and B are 2 and 4 respectively. The line L₃ passes through point B and the origin.
    
    1. Find the co-ordinates of A and B.                                                                    (3 marks)
    2. Find the equation of L₁.                                                                                    (2 marks)
    3. Find the equation of line L₂ in the form ^x/_a + ^y/_b = 1                                          (3 marks)
    4. Find the equation of the line L₃ in the form ax + by = c                               (2 marks)
18. The figure below represent a solid made up of a conical frustum and a hemispherical top. The slant height of the frustum is 8 cm and its base radius is 4.2 cm. Use π = ²²/₇ 
    
    If the radius of the hemispherical top is 3.5 cm
    1. Find the area of:
       1. The circular base.                                                                                      (2 marks)
       2. The curved surface of the frustum.                                                            (4 marks)
       3. The hemispherical surface.                                                                        (2 marks)
    2. A similar solid has a total surface area of 81.5cm². Determine the radius of the base.             (2 marks)
19.  
    1. Given the matrix A =   , find the inverse of A.         (2 marks)
    2. Two universities, TECK and KCT purchased beans and rice. TECK bought 90 bags of beans and 120 bags of rice for a total of sh 360, 000. KCT bought 200 bags of beans and 300 bags of rice for a total of sh 850,000. Use the matrix method to find the price of one bag of each item.                                                                                            (5 marks)
    3. The price of beans later decreased in the ratio 4∶ 5 while the price of rice increased by 20%. A businessman bought 20 bags of beans and 30 bags of rice. How much did he pay?                            (3 marks)
20. In the quadrilateral ABCD below, AB = 7.4 cm, BC = 9.1 cm AD = 12 cm, angle ABC = 133° and angle ACD = 48°
    
    1. Calculate the size of angle ADC. Give your answer correct to 3 significant figures. (5 marks)
    2. Calculate the shortest distance from D to the line AC.                                                (2 marks)
    3. Calculate the area of the quadrilateral ABCD. Give the answer correct to 2 significant figures.     (3 marks)
21. The football team in a school decided to raise Ksh. 3600 for a party. Each student was to contribute equal amount. However before the contributions were made, five members of the football team decided to transfer to other schools. This meant the remaining members had to pay Ksh. 24 more to meet the same target. Taking the original number of footballers to be x,
    1. Write an expression for the initial amount that each student should have contributed.             (1 mark)
    2. Write an expression for the contribution by each student after the transfer.          (1 mark)
    3. Form an equation hence find the number of members in the football team originally.         (5 marks)
    4. Calculate the percentage increase in the contribution per student caused by the transfer.            (3 marks)
22. Triangle OAB  is such that OA = a and OB = b. C lies on OB such that OC:CB =1:1 . D lies on AB such that AD:DB = 1:1 and E lies on OA such that OA:AE =3:1. Find in terms of a and b
    1.  →
       OC                                                                                                                   (1 mark)
    2.  →
       OD                                                                                                                  (2 marks)
    3.  →
       CD                                                                                                                    (2 marks)
    4.  →
       OE                                                                                                                    (2 marks)
    5.  →
       DE                                                                                                                  (2 marks)
    6.  →
       AB                                                                                                                (1 mark)
23. The lengths, in cm, of 37 leaves of a certain tree were recorded as follows:
    

    8.5	7.9	8.6	9.7	9.7	10.3	11.0	9,.0	8.5
    10.5	9.5	8.0	8.3	9.1	7.9	10.2	9.2	8.3
    10.1	10.2	9.9	9.4	9.9	8.7	.9.7	10.6	8.9
    9.9	8.9	9.9	10.9	9.5	10.1	10.0	9.2	9,5
    9.1

    1. Starting with 7.9 and using a class interval of 0.5 cm, prepare a frequency distribution table to represent the data.      (2 marks)
    2. Use the table in (a) above to determine the estimate of:
       1. The mean                                                   (3 marks)
       2. The median                                (3 marks)
    3. Draw a frequency polygon to represent the data.            (2 marks)
24. The displacement S metres of a moving particle from a point O after t seconds is given by S = t³ − 5t² + 3t + 10.
    1. Find S when t = 2 seconds.                                                                               (2 marks)
    2. Determine;
       1. The velocity of the particle when t = 5 seconds.                                 (3 marks)
       2. The value of t when the particle is momentarily at rest.                     (3 marks)
    3. Find the time when the velocity of the particle is maximum.                          (2 marks) 

MARKING SCHEME 

No.	Working	Comments
1.	LCM of 30, 45 and 54 = 2×33×5 = 270 Least value of x = 270 + 21 = 291	M1  M1A1
2.	N ⇒ (23/7 − 15/6) ÷ 5/6 (17/7−11/6) × 6/5 = 25/42 × 6/5 = 5/7 D ⇒ 2/3 × 9/4 = 3/2 3/2 − 8/7 = 5/14 N/D ⇒ 5/7 × 14/5 = 2	M1  M1  A1
3.		B1 – A’ B1 --- B’ B1 --- C’
4.	2x − 1 < 7 + x x < 8 7 + x < 3x + 2 5 <2x 2.5 2.5<x<8 Integral values are 3,4,5,6,7	M1  A1  A1
5.	10x − 2y = 180 ...... (i) 30x + 24y = 720 ..... (ii) From (i), y = 5x − 90 5x+4(5x−90) = 120 25x = 480  x = 480/25 = 19.2 y = 5(19.2) − 90 = 6	M1  M1 A1 both values
6.	Distance by bus in 2 hrs = 2×80 =160 km Time taken to catch up = 160/40 = 4 hrs Distance trave lled = 120 km/hr ×4=480 km	M1  M1A1
7.	N ⟹ 3x2 − 3xy − xy + y2 (3x − y)(x − y) D ⟹ (3x+y)(3x−y) N/D ⟹ (3x−y)(x−y)   =    (x−y)               (3x+y)(3x−y)      (3x+y)	M1 M1  A1
8.	Total charges = 250 + 130 + (1.5×898) = Ksh.1727 Fuel costs = 5/100 × 1727 = Ksh.86.35 Monthly bill = Ksh.1727 + 86.35 = Ksh.1 813.35	M1 M1 A1
9.		B1  B1  B1
10.	4x+3y=10 ......(i) 2x+3y=8 ......... (ii) Subtracting (ii) from (i) 2x=2  x=1   4(1)+3y=10  3y= 6 y=2 y=2 and x=1	M1 M1 A1
11.	sin⁡ A = cos⁡ B;A+B=90° (x−20)+(3x−50)=90° 4x-70=90° x=40°	M1  A1
12	DM = 2.6 ± 0.1cm	B1 – line AB and BC B1 for angle ABC B1 – parallelogram   B1
13	8.30 + 1hr + 35min + 6hrs    = 16.05p.m	M1  A1
14	CF = 0.5 × √(72 + 52) = 4.301cm  AF = √(132 − 4.3012)= 12.27cm Volume = 1/3 × base area × height              = 1/3 × 5 × 7 × 12.27=143.15 cm3	M1 M1A1
15	32(x²) = 33(2x+12)  2x2 − 6x − 36=0 x2 − 3x −18 = 0 x2 − 6x + 3x − 18 = 0 (x−6)(x+3) = 0 x=6 or x=−3	M1 M1   M1 A1 both values
16.	dy/dx ⟹ 12x2 − 4x − 3 At (2,1) m1 = 48−8−3=37 m2 = −1/37 −1/37 = (y−1)             (x-2) 37y − 37 = −x + 2 y = −x/37 + 39/37
17.	m1 = 1.2  1.2 = (y − 0)           (2 −0) y = 2.4 A(2,2.4) m2 = 2 2 = y − 2.4         4 − 2 y = 4 + 2.4 = 6.4 B(4,6.4) y = mx + c 2.4 = (1.2×2) + c  c = 0 l1 ⟹ y = 1.2x y−2.4 = 2   x−2 y−2.4 = 2x−4 2x−y = 1.6 ⟹ x/0.8 − y/1.6 = 1 m3 = 6.4/4 = 1.6 y−6.4 = 1.6  x −4 y − 6.4 = 1.6x − 6.4  y − 1.6x = 0	M1  M1  A1 both coordinates  M1  A1  M1 M1A1
18.	Area of circular base 22/7 × 4.2 × 4.2 = 55.44 cm2  (x+8) = 4.2    x        3.5 x = 40 cm 22/7 × 4.2 × 48 − 22/7×3.5×40 633.6 − 440 = 193.6 cm2  2×22/7×3.5×3.5 = 77cm2  Total S Area = 55.44 + 193.6 + 77 = 326.04 cm2 ASF = 326.04 = 4               81.5 LSF = 2 4.2/2 = 2.1 cm	M1A1  B1  M1  M1A1  M1A1  M1 A1
19	Det = 9 − 8 = 1 Let the cost of Beans = x and rice = y  90x + 120y = 360 000  200x + 300y = 850 000 x=Sh.2000 and y=Sh.1500 New price of beans = 4/5 × 2000 = Sh.1600 New price of rice = 120/100 × 1500 = Sh.1800 Total cost = (20×1600) + (30×1800)                 = Sh.36 000+54 000                 = Sh.90 000	M1  A1  M1  M1M1  M1  A1  M1  M1  A1
20.	Using cosine rule a2 = b2 + c2 − 2bc cos⁡A AC2 = 9.12 + 7.42 − 2×7.49.1 cos⁡133 = 82.81 + 54.76 + 91.85 = 229.42 AC = √229.42 = 15.15 cm Using sine rule    a    =     b    Sin A    Sin B    12    =   15.15  Sin 48      Sin θ θ = sin-1⁡(15.15 × sin⁡48) = 69.8°                         12 Angle DAC = 180 − (48+69.8) = 62.2° sin⁡ α = DF/12 DF = 12 × sin ⁡62.2 =10.61 cm Area of ∆ABC = 0.5×7.4×9.1×sin⁡ 133                        = 24.62 cm2 Area of ∆ADC= 0.5 × 15.15 × 12 × sin⁡62.2                        = 80.41 cm2 Total area = 24.62+80.41 = 105.03                 ≅110 cm2  (2 s.f)	M1  M1 M1 M1A1  M1 A1  M1  M1 A1
21.	3600     x 3600  x -5 3600 − 3600 =24   x−5        x 3600x − 3600x + 18000 = 24x2 − 120x x2 − 5x − 750 = 0 (x+25)(x−30)=0 x = −25 or x=30 Thus x = 30 students 3600 = sh.120   30 % change = 24/120 × 100 = 20%	B1   B1  M1  M1  M1  M1  A1  M1  M1A1
22.		B1  M1  A1  M1A1  M1A1  M1A1  B1
23.	a)    Class  x  f  fx  cf  7.9 - 8.3  8.1  5  40.5  5  8.4 - 8.8  8.6  4  34.4  9  8.9 - 9.3  9.1  7  63.7  16  9.4 - 9.8  9.6  7  67.2  23  9.9 - 10.3  10.1  10  101  33  10.4 - 10.8  10.6  3  31.8  36  10.9 - 11.3  11.1  1  11.1  37      37  349.7   b)  X̅ = 349.7           37     = 9.6 Median = 9.35 + (18.5 − 16) × 0.5                                     8                    = 9.35 + 0.4875                    = 9.5  c)	B1   B1         M1 M1 A1 M1 M1   A1     B1 – correct scale B1 --- correct frequency polygon
24.	S = 23 − 5(22 ) + (3×2) + 10     = 8 − 20 + 16     = 4 m   V = dS/dt = 3t2 −10t + 3 V = 3(52 ) − (10×5) + 3    = 75 − 50 + 3    = 28 m/s 3t2 − 10t + 3 = 0 3t2 − 9t − t + 3 = 0 (3t −1)(t −3) = 0 t =1/3 or t=3 dV/dt = 0 6t − 10 = 0 t = 10/6 = 12/3  sec	M1  A1  M1  M1  A1  M1  M1  A1 both values  M1 A1