Mathematics Paper 1 Questions and Answers - Form 3 Term 2 Opener Exams 2023

Instructions to candidates   

• The paper contains two sections: Section I and Section II.
• Answer All the questions in section I and strictly any five questions from Section II.
• All answers and working must be written on the question paper in the spaces provided below each question.
• 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.
• Non-programmable silent electronic calculators and KNEC mathematical tables may be used, unless stated otherwise.

SECTION A ( 50 MARKS )

Answer all the questions in this section

1. Without using mathematical tables or calculator, evaluate. 3 mks
      (³√(13824) − 4)     
     3 + 4 ÷ 2 − 5 × 7
2. A watch which looses a half a minute every hour was set read the correct time at 0445hr on Monday. Determine in twelve hour system the time the watch will show on Friday at 1845hr the same week. 3mks
3. Find the least whole number by which 2⁵ × 5⁴ × 7³  must be multiplied with to get a perfect cube. What is the cube root of the resulting number. 3mks
4. A woman went on a journey by walking, bus and matatu. She went by bus  ⁴/₅  of the distance , then by matatu for  ²/₃  of the rest of the distance. The distance by bus was 55km more than the distance walked. Find the total distance. 3mks.
5. Equity bank buys and sells foreign currencies as shown:
   

   Currency	Buying (ksh)	Selling (ksh)
   1 US Dollar	77.43	78.10
   1 S.A Rand	9.03	9.51

   A tourist arrived in Kenya with 5 600 US Dollars and changed the whole amount to Kenya shillings. While in Kenya he spent Sh. 201 376 and changed the balance to S.A Rand before leaving to South Africa. Calculate in SA Rand he received. 3 mks
6. The size of an interior angle of regular polygon is 3x° . While its exterior angle is  (x – 20)°. Find the number of sides of the polygon. 3mks
7. Use reciprocal, cosine and square tables only to evaluate to 4.s.f the expression.
        1      −  (cos 73.61)² 4mks
    15.79
8. Given that (x+ 20° )  = cos (2x+25)    find the value of  x  and hence find the value of tan x .            3mks
9. A rectangular room is 4m longer than it is wide. If its area is 12 m² ,find its dimensions. 3mks.
10. The masses of two similar building blocks are 2.7 kg and 800grams respectively. Find the surface area of the larger block if the surface area of the smaller block is 120 cm².     3 mks
11. By completing the square solve the following quadratic equation 4mks
    x² + 8x + 9 = 0
12. Without using a calculator, evaluate:      3 mks
     ( ³/₄ + 1²/₇  ÷ ³/₇  of 2¹/₃ )
          ²/₃  (1 ²/₇ − ³/₈)
13. Simplify the expression:    3mks.
          9t² − 25a²      
    6t² + 19at + 15a² 
14. A business bought 300 kg of tomatoes at Ksh. 30 per kg. He lost 20% due to waste. If he has to make a profit 20%, at how much per kilogram should he sell the tomatoes.    3mks.
15. Find the equation of the line through the point (2,3) and parallel to the line x – 8y − 2 = 0. Leave the equation in the form y = mx + c. 3 mks
16. A rectangular field measures 308m by 228m. Fence posts are placed along its sides at equal distance apart. If the posts are as far as possible, what is the distance between them.     3mks

SECTION II (50mks)

Answer only five questions in this section in the spaces provided.

17. Three points  P, Q and R are on a level ground. Q is 240 m from P on a bearing of 230°, R is 120m to the East of P.
    1. Using a scale of 1cm to represent 40m, draw a diagram to show the positions of  P, Q  and R in the space provided below. 4mks
    2. Determine:
       1. The distance of R from Q. 1mk
       2. The bearing of R from Q. 1mk
    3. A vertical post stands at P and another one at Q. A bird takes 18 seconds to fly directly from the top of the post at Q to the top of the post at P. Given that the angle of depression of the top of the post at P from the top of the post at Q is 9°, calculate.
       1. The distance to the nearest metre the bird covers. 2mks
       2. he speed of the bird in Km/h. 2mks
18. A and B are two towns. Tom left town A at 8:00 am travelling towards town B at an average speed of 90km/h. At 8:21 am on the same day, John left town A for town B travelling along the same road at an average speed of 97km/h. Determine;
    1. The time John caught up with Tom. 5mks
    2. The distance from town A to the point where John overtook Tom. 2mks
    3. On the same day, Paul left town B for A at 8:40am travelling at an average speed  of 80km/h. He met Tom after 2hours 30 minutes. Determine the distance between A and B. 3 mks
19. A surveyor recorded the following information in his field book after taking measurement in metres of a plot.
    

    	To E
    720 to F  240 to G	1000    880    640    480    400    200	320 to D  600 to C  400 to B
    	From A

    1. Sketch the layout of the plot. 4mks.
    2. Calculate the area of the plot in hectares. 6mks
20.  
    1. Complete the table for the function y = 1 – 2x − 3x² in the range −3 ≤ x ≤ 3 (2mks)
       

       X	−3	−2	−1	0	1	2	3
       −3x2	−27		−3	0		−12
       −2x		4		0			−6
       1	1	1	1	1	1	1	1
       Y	−20			1		−15

    2. Using the table above and the graph paper provided, draw the graph of
       y = 1 – 2x –3x²    (4mks)
    3. Use the graph in (b) above to solve
       1. 1 – 2x – 3x² = 0 (2mks)
       2. 2 – 5x – 3x² = 0 (2mks)
21. The diagram below shows two circles, centre A and B which intersect at points P and Q. Angle PAQ = 70°, angle PBQ = 40° and PA = AQ = 8cm.
    
    Calculate
    1. PQ to correct to 2 decimal places 2 Mks
    2. PB to correct to 2 decimal places 2 Mks
    3. Area of the minor segment of the circle whose centre is A 2 Mks
    4. Area of shaded region 4 Mks
22. The region marked R below is enclosed by three inequalities as shown.
            
    
    1. Determine the area of region R                      (2mks)
    2. determine the inequalities that enclose the region
       1. L₁ (2mks)
       2. L₂ (3mks)
       3. L₃ (3mks)
23. Three business partners, Bela ,Joan and Trinity contributed Kshs 112,000, Ksh,128,000 and ksh,210,000 respectively to start a business. They agreed to share their profit as follows:
    30% to be shared equally
    30% to be shared in the ratio of their contributions
    40%  to be retained for running the business.
    If  at the end of the year, the business realized a profit of ksh 1.35 Million. Calculate:
    1. The amount of money retained for the running of the business at the end of the year.        (1mk)
    2. The difference between the amounts received by Trinity and Bela                        (6mks)
    3. Express Joan’s share as a percentage of the total amount of money shared between the three partners.                        (3mks)
24. In the figure below POR is a diameter. PQT is a straight line,  and ∠QRT = 300 and RPS=∠350. O is the centre of the circle.
    
    Calculate
    1. Angle PRQ (2 Marks)
    2. Angle RPQ ( 2 Marks)
    3. Acute angle SOR (2 Marks )
    4. Angle RTQ (2 Marks)
    5. Angle PVS (2 Marks)

MARKING SCHEME

	WORKING
1.	(13824)¹/₃ = (24 × 33)¹/₃                  = 23 × 3 ✓      (23 × 3) − 4        =     24 − 4     ✓   3 + (4÷2) − (5×7)        3 + 2 − 35                                 =    20                                         6−35                                 = 20/−30                                 = − 2/3 ✓	M1  M1  A1
2	Monday hrs ⇒ 19hrs   15min  Tue, Wed & Thur  ⇒ 24 × 3 = 72hrs  Friday hrs ⇒ 18hrs 45min   19 : 15   72 : 00 +18 : 45   110 hrs  ✓  Total time lost = 110 × 30                                 60                        = 55 min ✓ Time on the watch     18:45  −     55      17 50 hrs or 5:50p.m	M1  M1  A1
3	(25 × 54 × 74) × (2 × 52)     2 × 52 = 50 ✓  (25 × 56 × 73)¹/₃ = 26÷3 × 56÷3 × 73÷3 ✓                            =22 × 52 × 7                           = 700 ✓	M1  M1  A1
4	Bus → 4/5x  Matatu  → 2/3 × 1/5x = 2/15x Walking  → 1 − (4/5 + 2/15)               = 1/15x ✓  4/5x = 1/15x + 55 ✓ 11/15x = 55 ⇒ x = 55 × 15                                 11                           = 75km ✓
5	Amount in ksh   5600 × 77.43 = Ksh. 433, 608  Reminder after spending   433608 − 201376 = Ksh. 232, 232  S.A Rand  232232 = 24 419.80 RAND                      9.51	M1  M1  A1
6	3x + (x − 20) = 180°  4x = 200  x = 50° ✓  Exterior Angle = 50 − 20°                           = 30° ✓  No. of sides =  360°                            30                     = 12 sides ✓	M1  M1  A1
7	1      =         1        15.79     15.79 × 10              = 1/1.579 × 1/10             = 0.6333 × 0.1             = 0.06333  ✓ (Cos 73.61)2 = (0.2822)2                      = 0.0796  ✓ 0.06333 − 0.0796 = − 0.01627  ✓	M1  M1  M1  A1
8	(x+20) + (2x+25) = 90°✓       3x + 45 = 90              3x = 45°✓               x = 15°  Tan 15° = 0.2679 ✓	M1  M1  A1
9	x (x + 4) =12  x2 + 4x = 12  x2 + 4x − 12 = 0 ✓  x2 + 6x −  2x − 12 = 0  x(x+6) − 2(x+6) = 0  (x − 2)(x+6) = 0   Either x − 2 = 0 ⇒ x = 2 ✓  OR  x + 6 = 0 ⇒ x = −6 ✓  width = 2m  length = 2+4 = 6m ✓	M1  M1 A1
10	v.s.f =   800g                2700g          = 8/27  l.s.f = 3√8/27 = 2/3   A.s.f = (2/3)2 = 4/9  4/9 = 120/A ⇒ A = 120 × 9/4                           = 270cm2	M1  M1  A1
11	x2 + 8x = −9  x2 + 8x + (4)2 = −9 + (4)2  (x+4)2 = 7  x + 4 = √7  x + 4 = ±2.65  x = − 4 ± 2.65  Either  x = −4 + 2.65    = − 1.35 ✓  OR  x = −4 −2.65    = − 6.65 ✓
12	Numerator:  3/4 + (9/7 ÷ 3/7 × 7/3)  3/4 + (9/7 × 1) = 3/4 + 9/7                         = 21 + 36                               28                       = 57/28 ✓  Denominator:  2/3 × (9/7 − 3/8) = 2/3 × (72−21)                                         56                         = 2/3 × 51/56                         = 17/28 ✓     57/28 × 28/17 = 36/17	M1  M1  A1
13	9t2 − 25a2 = (3t + 5a)(3t − 5a) ✓  6t2 + 19at + 15a2 = 6t2 + 9at + 10at + 15a2                               = 3t(2t + 3a) + 5a(2t + 3a)                               =(3t + 5a)(2t + 3a) ✓   (3t + 5a)(3t − 5a)   (3t + 5a)(2t + 3a)     3t − 5a    ✓    2t + 3a	M1  M1  A1
14	B.P = 300kg × Ksh 30         = 9000/= ✓ Remainder = 300 × 80/100 = 240kg ✓  S.P = 9000 × 120/100 = Sh. 10,800 ✓  S,P per kg = 10800 = Sh. 45                          240	M1  M1  A1
15	x − 8y − 2 = 0  x − 2 = 8y  ⇒ y = 1/8x − 1/4  ⇒ m = 1/8   y − 3 = 1/8  x − 2  8y − 24 = x − 2  y = 1/8x + 22/8
16	G.C.D = 2× 2 = 4m
17	RQ = 8.5cm × 40m  ✓       = 340km  ✓ Bearing = 065°  ✓     Cos 9° = 240/d d =   240    ✓       cos 9   = 243m  ✓ Speed = distance                   Time            = 243m               18sec             = 13.5m/s  ✓ 13.5 × 60 × 60        1000          = 48.6 km.h  ✓
18	Distance travelled by Tom for 21 min  D = S × T     = 90km/h × 21/60hrs  ✓     = 31.5km  ✓ Relative speed = 97km/h − 90km/h                         = 7km/h  ✓ Time taken by John to catch up  t = d/s     =  31.5km  = 4hrs 30 min  ✓        7kmk/h 8:21 a.m 4:30          12:51 p.m  ✓ d = s × t     = 97km/h ×4½hrs    = 436.5km Distance travelled by Paul from B d = s × t     = 80km/h × 2½hrs   = 200km ✓ Distance travelled by Tom by 11:10am D = 90km/h × 31/6hrs     = 285km ✓ Distance AB = 285 + 200= 485km  ✓
19	Area 1 = ½ × 400 × 200 = 40000m2  Area 2 = ½ × 280 × (400+600) = 140000 Area 3 = ½ × 400 × (600+320) = 184000 Area 4 = ½ × 120 × 320 = 19200 Area 5 = ½ × 360 × 720 = 129600 Area 6 = ½ × 240 × (240+720) = 115200 Area 7 = ½ × 400 × 240 = 48000                     Total = 676000m2  Area in ha = 676000 = 67.6ha                      10000
20	X  −3  −2  −1  0  1  2  3  −3x2  −27  −12  −3  0  −3  −12  −27  −2x  6  4  2  0  −2  −4  −6  1  1  1  1  1  1  1  1  Y  −20  −7  0  1  −4  −15  −32   B1  B1     x = −1  OR  x = 0.3 y = 1 − 2x − 3x2 0 = 2 − 5x − 3x2  y = −1 + 3x y = 3x − 1 x = −2   OR  x = 0.3
21	Sin 35° = x/8 ⇒ x = 8 sin 35°        = 4.59cm  ✓ PQ = 2 × 4.59       = 9.18cm  ✓ Sin 20° = 4.59                  PB ⇒ PB =  4.59  = 13.42cm   ✓              sin 20 Area of segment = (70/360 × 22/7  × 82) − (½ × 82 × sin 70°)                            = 39.1 − 30.1cm                            = 9cm2 Area of segment 2  = (40/360 × 22/7  × 13.422) − (½ × 13.422 × sin 40°)                            = 62.89 − 57.88                           = 5.01cm2  Area of APBQ = (½ × 82 × sin 70°) + (½ × 13.422 × sin 40°)                         = 30.1 + 57.88                         = 87.98cm2   ✓ Area of shaded R  = 87.98 − ( 9 + 5.01)                               = 73.97cm2	M1  A1  M1  A1  M1  A1  M1  A1  A1  A1
22	Area of R = ½ × 5 × 3  ✓                  = 7.5 square knots  ✓   Equation of L1  y = 0  ✓ Inequality y > 0 ✓ Equation of L2 (6, 0) and (0, 5) M = 5 − 0 = − 5/6 ✓        0 − 6 y − 5 = 5/6 x − 0 6y − 30 = 5x 5x + 6y = 30 ✓ 5x + 6y ≤ 30 ✓ Equation of L3  (3,0) and (0.5) M = 5 − 0 = −5/3 ✓        0 − 3 y − 5 = − 5/3 ✓ x − 0 3y − 15 = −57 5x + 3y = 15 5x + 3y ≥ 15  ✓	M1  M1  A1  A1  M1  M1  A1
23	40/100 × 1,350,000 = Sh. 540,000 Bela: Joan: Trinity = 112,000: 128,000: 210,000                                         56  :    64     :   105 Amount shared equally = 30/100 × 1,350,000 = 405,000   Sh. 135,000 Each partner Trinity's share = 135000 + (105/225 × 405000)                        = 135 000+ 189000                        = Sh. 324,000 Bela =135000 + (56/225 × 405000)         = 135000 + 100800          = Sh. 2358000 Difference = 324000− 235800 =  Sh. 88 200 Joan's share = 135000 +  (64/225 × 405000)                              = 135000 + 115200                      = Sh. 250,000 % of the total amount = 250200 × 100%                                       810000                                    = 30.8%
24	∠PRQ = 90 − 30           = 60° ∠RPQ = 90 − 60         = 30° ∠SOR = 2RPS          = 2 × 35          = 70° ∠RTQ = 180 − 90 − 30          = 60° ∠PRS = 90 − 35          = 55° ∠PVS = 180 − 55         = 125°