# MATHEMATICS Paper 2 Questions and Answers - Form 3 End Term 1 Exams 2023

## QUESTIONS

1. Evaluate without using tables or calculators. (3mks)
0.8064 X 6.048/1.008 X 0.1344
2.  Evaluate -4 of [(-4 + -5 ÷15) + -3 - 4 ÷ 6)]/84÷-7+3--5 (2mks)
3. Solve for θ without using table given that 0 ≤ θ ≤ 90o and that sin ( 2θ - 30o) –cos4θ =0 (3mks)
4. Solve for x given that 52x+ 2– 20 X 52x=625 (3mks)
5. The angle of a quadrilateral ABCD in order are 2(x -10),4(x + 5),5(x+4) and (x-20) in degrees. Find the exterior angles of the quadrilateral. (4mks)
6. A radio costing kshs. 1240 is marked to sell at a price calculated to give a profit of 40 %. What will be its selling price in sale when 25% is taken off the marked price? (3mks)
7. Show that if OA = - i+ 7j, OB = 3i – 5j and OC = 4j, then points AB and C are collinear. (4mks)
8. Four men can dig 2 acres of land in 3 days working 4 hour a day. How many men are required to dig 5 acres of land in 4 days working 3 hours a day at the same rate. (3mks)
9. The surface area of two similar bottles are 12cm2 and 108cm2 respectively. If the larger one has a volume of 810cm3. Find the volume of the smaller one. (3mks)
10. In the figure given below, AC is an arc of a circle centre B ABD = 60 o , AB = BC = 7cm and CD= 5 cm.

Calculate
1. The area of triangle ADB (2mks)
2. The area of the shaded region. (2mks)
11. Solve the inequalitites and represent the information on the number line. (3mks)
-3+2x&lt;3x+2&lt;4(x-5)
12. Make x the subject of the formula in 3s=2p√X/3X-5 (3mks)
13. Given x = 13.4cm and y = 4.3 cm. calculate the percentage error in x/Y correct to 4 d.p( 3mks)
14. A straight line through the point A (2, 1) and B (4,m) is perpendicular to the line whose equation is 3y = 5 – 2x, Determine the value of m. (3mks)
15. Okoth deposited some money at 10% compound interest compounded annually. How long will it take to double the amount to the nearest year? (3mks)
16. Chebet has 5 brown chicken and 3 black ones. She picks two of them for slaughter at random, one after the other. What is the probability that the two are of different colours. (3mks)

SECTION II

Answer only five questions.

1. A bus left Nairobi at 8.00am and traveled towards BUsia at an average speed of 80km/hr. At 8.30 am a car left Busia for Nairobi at an average speed of 120km/hr. Given that the distance between Nairobi and Busia is 400km.Calculate:
1. The time the car arrived in Nairobi. (2mks)
2. The time the two vehicles met. (4mks)
3. The distance from Nairobi to the meeting point. (2mks)
4. The distance of the bus from Busia when the car arrived in Nairobi. (2mks)
2. A triangle whose vertices are A (1,4) B ( 2,1) and C (5,2) is given the following transformation:
1. Reflection in the line y = -x to A 1 B 1 C 1
2. A1 B1 C1 is then given rotation of + 90 o about the origin to A11 B 11 C 11
3. A11 B11 C11 is then given a translation vector to A11 B11 C11
4. A 11 B 11 C 11 is then given an enlargement scale factor – 2 centre (0, 0) to AIV BIV CIV .
On the given grid plot a triangle ABC and it’s images A1 B1 C 1 , A 11 B 11 C 11 , A 111 B 111 C 111 and A IV B IV C IV . And give coordinates of A IV BIV CIV . (10mks)
3. A Post OT stand vertically on level ground John moves from O, the foot of the flag post to point R, on the level ground. The points T, O and R from a right angled isosceles triangle whose perimeter is 56m. S is another point on the level ground 35m from O calculate:
1. The angle of elevation of T from S. (6mks)
2. The distance ST. (2mks)
3. Find the maximum possible distance between R and S. (2mks)
4. A salesman received a basic salary of sh. 50,000 a year together with a commission of 6 % on the value of goods sold and a car allowance of sh. 2.50 per km.
1. Find the total amount he received in a year in which he sells goods worth sh. 625,000 and travels 10,000km. (4mks)
2. The next year he travels 12,000km and receives a total of shs. 134,000
1. Calculate the value of goods sold. (4mks)
2. Calculate the percentage increase in the value of the goods sold. (2mks)
5. Two airports a and B are such that B is 500km due east of A. two plane P and Q take off from A and B respectively and at the same time.
Plane P files at 360km/hr on a bearing of 030o
Plane Q flies at 240km/hr on a bearing of 315o
The two planes land after 90 minutes.
Using a scale of 1: 10,000,000
1. Show the positions of the planes after 90 min. (6mks)
2. Find the distance between the planes after 90 min. (2mks)
3. Find the bearing of plane Q from plane P after 90 minutes (2mks)
6. The figure below shows a container in form a frustrum of an open top radius 40cm and base radiu24cm. the depth is 56 cm. 56cm
1. Calculate the volume of the container in litres. (4mks)
2. Of the container is ¾ full of water by volume,
Calculate the radius of the meniscus. (6mks)
7. Use a ruler and compass only in this question.
1. Construct ∆ ABC such that AB = 6cm AC= 8.5 cm and < BAC = 120o (3mks)
2. Construct the locus ℓ, of points equidistant from A and B (2mks)
3. Construct the locus ℓzof points equidistant from AB and BC (3mks)
4. Find the points of intersection, P1 and P2 , of 11 and 12 and measure P1 P2 (2mks)
8. The diagram below shows the graph of a moving matatu from one bus stop to another.
1. Find the acceleration of the matatu. (2mks)
2. Find the deceleration of the matatu (2mks)
3. Calculate the distance the matatu while accelerating. (2mks)
4. Calculate the distance the matatu covered while traveling at an acceleration of 0m/s 2 (2mks)
5. Find the distance between the two bus stops.

## MARKING SCHEME

1. 0.8064 x 1000/1.008 x 10000    x 6.048 x 10000/0.134 x 1000
√80.644008 6048/1344
4 x 9  =   36
2. -4 (-4     ⅓ )
- 12    3     5
-4  X  -8/-4
-8
3. Sin (2θ  300)  Cos 4θ0
Sin (20   300)  Cos 4θ
Θ 1200/6
=200
4. Let 52x be U
52X(53)  20(52X) =625
25U     20U =   625
125
5U/5 = 625/5
U  = 125
52x   53
2x = 3
X = 3/2
5. The sum of n sides polygon in degrees (n -2) 1800
Angles in quadrilateral add up to 3600
5(x+4)+4(x+5)+2(x+10)+(x+20)+3600
(x+20) = 3600
12x = 360
X = 300
Interior angles
5(x+4)=1700
4(x+5)=1400
2(x+10)=400
X+20=100
Exterior angles
1800    1700   100
1800   1400   400
1800   400   1400
1800   100  1700
No mark for 1 or less correct angles
6. 140/100  x 1240 =1736
Selling price
75/100x1736
Ksh  1302 B  1
7. Co-odinates  A are (-1,7)
B  are (3,-5)
C are  (6,-4)
AB      3   -     -1                4
5           7              -1
BC      0   -      3             -3
4         -5             9
4BC    -3   -3  AB  hence BC parallel to AB.
They share a common point B hence collinear.
8. (30cm3)
ASF   108/12
LSF         ASF           3
VSF           (LSF3)     2
810/V1      27
V1  =   810/27
= 30cm
9.
1. Area of triangle ADB is  1/2 x 7 x 125 to 60°
=7 x 6 sin 600
= 36.37 cm2
2. Area of unshaded sector
60/360 x 22/7 x 7 x =25.6667
Shaded area   36.37    25.67
=10.7 cm2 B
10. -7 x X < 3x + 2  2x < 9,x7  4-5
3x+2<4(x   5), 3x + 2  < 4x    20
-x < -22, x > 22
If x > -5 and  X>22
22
11. (3.5) (2p√x/3X5)
(3x5)952   4p2x x3x  5
27x   52     4552      4p2x
27x52   -  4p2x    4552
X(2752    4p2)    4552
X      455²
2752   4p2
12. Max  val of  x 13.45 min 13.35
Max val of  y   4.35  min 4.25
Max value of x 13.45    3.1647
Y  4.25
Max value of x 13.35    3.069
Y  4.35
Actual value of x 13.4    3.1163
Y 4.3
Absolute error     3.1642     3.069    = 0.04785
2
Pere em  0.04785 x100  0.015355x100
3.1163
1.5355%
13. Gradient of AB is M-1     M- 1
4    2       2
Gradient product is -1
Gradient of second line
Y   5   2x    is 2
3    3
m-1/2        3/2
2m -2  =  6
2m = 8
M = 4
14. Principal  p amount   2p
2p   p(I  +  10)n   2p  -  p (1+0.1)n
100      p     p
2   (1   1)n
Log 2   n log 1.1
N   log 2         0.301
Log 1.1    0.0414
N =  7.27
Round upto 8
N = 8 yrs
15. P(black)  and p(brown) or p(brown and p(black)
5   x  3   x   3  x  4
8   x  7   x    8  x  7
15      3
56     14
27
56
16.
1.           400km
Nairobi → busia
Speed = 120 km/hr
Distance = 400 km
Time = 400/120
= 3hrs 20min
8.30 + 3 hrs 20min
11: 50 a.m
2. at 8.30 am distance covered by bus
½  x 80 = 40km
Distance left = 360 km
Speed  = 2000km/hr
Time  = 360/200 = 1 hr  48 min
They met at 8.30 + 1 hr 48min
10.18 am
3. 8 – 10.18 is 2 hrs 18min
Distance = 2 x 80  + 18/60 x 80
160 +24
184km from Nairobi.
4. Car arrived  in Nairobi after 3 hrs 20 min bus travelled at a time of
3hrs 20 min + 30 min
= 3hrs  50 min
Dist = 3 x 80 = 50/60  x 80
= 240 + 661/3
= 931/3km
Distance from Busia is 931/3 km
17. ABC
Y - = -x drawn
A1B1C1 draw (rotated)
A11B11C11 draw(translated)
A111B111C111 Enlarged
A1v(2,6) B1v(0,0)C1v(-6,2)
18.
1. x2+x2=(56 -2x)2
2x2 = 4x2– 224x + 3136
2x2- 224 x + 3136 = 0
X2 – 112x + 1568 = 0
X ( x –16) – 98 (x – 16) = 0
(x-16) ( x – 98) =0
X = 16 or 98
Height = 16 cm
Tan θ  = 16/35=0.4575
= 24.57o
2. ST =  √162 + 352
= √1481
= 38.48
3. Muximum distance
= 35 ± 16
= 51
19.
1. 50,00+6/100x62 5000 + 10000 x 250
50,000 + 37500 + 25,000
= 112,5000
2.
1. 12000 x 2.5+0.06x+50,000 =134 000
30,00 + 50,000 + 0.06x = 134 000
0.06x = 134 000 – 80 000
0.06/0.06x = 54000/0.06
x = 900,000
2. increase = 900 000 – 625 000
= 275,000
% increase  =  275 000/625 000  x 100
= 44 %
20. Distance  covered in 90 min
Bl P = 540km
Q = 360 km
Scale 1 cm = 10 000 000 cm
1 cm rep 100 km

Distance  btwn plane 2,1 cm  ± 0.1
2 10 ± 10
Bearing of plane Q from plane P
185 ± 1°
21.
1. 40/24 =x+56
40 =24x+1344
x = 84
vol = 27/3x1/3X402X140–22/7 x1/3 x242x84
= 183978.67/1000
= 183.978 litres
2. 40/140r/h
2/7=r/h
h = 7r/2
3/4 x22/7x1/3(402x140–242x 84) =22/7 x1(7r3– 242 x84)
3/4 (402x140–242 x84)=7r3/2– 242 x84)
3/4 x 175516 =7r3/2– 48384
180096 x 2/7 =r3
³√r3 = ³√51456
r = 37,19    M1
22.
1. ACC =  15–0/20
= 0.75m/s2
2. Dece = 0 – 15/20
= - 0.75
3. Area = 1/2x 20 x 15
= 150ml
4. Area=20x15
= 300m
5. Area = 1/2(30+60)X15
= 675m

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