# Mathematics Questions and Answers - Form 2 End Term 3 Exams 2021

INSTRUCTIONS TO CANDIDATES

1. This paper consists of two sections: Section l and Section II
2. Answer all questions in section l and Section II.
3. Show all the steps in your calculations, giving your answers at each stage in the spaces below each question.
4. Marks may be given for correct working even if the answer is wrong.
5. Non- programmable silent electronic calculators and KNEC Mathematical tables may be used.

SECTION I (40 MARKS)

ANSWER ALL QUESTIONS IN THIS SECTION

1. Find the equation of a line through point (5, -1) and perpendicular to line 4x + 2y – 3 = 0. (3mks)
2. Two spheres have surface areas of 36 cm2 and 49 cm2. If the volume of the smaller sphere is 20.2cm3. Calculate the volume of the larger one. (3 mks)
3. Find the integral values of x which satisfy the following inequality.
6 – 3x ≤ 2x – 4 < x + 1 (3 marks)
4. Factorize the quadratic expression below: x2 + 6x + 9 (3mks)
5. The interior angle of a regular polygon is 150o. Find the value of n. (3 mks)
6. Simplify the following expression (3 mks)
7. Calculate the area of the shaded region given that the radius is 27cm. (4 mks)
8. Simplify:(3 mks)
9.  Using a pair of compasses and a ruler only construct a triangle ABC such that AB= 4cm,
BC = 6cm and angle ABC = 135°. (3mks)
10. Calculate the volume of a sphere of radius 9cm. (3mks)
11. Solve for x. If 32X+3 +1 = 28. (3mks)
12. A straight line passing through A (-2,1) and B(2,-k). The line is perpendicular to a line 3y + 2x = 5. Determine the value of k and hence the equation passing through A and B. (3mks)
13. Use substitution method to solve:
3y+2x=13
2y-3x=0

SECTION II (30 MARKS)

ANSWER ALL QUESTIONS IN THIS SECTION

1. A soda depot had 30816 sodas which were packed in crates. Each crate contained 24 sodas. The mass of an empty crate was 2kg and that of a full crate is 12 kg.
1. How many crates were there? (2mks)
2. What was the total mass of empty crates? (2mks)
3. What was the total mass of sodas alone? (3mks)
4. A lorry was hired to transport the crates at a cost of sh. 5 per crate of soda per trip. The lorry could only carry 107 crates per trip. How much money was spent on transporting all the crates? (3mks)
2. Triangle PQR has vertices P(3,2), Q(-1,1) and R(-3,-1).
1. Draw PQR on the grid provided. (1mk)
2. Under a rotation the vertices of P1Q1R1 are P1(1,4), Q1(2,0) and R1(4,-1). Find the centre and angle of rotation using points P and Q. (4mks)
3. Triangle PQR is enlarged with scale factor 3 centre O(0,0) to give triangle P2Q2R2. Draw triangle P2Q2R2 and state its co-ordinates. (2mks)
4. Triangle P1Q1R1 undergoes reflection in line y = x to give triangle P3Q3R3. Draw P3Q3R3 and state its coordinates. (3mks)

3. A country bus left Nairobi at 10.45a.m and traveled towards Mombasa at an average speed of 60km/h. A matatu left Nairobi at 1:15p.m on the same day and traveled along the same road at an average speed of 100km/h. The distance between Nairobi and Mombasa is 500km.
1. Determine the time of the day when the matatu overtook the bus (5mks)
2. Both vehicles continued towards Mombasa at their original speeds. How long had the Matatu waited before the bus arrived? (5mks)

## Marking Scheme

1.   4x+2y-3=0
2y=-4+3
y=-2x+3
2
-2M2 = -1
M2=
2

y-1= 1
x-5   2

2y+2=x-5
2y-x=-7

2. Solution:
3. 6 – 3x ≤ 2x – 4 < x + 1
6 – 3x ≤ 2x – 4
-5x ≤ -10
2 ≤ x

2x – 4 < x + 1
x < 5

2 ≤ x < 5

4.    x2 + 6x + 9  (3,3)
x2(x+3)+3(x+3)
(x+3)(x+3)

5.  interior angles=150º
exterior angles= 30º
sum of all exterior angles= 360º
360º = 12
30º
=12 sides

6. ax-ay+bx-by
a+b
a(x-y)+b(x-y)
a+b
(a-b)(x-y)
a+b
= x-y

7. 360-35=325

325 x 27 x 27= 658.125cm2
360

8.  Answer:

9.  .
10. answer:
11. 32x + 3 + 1 = 28
32x + 3 = 28 -1=27
32x + 3 = 3 3
2x + 3 = 3
2x = 0
x = 0

12.  3y + 2x = 5
3y = -2x + 5
y  = -2x + 5
3     3
gradient M= -2/3
M= 3/2
change in y = -k - 1  = 3
change in x     2 + 2     2
-k - 1  =3
4           2
-k - 1 = x 4
2
-k -1 = 6
-k = 7
k= -7
13.   3y + 2x = 13
2y  - 3x = 0
2y = 3x
y = 3/2x
3(3/2x) + 2x = 13
9/2x + 2x =13
9x + 4x
2
13x
= 13
2
x = 2.
14. .
1.  30816 ÷ 24
= 1284 crates
2.  1284 x 2 = 25.68kg

3. 12 - 2 = 10kg
10 x 1284 = 12840kg

4. 1204 ÷ 107 = 12 trips
5 x 107 = 535 per trip
12 x 535 = Ksh. 6420

15. .
16. .
1. Time of day:
1:15pm - 10:45 am = 2 hours 30 minutes= 2.5 hrs
Distance= Speed x time
= 60km/h x 2.5 = 150km
Distance of overtaking means
distance for bus= distance for matatu
let distance that bus had covered when matatu left be x, then distance of bus that was covered will be 150+x
theerefore:   x  = 150+ x
60      100
100x = 9000 + 60x
100x - 60x = 9000
40x = 9000
x = 225 km
time for bus when overtaken = 225
60
= 3 hrs 45 minutes
1: 15
+ 3: 45
5:00
time = 5:00 pm

2.  Distance remaining= 500 - 150 - 225 = 125km
Time taken by matatu:  125  = 1.25 hours = 1 hour 15 minutes
100
Time taken by bus:  125  = 2hours 5 minutes
60
time waited: 2: 05
-1: 15
: 50
= 50 minutes

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Read 3668 times Last modified on Wednesday, 09 June 2021 06:36

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