# Mathematics Questions and Answers - Form 1 Opener Exam Term 3 2022

MATHEMATICS
FORM 1 OPENER EXAM
TERM 3 2022

INSTRUCTIONS TO CANDIDATES

• These papers consist of two sections; Section A and section B.
• Answer all questions in section A and Section B.
• Marks are awarded for steps which are correctly worked.
• Calculators must not be used.

SECTION A 50 (MARKS)
ANSWER ALL QUESTIONS IN THIS SECTION

1. Ayub travelled part of the journey by train and the rest of the journey by bus. The total fare was Ksh 4000. On return, the fare of the bus was hiked by a half of what he had paid and the total fare of the return journey hiked to Ksh 4800. Find the train fare. (3mks).
2. When 19346 is divided by a number, the quotient is 841 and the remainder is 3. What is the number? (3mks).
3. A square toilet is covered by a number of whole rectangular tiles of sides 60cm by 48cm. Calculate the least possible area of the room in square metres. (3mks).
4. Evaluate using prime factors and leave your answer in prime factors form. (4mks).
5. Evaluate (4mks).
6. Use number line to solve 6 – (- 4) (2mks).
7. Given that X= 4 and y = -6, evaluate the following. (3mks).
6x – 3xy
8. Simplify (3mks)
a + b  -  2a - b
2            3
9. The ratio of boys and girls in a mixed school is 2:3. If there are 160 boys, how many girls are there? (3mks).
10. Find the ratio a:c, given that;
a:b = 7: 1, b:d = 1: 2, d: c = 2: e (4mks).
11. The perimeter of a semicircular protractor is 14.28cm. Find its radius. (3mks)
12. The area of the sector of a circle is 38.5cm². Find the radius of the circle if the angle subtended at the centre is 900. (Take π= 22/7). (3mks).
13. A cylindrical column of fat has diameter 17.5cm and height 10cm. Calculate the density in g/cm³ of fat if the column has a mass of 2kg. (3mks)
14. Express the following measurements in 3 s.f (3mks)
1. 36.7892
2. 0.09854
3. 345204
15. Find the square root of 0.001952 using mathematical table. (3mks)
16. Arrange the following numbers in an ascending order. (3mks)
2 7/8, 14/5, 3, 17/9

SECTION B 50 MARKS
ANSWER ALL QUESTIONS IN THIS SECTION

1. A rectangle measures 18cm by 12cm.
1. If each dimension is reduced by 2cm, by what percentage is:
1. The perimeter reduced. (3mks)
2. The area of the rectangle reduced. (3mks)
2. If each dimension is reduced by 2%, by what percentage is the area of the rectangle reduced. (4mks)
2.
1. Study the equations below and complete the tables.
7x + 2y = 16
 x 0 1 2 y

2x-y = 3
 x 0 1 2 y
2. On the same axis, draw the graphs of y = 2x – 3 and y = 8 - 7/2x (4mks)
3. Use the graph to solve the simultaneous equations; (2mks)
y = 2x – 3
y = 8 - 7/2x
3.
1. Draw triangle ABC in which AB = 11cm, AC = 8cm and BC = 5.6cm. (2mks)
2. Construct the bisectors of any two angles of the triangle and let the bisectors meet at R. (2mks)
3. Draw the perpendicular from R to AB so that it cuts AB at M. (2mks)
4. With centre R and the radius RM, Draw a circle. (2mks)
5. Calculate the area of the circle. (2mks)
4. A carpenter constructed a closed wooden box with internal measurements 1.5m long by 0.8m wide and 0.4m high. The wood used in constructing the box was 1.0cm thick and had a density of 0.6 g/cm³.
1. Determine the:
1. Volume in cm³ of the wood used in constructing the box. (3mks)
2. Mass of the box in kg correct to 1dp. (2mks)
2. Identical cylindrical tins of diameter 10cm and height 20cm with a mass of 120g each were packed in the box. Calculate;
1. Maximum number of tins that were packed in the box. (3mks)
2. Total mass of the box with the tins. (2mks)
5. Find the size of the angles marked with letters.

1. (1mk)

2. (5mks)

3.
(4mks)
6.
1. The GCD of two numbers and their LCM is 360. If one of the numbers is 72, what is the other number? (3mks).
2. When a number u is divided by 34 or 24 or 40, the remainder is always 5. Find the least value of u. (3mks)
3. Three bells ring at intervals of 12minutes, 15minutes and 18minutes. If they sound together at 10:00 am;
1. After how long will they next sound together? (3mks).
2. What time will this be? (1mks)

MARKING SCHEME

1. 4800 – 4000 = 800
Let the bus fare be x.
1/2 x = 800
X = 800 x 2 = 1600
Train fare 4000 – 1600 = 2400

2. Let the number be x
19346 = 841 rem 3
x
19346 = x(841) + 3
19346 = 841x + 3
19346 – 3 = 841x
19343 = 841x
X = 23
3.
 2 48 60 2 24 30 2 12 15 2 6 15 3 3 15 5 1 5
L.C.M= 2⁴ x 3 x5
Area = 240 x 240 = 57600
57600   = 5.76 m²
10000
4.
 2 1470 3 735 5 245 7 49 7 7 1

 2 7056 2 3528 2 1764 2 882 3 441 3 147 7 49 7 7 1

1470² = ( 2 x 3 x 5 x 7²)²
7056= 2⁴ x 3² x 7²
( 2 x 3 x 5 x 7²)²
√2⁴ x 3² x 7²
2² x 3² x 5² x 7⁴
2² x 3 x 7
3 x 5² x 7³

5. Numerator
1/2 x 7/2 + 3/2 (5/2 - 2/3)
7/4 + 3/2(11/6)
7/4 + 11/4 = 18/4 = 9/2

Denomenator
3/4  of 21/2 ÷ 1/2
3/4 x 5/2 x 2/1 = 15/4
9/2÷15/4
9/2 x 4/15
= 11/5
6.

6 – (- 4) = 10
7. X = 4 y = -6
6x – 3xy
6(4) – 3 (4) (-6)
24 + 72
= 96
8. a+ b   - 2a - b
2         3
3(a+ b) – 2(2a – b)
6
3a+ 3b – 4a +2b
6
5b – a
6
9. B : G: T
2 : 3 :5
2/5 ⇒160
3/5 ⇒?
3 x – 160
2
= 3 x 80
= 240 girls

10. a : b = 7 : 1
b : d = 1 : 2
d : c = 2 : e

a : b : d : c = 7 : 1
1 : 2
2 : e
a : b : d : c = 7 : 1 : 2 : e
a : c = 7 : e

11. C = 2 πr
2πr + 2r = 14.28
2
πr + 2r = 14.28
r(π + 2) = 14.28
r(3.142 + 2) = 14.28
r(5.142) = 14.28
r = 14.28
5.142
r = 2.777cm

12. A = 38.5 cm²
ϴ = 90⁰
A =    ϴ    x πr
360
38.5 =    90     22   x r²
360        7
r² = 38.5 x 360 x 7
90 x 22
r² = 49
r = 7

13. V = πr²h
V =   22   17.5    17.5   x 10
7          2             2
V= 2406.25 cm³
m = 2kg
m = 2 x 1000 = 2000g
ρ =    m      2000
v        2406.25
ρ = 0.8312 g/cm³

14.
1. 36.7892
= 36.8

2. 0. 0 985 4
= 0.0985

3. 345204 = 345000

15. √0.001952
√0.001952 = √19.52 x10-4
√19.52 x10-4 = √19.52 x √10-4
= 4.4159
+23
4.4182
= 4.4182 x 10-2
= 4.8182 x 1/100
= 0.044182

16. 2 7/8, 14/5, 3, 17/9

23/8,14/5 3 , 17/9
L:C:M = 8 x 5 x 9 = 360
23/8 x 360 = 1035
14/5 x 360 = 1008
3 x 360 = 1080
17/9 x 360 = 680
17/9, 14/5, 27/8, 3
17.
1.

Reduced by 2cm
Length = 18 - 2 = 16cm
Width = 12 – 2 = 10cm
1. Perimeter = 2(18 + 12)
= 60cm
New Perimeter = 2 (16 + 10) = 52
% = (60 – 52) x 100
60
=   8   x 100 = 13.33%
10

2. Original area = 2 (18x 12)
= 2 x 216
= 432cm2
New = 2(16 x 10) = 320cm2
% change = (432 – 320) x 100
432
= 25.93%

2. Length =    98    x 18 = 17.64 cm
100
Width =     98    x 12 = 11.76 cm
100
Original area = 18 x 12 = 216cm2
New area = 17.64 x 11.76 = 207.4464 cm2
% change = (216 – 207.4464) x 100
216
= 3.96%
18.
1.  2x – y = 3
y= 2x – 3

7x + 2y = 16
2 y =  16  –  7 x
2        2        2
2.
 x 0 1 2 y -3 -1 1 x 0 2 4 y 8 1 -6
B1B1

3. B1B1
Graph

Point of intersection = (2,1)
X= 2, y = 1
19.

e. Area of the circle = 3.142 x 1.7 x 1.7
A= 9.08038cm2
A= 9.080cm2
20.
1.
1. Internal volume = 150 x 80 x 40 = 480000 cm3
External Volume = 152 x 82 x 42 = 523, 488 cm3
Volume of wood = 523,488 – 480,000 = = 43488cm3

2. Mass of the box
m = ρ x v
m = 0.6 x 4388
= 26092.8g
26092.8      = 26.0928kg
10000
= 26.1kg (1dp)
2.
1. V = πr2h
3.142 x 5 x 5 x 20
V = 1571cm3
Number of tins
= 480000
1571
= 305.54
= 305 tins

2. Total mass of the box with tins
= 26.0928 + (120 x 305)
1000
= 26.0928 + 36.6
= 62.6928kg
21.
1. n = 5
(2n - 4)90
(2(5) - 4)90
(10 – 4)90
6 x 90 = 540
2(180 – x) + 100 + 130 = 540
360 – 2x + 330 = 540
690 – 2x = 540
- 2x = -150
X = 75⁰

2. y = 180 – 60 = 1200
4x + 60 = 180
4x = 120⁰
x = 30⁰
2x = 2(30) = 60⁰
3x = 3(30) = 90⁰
4x = 4(30) = 120⁰

3. y = 87
w = 180 – 87 = 93⁰
z = 180 – (87+45) = 48⁰
x = 180 – (87 +45) = 45⁰

22.
1. G.C.D = 8 = 23
L.C.M = 360 = 23 x 32 x 5
X = 72 = 23 x 32

y = ? = 23 x 5
y = 40

2.
 2 24 34 40 2 12 17 20 2 6 17 10 3 3 17 5 5 1 17 5 17 1 17 1 1 1 1
L.C.M = 23 x 3 x 5 x 17 = 2040
Value of u= 240 + 5
u= 245
3.
1.
 2 12 15 18 2 6 15 9 3 3 15 9 3 1 5 3 5 1 5 1 1 1 1
2. L.C.M = 22 x 32 x5 = 180 mins = 3hrs
10: 00
+ 3. 00
13:00
- 12:00
1:00 pm

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