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Fractions Questions
1. Simplify 
(3 mks)
2. Simplify without using calculators and tables
(3mks)
3. Evaluate without using a calculator. (3mks)
4. A two digit number is such that the sum of the ones and the tens digit is ten. If the digits are reversed, the new number formed exceeds the original number by 54.
Find the number. (3 marks)
5. Evaluate 
(3mks)
6. Convert the recurring decimal l
into fraction (3 mks)
7. Simplify 
without using tables or calculator . (3mks)
8. Evaluate without using tables or calculators (4mks)
9. Mr. Saidi keeps turkeys and chickens. The number of turkeys exceeds the number of chickens by 6. During an outbreak of a disease, ¼ of the chicken and 1/3 of the turkeys died. If he lost total of 30 birds, how many birds did he have altogether? (2mks)
10. Work out 
(2mks)
11. Evaluate -4 of (-4 + -5 15) + -3 – 4 2)
84 -7 + 3 – -5
12. (a) The recurring decimal 0.
can be written as 
(i) Find the common ratio. (2mks)
(ii) Find an expression for the sum of n terms of this series. (3mks)
(iii) Find the 8th term of the series. (2mks)
(b) A ball is dropped from a height 30m above the ground and rebounds to ¾ of previous height continuously until it stops. Find the distance that the ball bounces when it hits the ground the 10th time. (ans to 2 d.p). (3mks)
13. Evaluate 
without using a calculator. (3mks)
14. Without using tables or calculators evaluate.
(4mks)
15. Without using tables or calculator, evaluate the following. (2 mks)
–8 + (–13) x 3 – (–5)
–1 + (–6) ÷ 2 x 2
16. Without using tables or calculator evaluate (3 marks)
17. Express 
as a single fraction (3 marks)
18. Simplify ½ of 3½ + 1½ (2½ – 2/3)
¾ of 2½ ½
19. Evaluate :
2/5
½ of 4/9 – 11/10
1/8 – 1/6 of 3/8
20. Without using a calculator or table, work out the following leaving the answer as a mixed
number in its simplest form:-
¾ + 12/7 ÷ 3/7 of 21/3
( 9/7–3/8 ) x 2/3
Fractions Answers
1 | = |
M1
M1 A1 | Simplifying numerator
simplify | |||||||
2. | Num
= 2 Denom. 8 x 1 x 25 = 200
|
M1
M1
A1 |
Or equivalent 0.01 | |||||||
03 | ||||||||||
3. |
|
M1
M1
A1 |
Application of bodmas
Simplification of both numerators and denominator | |||||||
03 | ||||||||||
4 | Let the digits be x and y The number becomes xy = 10x +y and x + y = 10 Reserved yx = 10y + x (10y + x) – (10x + y) = 54 10y + x – 10x – y = 54 9y – 9x = 54 y – x = 6 y – x = 6 y + x = 10 2y = 16 y = 8
x = 8-6 = 2 . . . The number is 28 |
M1
1M
A1 |
Splitting of ones & tens and the reverse
Solving of the simultaneous eqn.
Answer | |||||||
3 | ||||||||||
5 |
| M1
M1
A1 | ||||||||
6 |
Difference = 216 Difference of multipliers = 100 – 1 = 99 Fraction |
B1
M1A1 3 | ||||||||
7 |
| M1
M1
M1 A1 3 | For +ve index | |||||||
8. | |
M1
M1
M1
A1 | ||||||||
04 | ||||||||||
9. | Let the number of chicken be x Turkeys will be x + 6 ¼x + 1/3 (x + 6) = 30 ¼x + 1/3x + 2 = 30 7/12x = 28 = 48 Number of chickens = 48 Number of turkeys = 48 + 6 = 54 Total number of birds = 54 + 48 = 102 |
B1
B1 |
For 48
For 102 | |||||||
02 | ||||||||||
11. -4 of (-4 -3) + -3-2)
-12 + 3 + 5
-4 of (-7-3-2)
-4 M1 for -4
= 48
-4 M1 for 48
= -12 A1
3
12. | (a) (i) ratio
(b) 1st bounce 30m 2nd ¾ x 30 = 22.5m 3rd ¾ x 22.5 = 16.85m 4th ¾ x 16.85 = 12.64m 5th ¾ x 12.64 = 9.48m 6th ¾ x 9.48 = 7.11m 7th ¾ x 7.11 = 5.3325m 8th ¾ x 5.3325 = 3.9993m 9th ¾ x 3.9993 = 2.9995m 10th ¾ x 2.9995 = 2.2496m 2.25 Or using formula T10 = 30( ¾ )10-1 = 30( ¾ )9 = 30 x 0.07508 2.2524m 2.25m |
M1
A1
M1
M1
A1
M1
A1
M1
M1
M1 A1 2 d.p M1
A1 |
Every four
Every four | ||
13. |
| M1 M1 A1 | For num For den | ||
03 | |||||
14. |
| M1 M1
A1 | |||
03 | |||||
15 |
-8-39+5 -1-3×2
= -42 -7
= 6 |
A1 2 |
Numerators & Denominators | ||
16 |
|
M1
M1
A1 |
Simplified denominator
Show how to get factors of 13824 | ||
3 marks | |||||
17 | |
B1
B1
B1 | |||
3 | |||||
Method shown
18. ½ x 7 = 3 x 1 5 ¾ x 5 x X
2 2 6 2
7 + 3 x 11 = 15
4 2 2 4
7 + 11 = 18
4 4 4
18
15
4 4
18 x 4 = 6 = 1 1
4 15 5 5
19. 2/5 ÷ ½ 0f 4/9 – 11/10
= 2/5 ÷ ½ X 4/9 – 11/10
= 2/5 x 9/2 – 11/10
= 9/5 – 11/10 = 18 -11/ 10 = 7/10
1/8 – 1/6 X 3/8 = 1/8 – 1/16
= 2-1/16 = 1/16
2/5 ÷ ½ 0f 4/9 – 11/10 = 7/10
1/8 – 1/6 of 3/8 1/16
= 7/10 X 16/1
= 56/5 = 111/5
20. BODMAS
3/7 X 7/3 = 1
9/7 X 1 = 9/7
¾ + 9/7 = 21 + 36 = 57 M1
28 28
9/7 – 3/8 = 72 – 21 = 51 x 2/3 = 17/28 M1
57/ 28 x 28/17 = 3 6/17 A1
21. 2 x 9 – 11
5 2 10
1 – 1
8 16
= 7 x 16
10 1
= 56 = 11 1
5 5
22. 3/8 (38/5 – 55/36 x 12/5)
3/8 x 59/15 = 59/40 = 119/40
23. Numerator
(9/5 X 25/18 ) ÷ 5/2 X 24
7/3 – ( ¼ x 12 ) ÷ 5/3
9/5 x 25/18 = 5/2
5/3 x 24
5/2 x 3/5 x 24 = 36
7/3 – ¼ x 12 5/3
7/3 – 3 x 3/5
- 36 = 67.50
8/15 = 67 ½
7/3 – 3 x 3/5
24. Let X be money raised
Teachers house = 1/7 x
Classrooms = 2/3 x 6/7 = 4/7x
Remainder = 1/3 x 6/7 = 2/7x
2/7 x = 300000
x = Shs.1050000
21. Work out the following, giving the answer as a mixed number in its simplest form.
2/5
½ of
4/9 – 1 1/10
1/8 – 1/16
x
3/8
22. Evaluate
23. Without using a calculator, evaluate:
14/5 of 25/18
12/3 x 24
21/3 – ¼ of 12
5/3 Leaving the answer as a fraction in its simplest form
24. There was a fund-raising in Matisi high school. One seventh of the money that was raised
was used to construct a teacher’s house and two thirds of the remaining money was used
to construct classrooms. If shs.300,000 remained, how much money was raised
