## 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)

ecolebooks.com

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

1/81/6 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

( 9/73/8 ) x 2/3

 1 = M1 M1A1 Simplifying numerator simplify 2. Num= 2Denom. 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 yThe number becomes xy= 10x +yand x + y = 10Reserved yx = 10y + x(10y + x) – (10x + y) = 5410y + x – 10x – y = 549y – 9x = 54y – x = 6y – x = 6y + x = 102y = 16y = 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 = 216Difference of multipliers = 100 – 1 = 99Fraction B1  M1A13 7 M1 M1 M1A13 For +ve index 8. M1 M1   M1   A1 04 9. Let the number of chicken be xTurkeys will be x + 6¼x + 1/3 (x + 6) = 30¼x + 1/3x + 2 = 307/12x = 28  = 48Number of chickens = 48Number of turkeys = 48 + 6 = 54Total 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 30m2nd ¾ x 30 = 22.5m3rd ¾ x 22.5 = 16.85m4th ¾ x 16.85 = 12.64m5th ¾ x 12.64 = 9.48m6th ¾ x 9.48 = 7.11m7th ¾ x 7.11 = 5.3325m8th ¾ x 5.3325 = 3.9993m9th ¾ x 3.9993 = 2.9995m10th ¾ x 2.9995 = 2.2496m 2.25Or using formulaT10 = 30( ¾ )10-1 = 30( ¾ )9= 30 x 0.075082.2524m  2.25m M1 A1 M1  M1  A1  M1  A1 M1  M1     M1A12 d.pM1 A1 Every four     Every four 13. M1M1A1 For numFor den 03 14. M1M1 A1 03 15 -8-39+5-1-3×2 = -42 -7 = 6 M1 A1 2 Numerators & Denominators 16 M1 M1 A1 Simplified denominator  Method shownShow how to get factors of 13824 3 marks 17 B1             B1 B1 3

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 9/211/10

= 9/511/10 = 18 -11/ 10 = 7/10

1/8 1/6 X 3/8 = 1/81/16

= 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 (38/555/36 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

1/81/16
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

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