ADDING AND SUBTRACTING FRACTIONS PRACTICE QUESTIONS

Problem 1 :

Add :

2/5  +  1/2

Solution :

The given fractions are unlike fractions, to make it as like fractions. We have to find the least common multiple.

LCM(2, 5) = 10

= (2/5) x (2/2) + (1/2) x (5/5)

= 4/10 + 5/10

= (4 + 5)/10

= 9/10

Problem 2 :

Subtract  3/5  - 1/4

Solution :

The given fractions are unlike fractions, to make it as like fractions. We have to find the least common multiple.

LCM (5, 4) = 20

= (3/5) x (4/4) - (1/4) x (5/5)

= 12/20 - 5/20

= (12 - 5)/20

= 7/10

Problem 3 :

Subtract  3/7 - 1/2

Solution :

= (3/7) - (1/2)

LCM (7, 2) = 14

= (3/7) x (2/2) - (1/2) x (7/7)

= 6/14 - 7/14

= (6 - 7) / 14

= -1/14

Problem 4 :

Simplify  -1/2 + 3/4

Solution :

= -1/2 + 3/4

LCM (2, 4) = 4

= (-1/2) x (2/2) + (3/4)

= -2/4 + 3/4

= (-2 + 3)/4

= 1/4

Problem 5 :

Simplify 1/10  -  4/5

Solution :

= 1/10 - 4/5

LCM (5, 10) = 10

= 1/10 - (4/5) x (2/2)

= 1/10 - 8/10

= (1 - 8)/10

= -7/10

Problem 6 :

Simplify 7/9 + 2/3

Solution :

= 7/9 + 2/3

LCM (3, 9) = 9

= (7/9) + (2/3) x (3/3)

= 7/9 + 6/9

= (7 + 6) / 9

= 13/9

Converting into mixed fraction, we get

= 1  4/9

Problem 7 :

Find the sum of 1/3 and 2/5.

Solution :

To find the sum of the above fractions, we have to add these two fractions.

= 1/3 + 2/5

LCM (3, 5) = 15

= (1/3) x (5/5) + (2/5) x (3/3)

= 5/15 + 6/15

= (5 + 6)/15

= 11/15

Problem 8 :

The difference between 1/4 and 2/3.

Solution :

To find the difference between the given fractions, we have to subtract the above fractions.

= (1/4) - (2/3)

LCM (4, 3) = 12

= (1/4) x (3/3) - (2/3) x (4/4)

= 3/12 - 8/12

= (3 - 8)/12

= -5/12

Problem 9 :

The number 3 less than 2/3.

Solution :

= (2/3) - 3

= (2 - 9)/3

= -7/3

So, 3 less than 2/3 is -7/3.

Problem 10 :

What must 1/5 be increased by to get 2/3 ?

Solution :

Let x be the required number.

1/5 + x  =  2/3

x  =  (2/3) - (1/5)

lcm (3, 5)  =  15

x  =  (10/15) - (3/15)

x  =  (10-3)/15

x  =  7/15

Problem 11 :

What number is 3/4 less than -1  1/2 ?

Solution :

Let x be the required number.

x  =  -1  1/2 - (3/4)

x  =  -(3/2) - (3/4)

x  =  (-6-3)/4

x  =  -9/4

x  =  -2   1/4

So, the required number is -2  1/4.

Problem 12 :

Add 5/6, 3/5 and 1/3.

Solution :

= (5/6) + (3/5) + (1/3)

LCM (3, 5, 6) = 30

= (5/6) x (5/5) + (3/5) x (6/6) + (1/3) x (10/10)

= 25/30 + 18/30 + 10/30

= (25 + 18 + 10)/30

= 53/30

Converting into mixed fraction, we get

= 1  23/30

Problem 13 :

Over three successive days Colin builds 1/3 , 1/5 and 1/4 of the brickwork of his new garage. 

What fraction must he complete on the fourth and final day?

Solution :

Work completed on three consecutive days 

= (1/3) + (1/5) + (1/4)

LCM (3, 4, 5) = 60

= (1/3) x (20/20) + (1/5) x (12/12) + (1/4) x (15/15)

= 20/60 + 12/60 + 15/60

= (20 + 12 + 15)/60

= 47/60

Portion of work yet to be completed = 1 - (47/60)

= (60 - 47)/60

 = 13/60

Problem 14 :

Jamil spent 1/4 of his weekly salary on rent, 1/5 on food, and 1/6 on clothing and entertainment. The remaining money was banked.

a) What fraction of Jamil’s money was banked?

b) If he banked $138, what is his weekly salary?

c) How much did Jamil spend on food?

Solution :

Amount he spends for rent, food and clothing are

1/4, 1/5 and 1/6

= (1/4) + (1/5) + (1/6)

LCM (4, 6, 5) = 60

= (1/4) x (15/15) + (1/5) x (12/12) + (1/6) x (10/10)

= 15/60 + 12/60 + 10/60

= (15 + 12 + 10) / 60

= 37/60

a) Amount remaining = 1 - 37/60

= (60 - 37)/60

= 23/60

b)  Amount he has banked = 23/60

Let his weekly salary be x

23/60 of x = 138

x = 138 x (60/23)

= 6(60)

= 360

So, his weekly salary is $360.

c) Amount spent for food = 1/5 of 360

= (1/5) x 360

= $72

So, $72 is the amount he has spent for food.

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