MULTIPLY FRACTIONS WORD PROBLEMS

Problem 1 :

The denominator of a fraction is 9/4 of the numerator. Find the fraction.

Solution :

Let x be the numerator.

Given : The denominator of the fraction is 9/4 of the numerator.

Then, the denominator is 9x/4.

Therefore, the required fraction is

= x/(9x/4)

= x ⋅ (4/9x)

= (x ⋅ 4)/9x

= 4x/9x

= 4/9

Problem 2 :

In a school, there are 450 students in total. If 2/3 of the total strength is boys, find the number of girls in the school.

Solution :

Given : 2/3 of the total strength is boys.

Then, the number of boys in the school is

= 2/3 ⋅ 450

= (2 ⋅ 450)/3

= 900/3

= 300

Therefore, the number of girls in school is 

= Total strength - No. of boys

= 450 - 300

= 150

Problem 3 :

If the fraction is multiplied 3, it becomes 5/4. And sum of the numerator and denominator is 17. Find the fraction.

Solution :

Let x/y be the required fraction.

Given : If the fraction is multiplied 3, it becomes 5/4.

Then, we have

3 ⋅ x/y = 5/4

3x/y = 5/4

12x = 5y

12x - 5y = 0 -----(1)

Given : Sum of the numerator and denominator is 17.

x + y = 17 ----(2)

Solving (1) and (2), we get

x = 5 and y = 12

x/y = 5/12

So, the required fraction is 5/12.

Problem 4 :

Of two numbers, 1/5th of a the greater equal to 1/3rd of the smaller and their sum is 16. Find the numbers.

Solution :

Let x and y be the required two numbers such that x > y.

Given : 1/5th of a the greater equal to 1/3rd of the smaller.

Then, we have

x/5 = y/3

3x = 5y

3x - 5y = 0 ----(1)

Given : Sum of the numbers is 16.

Then, we have

x + y = 16 ----(2)

Solving (1) and (2), we get

x = 10 and y = 6

So, the numbers are 6 and 10.

Problem 5 :

The fourth part of a number exceeds the sixth part by 4. Find the number.

Solution :

Let x be the required number.

Then,

Fourth part of the number = x/4

Sixth part of the number = x/6

Given : The fourth part of a number exceeds the sixth part by 4.

Then, we have

x/4 = x/6 + 4

x/4 - x/6 = 4

3x/12 - 2x/12 = 4 

(3x - 2x)/12 = 4

x/12 = 4

x = 48

So, the number is 48.

Problem 6 :

A work was assigned to A and B. A completed 2/3 of the work and B completed 5/8 of the remaining work. What fraction of the original word completed by B?

Solution :

Given : A completed 2/3 of the work.

Remaining = 1/3

Given : B completed 5/8 of the remaining work.

Then, the part of the work completed by B is

= 5/8 ⋅ 1/3

= (5 ⋅ 1)/(8 ⋅ 3)

= 5/24

= 5/24

So, B completed 5/24 of the original work.

Problem 7 :

Rachel bought a pizza and ate 2/5 of it. If he had given 2/3 of the remaining to his friend, what fraction of the original pizza will be remaining now ?

Solution :

Given : Rachel 2/5 of a pizza.

Remaining = 3/5

Given : He gave 2/3 of the remaining to his friend.

Then, Amount of pizza given to his friend is

= 2/3 ⋅ 3/5

= (2 ⋅ 3)/(3 ⋅ 5)

= (2 ⋅ 1)/(1 ⋅ 5)

= 2/5

Now, we have

Then, we have

Remaining pizza = 1 - (2/5 + 2/5)

Remaining pizza = 1 - 4/5

Remaining pizza = 5/5 - 4/5

Remaining pizza = 1/5

So, 1/5 of the original pizza will be remaining.

Problem 8 :

A work was  assigned to David. He completes 1/9 of the work each day. If the total work is considered to be 450 units, how many units will he complete in 5 days?

Solution :

Given : David completes 1/9 of work each day.

Then, the fraction of work completed by him in 5 days is 

= 5 ⋅ 1/9

= (5 ⋅ 1)/9

= 5/9

Given : The total work is considered to be 450 units.

Then, the number of units completed in 5 days is

= 5/9 ⋅ 450

= (5 ⋅ 450)/9

= 2250/9

= 250

So, David will complete 250 units of work in 5 days.

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