WORD PROBLEMS ON FRACTIONS

About "Word problems on fractions"

Here we are going to see, how to solve word problems on fractions through some examples.

Word problems on fractions - Examples

To learn solving word problems on fractions, let us look at some examples.

Example 1 :

The denominator of a fraction exceeds the numerator by 5. If 3 be added to both, the fraction becomes 3/4. Find the fraction.

Solution :

Let "x" be the numerator.

"The denominator of the fraction exceeds the numerator"

From the above information, fraction = x / (x+5) ----------(1)

"If 3 be added to both, the fraction becomes 3/4"

From the above information, we have (x+3) / (x+5+3) = 3/4

(x+3)/(x+8) = 3/4 ---------> 4(x+3) = 3(x+8)

4x + 12 = 3x + 24 ---------> x = 12

(1)--------> x / (x+5) = 12 / (12+5) = 12/27

Hence, the required fraction is 12/27

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Example 2 :

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

Solution :

No. of boys in the school = 450 x 2/3 = 300

Total no. of students = 450.

Then no. of girls = 450 - 300 = 150

Hence the numbers girls in the school = 150

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Example 3 :

If the numerator of a fraction is increased by 2 and the denominator by 1, it becomes 1. In case, the numerator is decreased by 4 and the denominator by 2, it becomes 1/2. Find the fraction.

Solution :

Let "x/y" be the required fraction.

"If the numerator is increased by 2 and the denominator by 1, the fraction becomes 1"

From the above information, we have (x+2) / (y+1) = 1

(x+2) / (y+1) = 1 -----> x+2 = y+1 -----> x - y = -1 ----------(1)

"In case the numerator is decreased by 4 and the denominator by 2, the fraction becomes 1/2"

From the above information, we have (x-4) / (y-2) = 1/2

(x-4) / (y-2) = 1/2 -----> 2(x-4) = y-2 -----> 2x - y = 6----------(1)

Solving (1) and (2), we get x = 7 and y = 8

So, x/y = 7/8

Hence, the required fraction is 7/8

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Example 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.

From the information given in the question, we have

x + y = 16 ----------(1)

and 1/5(x) = (1/3)y ---------> 3x = 5y -------> 3x - 5 y = 0 --------(2)

Solving (1) and (2), we get x = 10 and y = 6.

Hence, the two numbers are 10 and 6

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Example 5 :

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

Solution :

Let "x" be the required number.

Fourth part of the number = x/4

Sixth part of the number = x/6

According to the question, we have x/4 - x/6 = 4

3x/12 - 2x/12 = 4 ------> (3x - 2x) / 12 = 4 -----> x / 12 = 4 ------> x = 48

Hence, the required number is 48

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Example 6 :

The width of the rectangle is 2/3 of its length. If the perimeter of the rectangle is 80 cm. Find its area.

Solution :

Let "x" be the length of the rectangle.

Then, width of the rectangle = (2/3)x

Perimeter = 80 cm -----> 2(l + w) = 80 -------> l + w = 40

l + w = 40 -------> x + (2/3)x = 40 -------> (3x+2x) / 3 = 40

(3x+2x) / 3 = 40 ------> 5x = 120 ------> x = 24

So, length = x = 24 cm

and width = (2/3)x = (2/3)24 = 16 cm

Area = l x w = 24x16 = 384 square cm.

Hence, area of the rectangle is 384 square cm

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Example 7 :

If a number of which the half is greater than 1/5 th of the number by 15, find the number.

Solution :

Let "x" be the required number.

Half of the number = (1/2)x

1/5 the of the number = (1/5)x

According to the question, we have (1/2)x - (1/5)x = 15

(1/2)x - (1/5)x = 15 ------> (5x - 2x) / 10 = 15 -------> 3x = 150

3x = 150 ----------> x = 50

Hence, the required number is 50

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Example 8 :

David's salary is $1800. David spent 2/3 of the total money for salary. He spent 1/2 of the remaining for his kids education and saved the rest. How much did he save ?

Solution :

Money spent on food = 1800x2/3 = 1200

Remaining = 1800 - 1200 = 600 -------(1)

Money spent on kids education = 1/2 of remaining = (1/2)x600

(1/2) x 600 = 300 --------(2)

Then, his savings = (1) - (2) = 600 - 300 = 300

Hence, his savings is $300

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Example 9 :

A, B and C are friends. A has 1/3 of money that B has. C has 1/2 of money that A has. If they all together have $450. How much money do A, B and C have separately ?

Solution :

Let us assume that B has the money "x".

Then, A = (1/3)x = x/3

C = 1/2 of A = (1/2) x (x/3) = x/6

Given : A + B + C = 300 -------> x/3 + x + x/6 = 450

4x/12 + 12x/12 + 2x/12 = 450 -------> (4x+12x+2x)/12 = 450

(4x+12x+2x)/12 = 450 -------->18x/12 = 450-------> x = 300

Now, A = x/3 = 300/3 = 100

B = x = 300

C = x/6 = 300/6 = 50

Hence, A has $100, B has $450 and C has $50

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Example 10 :

John's present age is 1/3 of David's age 5 years back.If David is 20 years old now, find the present age of John.

Solution :

Present age of David = 20 years

David's age 5 years back = 20 - 5 = 15 years

John's present age = 1/3 of David's age 5 years back

John's present age = 1/3 of 15 years = (1/3)x15 = 5 years.

Hence, John's present age is 5 years.

Problem 11 :

If good are purchased for $ 1500 and one fifth of them sold at a loss of 15%. Then at what profit percentage should the rest be sold to obtain a profit of 15%?

Solution :

As per the question, we need 15% profit on $1500.

Selling price for 15% on 1500

S.P =115% x 1500 = 1.15x1500 = 1725

When all the good sold, we must have received $1725 for 15% profit.

When we look at the above picture, in order to reach 15% profit overall, the rest of the goods ($1200) has to be sold for $1470.

That is,

C.P = $1200, S.P = $1470, Profit = $270

Profit percentage = (270/1200) x 100

Profit percentage = 22.5 %

Hence, the rest of the goods to be sold at 22.5% profit in order to obtain 15% profit overall.

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Example 12 :

I purchased 120 books at the rate of $3 each and sold 1/3 of them at the rate of $4 each. 1/2 of them at the rate of $ 5 each and rest at the cost price. Find my profit percentage.

Solution :

Total money invested = 120x3 = $360 -------(1)

Let us see, how 120 books are sold in different prices.