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

Let us look at the next problem on "word problems on fractions"

**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**

Let us look at the next problem on "word problems on fractions"

**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**

Let us look at the next problem on "word problems on fractions"

**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**

Let us look at the next problem on "word problems on fractions"

**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**

Let us look at the next problem on "word problems on fractions"

**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**

Let us look at the next problem on "word problems on fractions"

**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**

Let us look at the next problem on "word problems on fractions"

**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**

Let us look at the next problem on "word problems on fractions"

**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**

Let us look at the next problem on "word problems on fractions"

**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.**

Let us look at the next problem on "Word problems on fractions"

**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.

From the above picture,

Total money received = 160 + 300 +60 = $ 520 --------(2)

Profit = (2) - (1) = 520 - 360 = $160

Profit percentage = (160/360)x100 % = 44.44%

**Hence the profit percentage is 44.44**

After having gone through the examples explained above, we hope that students would have understood "Word problems on simultaneous linear equations".

Apart from the examples, if you want to know more about "Word problems on fractions", please click here.

**Please click the below links to know "How to solve word problems in each of the given topics"**

**1. Solving Word Problems on Simple Equations**

**2. Solving Word Problems on Simultaneous Equations**

**3. Solving Word Problems on Quadratic Equations**

**4. Solving Word Problems on Permutations and Combinations**

**5. Solving Word Problems on HCF and LCM**

**6. Solving Word Problems on Numbers**

**7. Solving Word Problems on Time and Work**

**8. Solving Word Problems on Trains**

**9. Solving Word Problems on Time and Work.**

**10. Solving Word Problems on Ages.**

**11.Solving Word Problems on Ratio and Proportion**

**12.Solving Word Problems on Allegation and Mixtures.**

**13. Solving Word Problems on Percentage**

**14. Solving Word Problems on Profit and Loss**

**15. Solving Word Problems Partnership**

**16. Solving Word Problems on Simple Interest**

**17. Solving Word Problems on Compound Interest**

**18. Solving Word Problems on Calendar**

**19. Solving Word Problems on Clock**

**20. Solving Word Problems on Pipes and Cisterns**

**21. Solving Word Problems on Modular Arithmetic**

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