WORD PROBLEMS ON RATIO AND PROPORTION

About "Word Problems on Ratio and Proportion"

Word Problems on Ratio and Proportion :

In this section, we are going to see, how ratio and proportion word problems can be solved easily. 

Before we look at the problems, first let us come to know the shortcuts in ratio and proportion which are required to solve the problems.

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Word problems on Ratio and Proportion

Problem 1 :

The average age of three boys is 25 years and their ages are in the proportion 3:5:7. Find the age of the youngest boy.  

Solution :

From the ratio 3 : 5 : 7, the ages of three boys are 3x, 5x and 7x.

Average age of three boys = 25

(3x+5x+7x)/3 = 25 ----------> 15x = 75 -----------> x = 5

Age of the first boy = 3x = 3(5) = 15

Age of the first boy = 5x = 5(5) = 25

Age of the first boy = 7x = 7(5) = 105

Hence the age of the youngest boy is 15 years.

Problem 2 :

John weighs 56.7 kilograms. If he is going to reduce his weight in the ratio 7:6, find his new weight.

Solution :

Original weight of John = 56.7 kg (given)

He is going to reduce his weight in the ratio 7:6

His new weight = (6x56.7)/7 = 6x8.1 = 48.6 kg.

Hence his new weight = 48.6 kg

Problem 3 :

The ratio of the no. of boys to the no. of girls in a school of 720 students is 3:5. If 18 new girls are admitted in the school, find how many new boys may be admitted so that the ratio of the no. of boys to the no. of girls may change to 2:3.

Solution :

Sum of the terms in the given ratio  =  3 + 5  =  8

So, no. of boys in the school  =  720  (3/8)  =  270 

No. of girls in the school  =  720  (5/8)  =  450 

Let "x" be the no. of new boys admitted in the school.

No. of new girls admitted  =  18  (given)

After the above new admissions,

No. of boys in the school  =  270 + x

No. of girls in the school  =  450 + 18  =  468

The ratio after the new admission is 2 : 3   (given)

So, we have

(270 + x) : 468  =  2 : 3

Use cross product rule. 

3(270 + x)  =  468 ⋅ 2    

810 + 3x  =  936

3x  =  126

x  =  42

Hence the no. of new boys admitted in the school is 42.

Problem 4 :

The monthly incomes of two persons are in the ratio 4:5 and their monthly expenditures are in the ratio 7:9. If each saves $50 per month, find the monthly income of the second person. 

Solution :

From the given ratio of incomes ( 4 : 5 ), 

Income of the 1st person = 4x

Income of the 2nd person = 5x

(Expenditure  =  Income - Savings)

Then, expenditure of the 1st person  =  4x - 50

Expenditure of the 2nd person  =  5x - 50

Expenditure ratio  =  7 : 9  (given)

So, we have

(4x - 50) : (5x - 50)  =  7 : 9

Use cross product rule.

9(4x - 50)  =  7(5x - 50)

36x - 450  =  35x - 350 

x  =  100

Then, the income of the second person is 

=  5x 

=  5(100) 

=  500.

Hence, income of the second person is $500.

Problem 5 :

The ratio of the prices of two houses was 16:23. Two years later when the price of the first has increased by 10% and that of the second by $477, the ratio of the prices becomes 11:20. Find the original price of the first house.   

Solution :

From the given ratio 16:23,

Original price of the 1st house  =  16x

Original price of the 2nd house  =  23x

After increment in prices,

Price of the 1st house  =  16x + 10% of 16x 

=  16x + 1.6x

=  17.6x

Price of the 2nd house  =  23x + 477

After increment in prices, the ratio of prices becomes 11:20.

Then, we have

17.6x : (23x + 477)  =  11 : 20

Use cross product rule.

20(17.6x)  =  11(23x + 477) 

352x  =  253x + 5247

99x  =  5247

x  =  53

Then, original price of the first house is

=  16x

=  16(53)

=  848

Hence, original price of the first house is $848.

Apart from the problems on "Word problems on ratio and proportion", if you need more problems on ratio and proportion, please click the following links. 

Ratio and Proportion Word Problems - 1

Ratio and Proportion Word Problems - 2

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