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
So, 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 :
Given : Original weight of John = 56.7 kg.
He is going to reduce his weight in the ratio 7:6.
We can use the following hint to find his new weight, after it is reduced in the ratio 7 : 6.
His new weight is
= (6 ⋅ 56.7) / 7
= 48.6 kg.
So, John's new weight is 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 is
= 3 + 5
= 8
So, no. of boys in the school is
= 720 ⋅ (3/8)
= 270
No. of girls in the school is
= 720 ⋅ (5/8)
= 450
Given : Number of new girls admitted in the school is 18.
Let x be the no. of new boys admitted in the school.
After the above new admissions,
No. of boys in the school = 270 + x
No. of girls in the school = 450 + 18 = 468
Given : The ratio after the new admission is 2 : 3.
Then, we have
(270 + x) : 468 = 2 : 3
Use cross product rule.
3(270 + x) = 468 ⋅ 2
810 + 3x = 936
3x = 126
x = 42
So, the number 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.
So, 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
So, original price of the first house is $848.
Apart from the word problems given above, if you need more word 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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