Problems based on age Solved with shortcut methods
Problems Based on Age
Equation problems of age are a part of the quantitative aptitude section. In these problems the questions result in equations, which may be linear or non-linear, and their solutions represent the ages of the people in the question. In the following sections we solve some examples of these problems and learn the shortcuts that can be used to solve them.
The three situations
- Age some years ago
- Present age
- Age some years hence
Age some years ago
Ex.1Ram is 40 years old and Mohan is 60 years old. How many years ago was the ratio of their ages 3:5?
Let the required number of years be x.
(40 − x)/(60 − x) = 3/5
200 − 5x = 180 − 3x, so 2x = 20 and x = 10.
Check: 30 : 50 = 3 : 5
Ans: 10 years ago
Present age
Ex.2The ratio of the present ages of Rahul and Vikram is 2:1. The ratio of their ages after 30 years will be 7:6. What is the present age of Rahul?
Conventional method
Let the present ages of Rahul and Vikram be 2x and x.
After 30 years: (2x + 30)/(x + 30) = 7/6
12x + 180 = 7x + 210, so 5x = 30 and x = 6.
Rahul's present age = 2 × 6 = 12 years.
Short trick
Rahul : Vikram now = 2 : 1, and after 30 years = 7 : 6
Change in Rahul's ratio = 7 − 2 = 5, and change in Vikram's ratio = 6 − 1 = 5.
So 5 ratio units = 30 years, and 1 unit = 6 years.
Present ages = 6 × 2 : 6 × 1 = 12 years and 6 years.
Ans: 12 years
Ex.3Neeraj is as much younger than Gopal as he is older than Deepak. If the sum of the ages of Gopal and Deepak is 58 years, what is Neeraj's age?
The differences are equal: G − N = N − D
So G + D = 2N
58 = 2N, so N = 29.
Ans: 29 years
Type 1: ratio some years ago and now
If t1 years earlier the father's age was x times that of his son, and at present the father's age is y times that of his son, then:
Son's age = t1(x − 1)/(x − y)
Ex.4The age of a father 3 years ago was 7 times that of his son. At present the father's age is 5 times that of his son. What are the present ages of the father and the son?
Conventional method
Let the present age of the son = x years, so the father's present age = 5x.
3 years ago: 7(x − 3) = 5x − 3
7x − 21 = 5x − 3, so 2x = 18 and x = 9.
Son = 9 years, father = 5 × 9 = 45 years.
Short trick, Method 1
Son's age = t1(x − 1)/(x − y) = 3(7 − 1)/(7 − 5) = 9 years
Father's age = 9 × 5 = 45 years.
Short trick, Method 2
Son : Father 3 years ago = 1 : 7 (difference 6), and now = 1 : 5 (difference 4).
Make the differences equal: (1 : 7) × 4 = 4 : 28 and (1 : 5) × 6 = 6 : 30 (both differences are 24).
The son's ratio rose from 4 to 6, so 2 units = 3 years and 1 unit = 3/2 years.
Present ages = (6 : 30) × 3/2 = 9 : 45
Ans: Son 9 years, father 45 years
Type 2: ratio now and some years hence
If the present age of the father is y times the age of his son, and T2 years hence the father's age becomes z times the age of his son, then:
Son's age = (z − 1)T2/(y − z)
Ex.5At present the age of a father is five times the age of his son. Three years hence, the father's age will be four times that of his son. Find the present ages of the father and the son.
Conventional method
Let the present age of the son = x, so the father's present age = 5x.
3 years hence: 4(x + 3) = 5x + 3
4x + 12 = 5x + 3, so x = 9.
Son = 9 years, father = 45 years.
Short trick, Method 1
Son's age = (z − 1)T2/(y − z) = (4 − 1) × 3/(5 − 4) = 9 years
Father's age = 9 × 5 = 45 years.
Short trick, Method 2
Son : Father now = 1 : 5 (difference 4), and 3 years hence = 1 : 4 (difference 3).
Make the differences equal: (1 : 5) × 3 = 3 : 15 and (1 : 4) × 4 = 4 : 16 (both differences are 12).
The son's ratio rose from 3 to 4, so 1 unit = 3 years.
Present ages = (3 : 15) × 3 = 9 years and 45 years.
Ans: Son 9 years, father 45 years
Type 3: ratio some years ago and some years hence
If t1 years earlier the age of the father was x times the age of his son, and t2 years hence the age of the father becomes z times the age of his son, then:
Son's age = [t2(z − 1) + t1(x − 1)]/(x − z)
Ex.6Three years earlier, a father was 7 times as old as his son. Three years hence, the father's age will be 4 times that of his son. What are the present ages of the father and his son?
Conventional method
Let the present age of the son = x years and of the father = y years.
3 years earlier: 7(x − 3) = y − 3
3 years hence: 4(x + 3) = y + 3
Subtracting: 7x − 21 − 4x − 12 = −6, so 3x = 27 and x = 9. Then y = 7 × 6 + 3 = 45.
Short trick, Method 1
Son's age = [t2(z − 1) + t1(x − 1)]/(x − z) = [3(4 − 1) + 3(7 − 1)]/(7 − 4) = 27/3 = 9 years
Father's age = 45 years.
Short trick, Method 2
Son : Father 3 years ago = 1 : 7 (difference 6), and 3 years hence = 1 : 4 (difference 3).
Make the differences equal: (1 : 7) × 3 = 3 : 21 and (1 : 4) × 6 = 6 : 24 (both differences are 18).
The son's ratio rose from 3 to 6, so 3 units = 6 years (3 years ago to 3 years hence) and 1 unit = 2 years.
The present age is 3 years after '3 years ago', that is 1.5 units: son = 3 + 1.5 = 4.5 and father = 21 + 1.5 = 22.5 units.
Present ages = (4.5 : 22.5) × 2 = 9 years and 45 years.
Ans: Son 9 years, father 45 years
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