Important Formulas on "Problems on Ages”:
- 1. If the current age is ‘x’, then ‘n’ times the age is ‘nx’.
- 2. If the current age is x, then age n years later/hence = (x + n).
- 3. If the current age is ‘x’, then age ‘n’ years ago = (x - n).
- 4. The ages in a ratio ‘a : b’ will be ‘ax’ and ‘bx’.
- 5. If the current age is x, then 1/n of the age is x/n.
Example. 01
A father was 4 times as old as his son 8 years ago. Eight years hence, father will be twice as old as his son. Find their present ages.
Solution:
Let son's age 8 years ago be x years.
Thus, father's age at that time = 4x years
After 8 years, son's age = (x + 8) + 8 = (x+16) years
After 8 years, father's age = (4x + 8) + 8 = (4x+16) years
So, According to Question, 2(x + 16) = 4x + 16 or x = 8
Therefore, The present age of the son = x + 8 = 16 years
The present age of the father = 4x + 8 = 32 + 8= 40 years
Example. 02
Father is aged three times more than his son. After 8 years, he would be two and a half times of his son's age. After further 8 years, how many times would he be of his son's age?
Solution:
Let Son's present age be ‘x’ years.
Then, father's present age =(x + 3x) years = 4x years.
Therefore, (4x + 8) = (2/5) (x + 8)
=> 8x + 16 = 5x + 40
=> 3x = 24 => x = 8.
After further 8 years, Son's age will be (x + 16) = 24 years.
And father's age will be (4x +16) = 48 years.
Hence, the required ratio is (4x +16)/(x+16) = 48/24 = 2.
Example. 03
A man is 24
years older than his son. In two years, his age will be twice the age of his
son. The present age of his son is:
(A) 14 years
(B) 18 years
(C) 20 years
(D) 22 years
Answer:
Option D
Solution:
Let the
son's present age be x years. Then, man's present age = (x +
24) years.
∴ (x +
24) + 2 = 2(x + 2)
=> x + 26 = 2x +
4
=> x = 22.
=> x = 22.