Problems on Ages
Problems based on ages are generally asked in most of the competitive examinations. To solve these problems, the knowledge of linear equations is essential. In such problems, there may be three situations:-
- Age some years ago
- Present age
- Age some years hence
Some Useful Shortcut Method :-
Rule- If ratio of present ages of A and B is p : q and after t years, the ratio of their ages will be m : n then
= | |||
( q + t ) | n |
Example- I am four times as old as my son. 5 years hence, I will be thrice as old as my son. What is the sum of our present ages.
Solution :- Let the present ages of man and his son be 4k and k years. p = 4k , q = k , t = 5
5 years hence ratio is 3 : 1 ,
⇒ | = | ||
n | 1 |
= | |||
( q + t ) | n |
⇒ | = | ||
k + 5 | 1 |
⇒ 4k + 5 = 3k + 15
⇒ 4k - 3k = 15 - 5
∴ k = 10
∴ Sum of their ages = 4k + k = 5k
= 5 × 10 = 50 years
Rule- If the ratio of present ages of A and B is p : q and t years ago, the ratio of their ages was m : n then
= | |||
( q - t ) | n |
Example- A father is thrice as old as his son. 5 years back, he was four times as old as his son. What is the present age of father?
Solution:- Let the present age of father be 3k years and that of son is k years.
p = 3k , q = k , t = 5
5 years back the ratio of their ages was 4 : 1 | = | ||
n | 1 |
then , | = | ||
q - t | n |
⇒ | = | ||
k - 5 | 1 |
⇒ 3k - 5 = 4k - 20
⇒ 20 - 5 = 4k - 3k
∴ k = 15
∴ Present age of father = 3k = 3 × 15 = 45 years
Rule- If t years after age of one person is p times of that of another person and at present. the age of first person is q times that of another person, then
Age of first person = | |
( q - p ) |
Age of second person = | |
( q - p ) |
Example– Present age of Ram is 5 times the age of Mohan. After 10 years, Ram will be 3 times as old as Mohan. What are the present ages of Ram and Mohan?
Solution:- Let the present age of Mohan be x years.
then, present age of Ram = 5x years
After 10 years, the ratio of ages will be 3 : 1
According to the question , | = | ||
( x + 10 ) | 1 |
⇒ 5x + 10 = 3x + 30
⇒ 5x – 3x = 30 – 10
⇒ 2x = 20 ⇒ x = 10
∴ Mohan’s present age = 10 years
and Ram’s present age = 5 × 10 = 50 years
By Rule,
Hence, t = 10, q = 5 and p = 3
Ram’s present age = | years | |
( q - p ) |
= | years | |
( 5 – 3 ) |
Ram’s present age = 50 × | = 50 years | |
2 |
Mohan’s present age = | years | |
( q - p ) |
= | = 10 × 1 = 10 years | |
( 5 – 3 ) |
Rule- If t1 years before, age of a person was p times of age of another person. After t2 years, age of a person will be n times of age of second person, then
Age of first person = | years | |
( p - q ) |
Age of second person = | years | |
( p - q ) |
Example- 5 years ago, age of A was 4 times the age of B and after 10 years, A will be twice as old as B. Find the present ages of A and B.
Solution:- Let the present ages of A and B be x years and y years respectively.
5 years ago , | = | ||
( y - 5 ) | 1 |
⇒ x - 5 = 4y - 20
⇒ x - 4y = 5 - 20
⇒ x - 4y = - 15 …………. (1)
After 10 years , | = | ||
( y + 10 ) | 1 |
⇒ x + 10 = 2y + 20
⇒ x - 2y = 10 ………… (2)
On subtracting Eq. (2) fron Eq. (1), we get
x - 4y - ( x - 2y ) = -15 - 10
⇒ x - 4y - x + 2y = -25
⇒ x - 4y - x + 2y = -25
⇒ -2y = -25
∴ y = | = 12 | years | ||
2 | 2 |
⇒ x - 4 × | = -25 | |
2 |
⇒ x = -25 + 50
∴ x = 25
Hence, present ages of A is 25 years and present age of B is 12 1 / 2 years.
By Rule,
Here,
t1 = 5 , t2 = 10 , p = 4 and q = 2
A's present age = | years | |
( p - q ) |
= | years | |
( 4 - 2 ) |
A's present age = | = | years | ||
2 | 2 |
A's present age = | = 35 years | |
2 |
B's present age = | years | |
( p - q ) |
= | years | |
( 4 - 2 ) |
B's present age = | = 12 | years | ||
2 | 2 |