Problems on ages basically use formulae from Algebra and do not require any special formula. It is like a question type of algebra.

**Question 1 : ** A’s age after 15 years would be equal to 5 times his age 5 years ago. Find his age 3 years hence.

**Solution : ** Let A’s present age be ‘n’ years.

According to the question,

n + 15 = 5 (n – 5)

=> n + 15 = 5 n – 25

=> 4n = 40

=> n = 10

=> A’s present age = 10 years

Therefore, A’s age 3 years hence = 10 + 3 = 13 years

**Question 2 : ** The product of the ages of A and B is 240. If twice the age of B is more than A’s age by 4 years, what was B’s age 2 years ago?

**Solution : ** Let A’s present age be x years. Then, B’s present age = 240 / x years

So, according to question

2 (240 / x ) – x = 4

=> 480 – x^{2} = 4 x

=> x^{2} + 4 x – 480 = 0

=> (x + 24) (x – 20) = 0

=> x = 20

=> B’s present age = 240 / 20 = 12 years

Thus, B’s age 2 years ago = 12 – 2 = 10 years

**Question 3 : ** The present age of a mother is 3 years more than three times the age of her daughter. Three years hence, mother’s age will be 10 years more than twice the age of the daughter. Find the present age of the mother.

**Solution : ** Let the daughter’s present age be ‘n’ years.

=> Mother’s present age = (3n + 3) years

So, according to the question

(3n + 3 + 3) = 2 (n + 3) + 10

=> 3n + 6 = 2n + 16

=> n = 10

Hence, mother’s present age = (3n + 3) = ((3 x 10) + 3) years = 33 years

**Question 4 : ** The ratio of present ages of A and B is 6 : 7. Five years hence, this ratio would become 7 : 8. Find the present age of A and B.

**Solution : ** Let the common ratio be ‘n’.

=> A’s present age = 6 n years

=> B’s present age = 7 n years

So, according to the question

(6 n + 5) / (7 n + 5) = 7 / 8

=> 48 n + 40 = 49 n + 35

=> n = 5

Thus, A’s present age = 6 n = 30 years and B’s present age = 7 n = 35 years

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