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## Age

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

### Sample Problems

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 – x2 = 4 x
=> x2 + 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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