Problems on Ages

1.Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit's age. After further 8 years, how many times would he be of Ronit's age?

Ans: Let Ronit's present age be x years. Then, father's present age =(x + 3x) years = 4x years.

	(4x + 8) =	5	(x + 8)
		2

 8x + 16 = 5x + 40

 3x = 24

 x = 8.

Hence, required ratio =

(4x+16)

=

48

= 2.

2. The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?

Ans: Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.

Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50

 5x = 20

 x = 4.

 Age of the youngest child = x = 4 years.


3. A father said to his son, "I was as old as you are at the present at the time of your birth". If the father's age is 38 years now, the son's age five years back was:

Ans: Let the son's present age be x years. Then, (38 - x) = x
 2x = 38.
 x = 19.
 Son's age 5 years back (19 - 5) = 14 years.