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?
Solution:
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:
Solution:
Let C's age be x years. Then, B's age = 2x years. A's age = (2x + 2) years.
∴ (2x + 2) + 2x + x = 27
⇒ 5x = 25
⇒ x = 5.
Hence, B's age = 2x = 10 years.
5.
Present ages of Sameer and Anand are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Anand's present age in years?
