Father A is twice the age of the difference in years of the ages of Father B and Son A, who is one and a half times older than Son B.

Father B is currently twice the age of Son A is going to be when Son B will be double the age he is now.

All of the ages are multiples of five.

How old is Father A?

Let Fa, Fb, Sa, and Sb represent the ages of the four people in question. Sa=1.5*Sb and all ages are multiples of 5, so assume Sb=10 and Sa = 15. Son B will be twice his age in Sb years, at which time Son A's age will be Sa+Sb, so
Fb=2*(Sa+Sb)=2*25=50. Fa=2*(Fb-Sa)=2*(50-15)=70. So the four ages are:

Father A is 70
Father B is 50
Son A is 15
Son B is 10

For 1.5*Sb to be a multiple of 5, all ages could be zero, which is not possible, or all ages could be multiplied by the same integer factor, which would make the Fathers too old to have fathered.

Posted by Bryan on 2003-06-11 08:13:52