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?

(In reply to Answer by K Sengupta)

Let the respective ages of son A, Son B, Father A and Father B (in years) be P, Q, R and S.

By the given conditions, we have:

(i) R = 2(S-P); (ii) P = (3/2)*Q

Since each of P and Q are positive integers, it follows from (ii) that:
(P, Q) = (2x, 3x), for some positive integer x.
Also, all ages are multiples of 5, and since gcd(2,5) = gcd(3,5) = 1, we must have:

(p, Q) = (10y, 15y), for x=5y where y is a positive integer.

Now, twice the curent age of Son B = 20y years, which he will attain in 10y years in the future, whence the age of Son B will be 15y + 10y = 25y years.

Thus, the cureent age of father B = S = 2*(25y) = 50y years.

Accordingly, from (i), we must have:
R = 2(50y - 15y) = 70y

Therefore, (P, Q, R, S) = (10y, 15y, 70y, 50y)

If y>=2, then it follows that R >= 140 and (R-P)>=120. It is highly unusual that the age (in years) of a human being should be 140, let  alone exceed that. Moreover, it is inconceivable that any (human) father will be more than 120 years older than his son. Thus, y>=2 is a contradiction.

y=0 would give the age of each of the four individuals as 0 years. This is a contradiction, since the age of any father cannot be 0.

Therefore, y=1, so that:

(P, Q, R, S) = (10, 15, 70, 50)

Consequently, the respective ages of Son A, Son B, Father A and Father B are 10 years, 15 years, 70 years and 50 years.

Note:

Usually for  problems dealing with ages of individuals inclusive of one or more families, the said individuals are deemed human beings unless stated otherwise.  

Posted by K Sengupta on 2008-05-14 15:57:51