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