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?

Here's the solution I meant to post previously.

Let x = age of Son B ,

then 1.5 x = age of Son A .

Now, Son B will be double his present age x years from now, at which time Son A will be 2.5 x years old.

Hence, the current age of Father B = 2(2.5 x) = 5x

It follows then that the age of Father A is :

2( 5x - 1.5 x) = 7x

In summary, the current ages are:

Son B = x

Son A = 1.5 x

Father B = 5x

Father A = 7x

Since all of the ages are multiples of five , then x must be a multiple of 10 ( i.e., x = 10, 20, 30, ...) . But, unless these fathers and sons lived during the time of Abraham and Noah, the only realistic solution occurs when x = 10.

Therefore, the ages are as follows:

SB = 10

SA = 15

FB = 50

FA = 70

Posted by mathemagician on 2003-07-05 01:34:59