Father A is twice the difference in years of the ages of Father B and Son A, who is one and a half times the age of Son B.
Father B is currently twice the age Son A is going to be when Son B will be double the age he is now.
All of the ages are the multiples of five, and they are all under 100. How old are all four people?
Answer
Son A = 15 years old
Son B = 10 years old
Father B = 50 years old
Father A = 70 years old
You know that Son A is 1.5 times older than Son B. Therefore : Sa = 1.5*Sb, or
Sb = 2/3 * Sa (call this equation 1)
You also know that Father B is twice the age Son A will be when Son B’s age doubles. You can set up the equation:
Fb = 2(Sa + Sb) (call this equation 2) Plugging equation (1) into equation (2) and reducing, you get : Fb = 10/3Sa
Because of the fraction, Fa must be evenly divisible by 3. The only multiple of 5 that satisfies this and keeps Father B’s age below 100 is 15, which is Son A’s age. Once this is known, the other ages are pretty straightforward to figure out.