### Brain Teasers

# Fathers and Sons

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 multiples of five, and they are all under 100.

How old are all four people?

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 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 = 10/3*Sa

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.

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