Henry Ernest Dudeney/Puzzles and Curious Problems/42 - Family Ages/Solution

by : $42$

 * Family Ages

Solution
The children are triplets, all aged $6$.

The mother and father are both aged $36$.

Proof
Let $j$, $b$ and $m$ denote the ages of John, Ben and Mary respectively.

Let $F$ and $M$ denote the ages of the father and mother respectively.

We have that:


 * $\size {F - M} = \size {j - b} = \size {b - m}$

It is immediately apparent that this can be solved simply by:
 * $F = M$

and so:
 * $j - b - m$

where:


 * $j^2 = b^2 = m^2 = F = M$

such that:
 * $3 j + 2 F = 90$

The very $3$-ish nature of the final equation tempts us to try $j = 6$, and we see:


 * $3 \times 6 + 2 \times 36 = 90$

Note that we are asked to specify the age of each person.

But the only information we have is the difference between the ages, and not which is older than the other.

So if the children have different ages, we cannot tell which is the eldest.

Hence it is impossible in that case to state which child is of which age.

When they are all the same age, the age of each is obvious.