A father's age today is 4 times his child's age. Twelve years ago he was 10 times as old as the child. How old is the child now?

Step-by-step solution

1. 1

   Let the child's present age be the variable. Then the father's present age comes free from the first sentence:

   $x=\text{child now},\qquad 4x=\text{father now}$

2. 2

   Roll both ages back by 12 years. Time passes at the same rate for both people, so subtract $12$ from each present age:

   $x-12=\text{child then},\qquad 4x-12=\text{father then}$

3. 3

   Impose the second ratio. Twelve years ago the father was $10$ times as old:

   $4x-12=10(x-12)$

4. 4

   Expand and isolate $x$.

   $4x-12=10x-120$

   $120-12=10x-4x$

   $108=6x$

   $x=18$

5. 5

   Verify against both sentences. Now: child $18$, father $4\cdot 18=72$, and $72=4\times 18$ $\checkmark$. Twelve years ago: child $6$, father $60$, and

   $60=10\times 6\quad\checkmark$

   So the child is $18$ and the father is $72$. The larger the multiplier gap, the older the child turns out to be — compare this with the $3 imes/5 imes$ version, where the answer was $20$ with only a $10$-year gap.