2015 AMC 10A Problem 8

First find their ages today, then find when the ratio becomes $2 : 1$.

1. Let Claire's age two years ago be $c$. Then Pete's age two years ago was $3c$.
2. Two years before that (four years ago), each was $2$ years younger: Claire was $c - 2$ and Pete was $3c - 2$. At that time Pete was four times as old:
$$3c - 2 = 4(c - 2) = 4c - 8.$$
3. Solve: add $8$ to both sides and subtract $3c$ from both sides to get $c = 6$.
4. So two years ago Claire was $6$ and Pete was $18$. Check the older fact: four years ago they were $4$ and $16$, and $16 = 4 \times 4$. Good.
5. Today Claire is $8$ and Pete is $20$.
6. Now count years forward:

In $4$ years Pete is exactly twice Claire's age.

The answer is $\boxed{\textbf{(B)}\ 4}$.

In age problems both people get older at the same rate, so the ratio of their ages keeps changing. Pick the moment that makes the first equation simplest (here, two years ago), write every other moment by adding or subtracting years, and each fact becomes one easy equation.

Use the age gap, which never changes. Two years ago Pete was $3c$ and Claire $c$, so the gap is $2c$. Four years ago Claire was $c - 2$ and Pete four times that, so the gap was $3(c - 2)$. The gaps are equal: $2c = 3c - 6$, so $c = 6$ and the gap is $12$. Pete is twice Claire's age exactly when Claire's age equals the gap, $12$. Claire is $8$ now, so that is $4$ years away.