2015 AMC 10A Problem 8

Let Pete's and Claire's ages two years ago be $p$ and $c$, so $p = 3c$. Two years before that their ages were $p - 2$ and $c - 2$, with
$$p - 2 = 4(c - 2) = 4c - 8.$$
Substituting $p = 3c$: $3c - 2 = 4c - 8$, so $c = 6$ and $p = 18$. Today Claire is $8$ and Pete is $20$.

In $x$ years the ratio is $2 : 1$ when
$$20 + x = 2(8 + x) = 16 + 2x \quad\Longrightarrow\quad x = 4.$$
Check: in $4$ years Pete is $24$ and Claire is $12$.

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

In age problems the difference of two ages never changes, only the ratio does. Two snapshots of the ratio pin down the ages at those times; then the future condition is one more linear equation. Anchoring the variables at "two years ago" keeps the arithmetic smallest.

Use the constant age gap. From "three times" two years ago, the gap is $2c$; from "four times" four years ago, the gap is $3(c - 2)$. Equal gaps give $2c = 3c - 6$, so $c = 6$ and the gap is $12$. The ratio is $2 : 1$ exactly when Claire's age equals the gap, $12$, which is $4$ years from now (she is $8$ today).

• Using "two years ago" for both conditions instead of shifting the second condition four years back, or stopping after finding the current ages.
• Answering $6$, the number of years from the first snapshot (two years ago) until the ratio becomes $2 : 1$, rather than from today.