2015 AMC 10A Problems/Problem 8

Revision as of 15:58, 4 February 2015 by Ariel86 (talk | contribs) (Problem)


Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be 2 : 1 ?

$\textbf{(A)}\ 2 \qquad\textbf{(B)} \ 4 \qquad\textbf{(C)} \ 5 \qquad\textbf{(D)} \ 6 \qquad\textbf{(E)} \ 8$


This problem can be converted to a system of equations. Let $p$ be Pete's current age and $c$ be Claire's current age.

The first statement can be written as $p-2=3(c-2)$. The second statement can be written as $p-4=4(c-4)$

To solve the system of equations:






Let $x$ be the number of years until Pete is twice as old as Claire.




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

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