2011 AIME II Problems/Problem 1

Revision as of 09:35, 23 August 2011 by Azjps (talk | contribs) (Problem 1)

Problem

Gary purchased a large beverage, but only drank $m/n$ of it, where $m$ and $n$ are relatively prime positive integers. If he had purchased half as much and drunk twice as much, he would have wasted only $2/9$ as much beverage. Find $m+n$.

Solution

Let $x$ be the fraction consumed, then $(1-x)$ is the fraction wasted. We have $1/2 - 2x = 2/9(1-x)$, or $9 - 36x = 4 - 4x$, or $32x = 5$ or $x = 5/32$. Therefore, $m + n = 5 + 32 = \boxed{037.}$

See also

2011 AIME II (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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All AIME Problems and Solutions