2011 AIME II Problems/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 $\frac{1}{2} - 2x =\frac{2}{9} (1-x)$ , or $9 - 36x = 4 - 4x$ , or $32x = 5$ or $x = 5/32$ . Therefore, $m + n = 5 + 32 = \boxed{037}$ .

Solution 2

$WLOG, Gary purchased $ n $ liters and consumed $ m $ liters. After this, he purchased $ \frac{n}{2} $ liters, and consumed $ 2m $ liters. He originally wasted $ n-m $ liters originally, but now he wasted $ \frac{n}{2} - 2m $. \[ \frac{n}{2} - 2m = \frac{4}{18} \cdot (n-m) \] \[ 9n - 36m = 4n - 4m \implies 5n = 32m \implies \frac{m}{n} = \frac{5}{32}. \]$

2011 AIME II (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

2011 AIME II Problems/Problem 1

Contents

Problem

Solution

Solution 2

See also