# Difference between revisions of "2011 AIME II Problems/Problem 1"

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− | Problem | + | == Problem 1 == |

+ | 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 <math>x</math> be the fraction consumed, then <math>(1-x)</math> is the fraction wasted. We have <math>1/2 - 2x = 2/9(1-x)</math>, or <math>9 - 36x = 4 - 4x</math>, or <math>32x = 5</math> or <math>x = 5/32</math>. Therefore, <math>m + n = 5 + 32 = \fbox{37.}</math> | |

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## Revision as of 00:20, 2 April 2011

## Problem 1

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 be the fraction consumed, then is the fraction wasted. We have , or , or or . Therefore,