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

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Problem:
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== Problem 1 ==
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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''.
  
Gary purchased a large bevarage, but only drank ''m''/''n'' of it, where ''m'' and ''n'' are relatively prime positive integers. If he had purchased half as much and drank twice as much, he would have wasted only 2/9 as much bevarage. Find ''m''+''n''.
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== Solution ==
 
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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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Solution:
 
 
 
Set up a system of equations, which I will leave to somebody with more free time, and you get that ''m''=5 and ''n''=32. ''m''+''n''=37
 

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 $x$ 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, $m + n = 5 + 32 = \fbox{37.}$

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