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

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== Problem 1 ==
 
== 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''.
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Gary purchased a large beverage, but only drank <math>m/n</math> of it, where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. If he had purchased half as much and drunk twice as much, he would have wasted only <math>2/9</math> as much beverage. Find <math>m+n</math>.
  
 
== Solution ==
 
== 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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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 = \boxed{037.}</math>
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==See also==
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{{AIME box|year=2011|n=II|before=First Problem|num-a=2}}
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[[Category:Intermediate Geometry Problems]]

Revision as of 21:34, 22 August 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, $m + n = 5 + 32 = \boxed{037.}$

See also

2011 AIME II (ProblemsAnswer KeyResources)
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