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Difference between revisions of "1992 AIME Problems/Problem 3"

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== Problem ==
 
== Problem ==
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A tennis player computes her win [[ratio]] by dividing the number of matches she has won by the total number of matches she has played. At the start of a weekend, her win ratio is exactly <math>\displaystyle.500</math>. During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater than <math>\displaystyle .503</math>. What's the largest number of matches she could've won before the weekend began?
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A tennis player computes her win [[ratio]] by dividing the number of matches she has won by the total number of matches she has played. At the start of a weekend, her win ratio is exactly <math>.500</math>. During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater than <math>.503</math>. What's the largest number of matches she could've won before the weekend began?
  
 
== Solution ==
 
== Solution ==
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Let <math>\displaystyle n</math> be the number of matches won, so that <math>\displaystyle \frac{n}{2n}=\frac{1}{2}</math>, and <math>\displaystyle \frac{n+3}{2n+4}>\frac{503}{1000}</math>. [[Cross multiply]]ing, <math>\displaystyle 1000n+3000>1006n+2012</math>, and <math>\displaystyle n<\frac{988}{6}</math>. Thus, the answer is <math>\displaystyle 164</math>.
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Let <math>n</math> be the number of matches won, so that <math>\frac{n}{2n}=\frac{1}{2}</math>, and <math>\frac{n+3}{2n+4}>\frac{503}{1000}</math>. Cross [[multiply]]ing, <math>1000n+3000>1006n+2012</math>, and <math>n<\frac{988}{6}</math>. Thus, the answer is <math>164</math>.
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== See also ==
 
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* [[1992 AIME Problems/Problem 2 | Previous problem]]
 
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* [[1992 AIME Problems/Problem 4 | Next problem]]
 
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* [[1992 AIME Problems]]
 
  
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{{AIME box|year=1992|num-b=1|num-a=3}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 21:56, 11 November 2007

Problem

A tennis player computes her win ratio by dividing the number of matches she has won by the total number of matches she has played. At the start of a weekend, her win ratio is exactly $.500$. During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater than $.503$. What's the largest number of matches she could've won before the weekend began?

Solution

Let $n$ be the number of matches won, so that $\frac{n}{2n}=\frac{1}{2}$, and $\frac{n+3}{2n+4}>\frac{503}{1000}$. Cross multiplying, $1000n+3000>1006n+2012$, and $n<\frac{988}{6}$. Thus, the answer is $164$.

1992 AIME (Problems • Answer Key • Resources)
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Problem 3
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All AIME Problems and Solutions
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