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

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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?
 
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?
  
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== Solution ==
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== Solution 1 ==
 
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>.  
 
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>, so <math>n<\frac{988}{6}=164 \dfrac {4}{6}=164 \dfrac{2}{3}</math>. Thus, the answer is <math>\boxed{164}</math>.
 
Cross [[multiply]]ing, <math>1000n+3000>1006n+2012</math>, so <math>n<\frac{988}{6}=164 \dfrac {4}{6}=164 \dfrac{2}{3}</math>. Thus, the answer is <math>\boxed{164}</math>.
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== Solution 2 ==
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Let <math>n</math> be the number of matches she won before the weekend began. Since her win ratio started at exactly .<math>500 = \tfrac{1}{2},</math> she must have played exactly <math>2n</math> games total before the weekend began. After the weekend, she would have won <math>n+3</math> games out of <math>2n+4</math> total. Therefore, her win ratio would be <math>(n+3)/(2n+4).</math> This means that<cmath>\frac{n+3}{2n+4} > .503 = \frac{503}{1000}.</cmath>Cross-multiplying, we get <math>1000(n+3) > 503(2n+4),</math> which is equivalent to <math>n < \frac{988}{6} = 164.\overline{6}.</math> Since <math>n</math> must be an integer, the largest possible value for <math>n</math> is <math>\boxed{164}.</math>
  
 
{{AIME box|year=1992|num-b=2|num-a=4}}
 
{{AIME box|year=1992|num-b=2|num-a=4}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 07:04, 2 January 2024

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 1

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, $1000n+3000>1006n+2012$, so $n<\frac{988}{6}=164 \dfrac {4}{6}=164 \dfrac{2}{3}$. Thus, the answer is $\boxed{164}$.

Solution 2

Let $n$ be the number of matches she won before the weekend began. Since her win ratio started at exactly .$500 = \tfrac{1}{2},$ she must have played exactly $2n$ games total before the weekend began. After the weekend, she would have won $n+3$ games out of $2n+4$ total. Therefore, her win ratio would be $(n+3)/(2n+4).$ This means that\[\frac{n+3}{2n+4} > .503 = \frac{503}{1000}.\]Cross-multiplying, we get $1000(n+3) > 503(2n+4),$ which is equivalent to $n < \frac{988}{6} = 164.\overline{6}.$ Since $n$ must be an integer, the largest possible value for $n$ is $\boxed{164}.$

1992 AIME (Problems • Answer Key • Resources)
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Problem 4
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