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

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== Solution ==
 
== Solution ==
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>\boxed{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>.  
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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>.
  
 
{{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]]
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{{MAA Notice}}

Latest revision as of 21:29, 24 May 2021

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

1992 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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