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

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== 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 multiplying, <math>\displaystyle 1000n+3000>1006n+2012</math>, and <math>\displaystyle n<\frac{988}{6}</math>. Thus, the answer is <math>\displaystyle 164</math>.
  
 
== See also ==
 
== See also ==
 
* [[1992 AIME Problems]]
 
* [[1992 AIME Problems]]

Revision as of 19:21, 26 July 2006

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 $\displaystyle.500$. During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater than $\displaystyle .503$. What's the largest number of matches she could've won before the weekend began?

Solution

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

See also

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