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>, | + | 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]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:29, 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 
. During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater than 
. What's the largest number of matches she could've won before the weekend began?
Solution
Let 
be the number of matches won, so that 
, and 
.
Cross multiplying, 
, so 
. Thus, the answer is 
.
| 1992 AIME (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
