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? | ||
− | == Solution == | + | == 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>. | ||
+ | |||
+ | == Solution 2 (same as solution 1) == | ||
+ | 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}} |
Latest revision as of 08:05, 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 . 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 1
Let be the number of matches won, so that
, and
.
Cross multiplying, , so
. Thus, the answer is
.
Solution 2 (same as solution 1)
Let be the number of matches she won before the weekend began. Since her win ratio started at exactly .
she must have played exactly
games total before the weekend began. After the weekend, she would have won
games out of
total. Therefore, her win ratio would be
This means that
Cross-multiplying, we get
which is equivalent to
Since
must be an integer, the largest possible value for
is
1992 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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