Difference between revisions of "1992 AIME Problems/Problem 3"
(added 1 solution from alcumus) |
m (→Solution 2) |
||
| Line 7: | Line 7: | ||
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 == | + | == 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> | 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> | ||
Latest revision as of 07:05, 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![\[\frac{n+3}{2n+4} > .503 = \frac{503}{1000}.\]](http://latex.artofproblemsolving.com/4/7/4/474959c440b3d3ce9e2450a323a5954abf4970c8.png)
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 | |
| 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. 
