1992 AIME Problems/Problem 3

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 1

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 $\boxed{164}$ .

Solution 2 (same as solution 1)

Let $n$ be the number of matches she won before the weekend began. Since her win ratio started at exactly . $500 = \tfrac{1}{2},$ she must have played exactly $2n$ games total before the weekend began. After the weekend, she would have won $n+3$ games out of $2n+4$ total. Therefore, her win ratio would be $(n+3)/(2n+4).$ This means that $\frac{n+3}{2n+4} > .503 = \frac{503}{1000}.$ Cross-multiplying, we get $1000(n+3) > 503(2n+4),$ which is equivalent to $n < \frac{988}{6} = 164.\overline{6}.$ Since $n$ must be an integer, the largest possible value for $n$ is $\boxed{164}.$

1992 AIME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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1992 AIME Problems/Problem 3

Problem

Solution 1

Solution 2 (same as solution 1)