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1978 AHSME Problems/Problem 30

Revision as of 08:24, 27 August 2022 by Jccc (talk | contribs) (Add alternative solution)

Problem 30

In a tennis tournament, $n$ women and $2n$ men play, and each player plays exactly one match with every other player. If there are no ties and the ratio of the number of matches won by women to the number of matches won by men is $7/5$, then $n$ equals

$\textbf{(A) }2\qquad \textbf{(B) }4\qquad \textbf{(C) }6\qquad \textbf{(D) }7\qquad \textbf{(E) }\text{none of these}$

Solution 1

Since there are $n$ women, the number of matches between only women is $\frac{n(n-1)}{2},$ and similarly, there are $(2n - 1)n$ matches between only men. Since every woman plays every man exactly once, there are $2n\cdot n = 2n^2$ matches which are between a man and a woman. Call these $2n^2$ matches co-ed matches, and let $w$ be the number of co-ed matches won by women.

Then it follows that \[\frac 75 = \frac{\frac{n(n-1)}{2} + w}{(2n - 1)n + 2n^2 - w},\] which can be simplified to \[w = \frac{17}{8}n^2 - \frac38 n.\]

The number of matches won by women must be less than the total number of matches, so we obtain the inequality \[\frac{17}{8}n^2 - \frac38 n\leq 2n^2.\] Rearranging and factoring gives \[0\leq -n(n-3),\] and the only integers which satisfy this inequality are $n = 0,1,2,$ and $3.$

Clearly, there could not have been $0$ people in the tournament, so $n\neq 0.$ If $n = 1,$ then there would have been only one woman and two men in the tournament, in which case the woman could not have won the majority of the matches.

We can now plug $n = 2$ back into the equation \[\frac 75 = \frac{\frac{n(n-1)}{2} + w}{(2n - 1)n + 2n^2 - w},\] and solving for $w$ gives $w = \frac{31}{4}.$ Since $w$ must be an integer, $n$ cannot be $2.$ It follows that $n = 3,$ so the answer is (E). (When $n = 3,$ solving gives $w = 18.$)

Solution 2

Since there are $n$ women and $2n$ men, there are a total of $n+2n=3n$ players. Hence, the number of total matches must be $\binom{3n}{2}=\frac{3n(3n-1)}{2}$. We also know that the ratio of the number of matches won by women to the number of matches won by men is $\frac{7}{5}$ and that there were no draws, so the total number of matches must be $(7+5)x=12x$ for some value $x$. This gives \[\frac{3n(3n-1)}{2}=12x\] So \[n(3n-1)=8x\] It follows that either $n=8$ or $3n-1=8$, giving solutions $n=8$ and $n=3$. Hence, the answer must be $\boxed{\textbf{(E) }\text{none of these}}$.

Note: although unnecessary to show, $n=8$ is invalid. This can be shown as when $n=8$, \[\frac{24(23)}{2}=12x\] \[x=23\] The ratio of matches won by gender is $\frac{7x}{5x}=\frac{161}{115}$, so the number of matches won by women must be at least $161$. The maximum number of matches won by women is equivalent to the number of matches between men (generates 1 match won by men no matter the outcome) subtracted from the total number of matches, which is $\binom{24}{2}-\binom{16}{2}=156$. As the maximum number of matches won is smaller than $161$, the solution is extraneous/invalid.

~JcCC

See also

1978 AHSME (Problems • Answer Key • Resources)
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