Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1978 AHSME Problems/Problem 30"

m (→‎Solution)
m (Changes to solution 2)
 
(3 intermediate revisions by 2 users not shown)
Line 10: Line 10:
 
\textbf{(E) }\text{none of these}  </math>
 
\textbf{(E) }\text{none of these}  </math>
  
−
 
+
==Solution 1==
−
==Solution==
 
  
 
Since there are <math>n</math> women, the number of matches between only women is <math>\frac{n(n-1)}{2},</math> and similarly, there are <math>(2n - 1)n</math> matches between only men. Since every woman plays every man exactly once, there are <math>2n\cdot n = 2n^2</math> matches which are between a man and a woman. Call these <math>2n^2</math> matches co-ed matches, and let <math>w</math> be the number of co-ed matches won by women.
 
Since there are <math>n</math> women, the number of matches between only women is <math>\frac{n(n-1)}{2},</math> and similarly, there are <math>(2n - 1)n</math> matches between only men. Since every woman plays every man exactly once, there are <math>2n\cdot n = 2n^2</math> matches which are between a man and a woman. Call these <math>2n^2</math> matches co-ed matches, and let <math>w</math> be the number of co-ed matches won by women.
Line 41: Line 40:
 
</cmath>
 
</cmath>
 
and solving for <math>w</math> gives <math>w = \frac{31}{4}.</math> Since <math>w</math> must be an integer, <math>n</math> cannot be <math>2.</math> It follows that <math>n = 3,</math> so the answer is (E). (When <math>n = 3,</math> solving gives <math>w = 18.</math>)
 
and solving for <math>w</math> gives <math>w = \frac{31}{4}.</math> Since <math>w</math> must be an integer, <math>n</math> cannot be <math>2.</math> It follows that <math>n = 3,</math> so the answer is (E). (When <math>n = 3,</math> solving gives <math>w = 18.</math>)
 +
 +
==Solution 2==
 +
 +
Since there are <math>n</math> women and <math>2n</math> men, there are a total of <math>n+2n=3n</math> players. Hence, the number of total matches must be <math>\binom{3n}{2}=\frac{3n(3n-1)}{2}</math>. We also know that the ratio of the number of matches won by women to the number of matches won by men is <math>\frac{7}{5}</math> and that there were no draws, so the total number of matches must be <math>(7+5)x=12x</math> for some value <math>x</math>. This gives<cmath>\frac{3n(3n-1)}{2}=12x</cmath>So<cmath>n(3n-1)=8x</cmath>It follows that <math>n=8,4,3,2,1</math> (various ways of splitting <math>8x</math> into two factors). However, for solutions <math>n=4,2,1</math>, <math>x</math> isn't an integer, therefore leaving solutions <math>n=8</math> and <math>n=3</math>.
 +
 +
<math>n=8</math> is invalid. This can be shown as when <math>n=8</math>,<cmath>\frac{24(23)}{2}=12x</cmath><cmath>x=23</cmath>The ratio of matches won by gender is <math>\frac{7x}{5x}=\frac{161}{115}</math>, so the number of matches won by women must be at least <math>161</math>. 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 <math>\binom{24}{2}-\binom{16}{2}=156</math>. As the maximum number of matches won is smaller than <math>161</math>, the solution is extraneous/invalid.
 +
 +
Hence, the answer must be <math>n=3</math>, or <math>\boxed{\textbf{(E) }\text{none of these}}</math>
 +
 +
~JcCC
 +
 +
== See also ==
 +
{{AHSME box|year=1978|n=I|num-b=29|after=Last Problem}}
 +
{{MAA Notice}}

Latest revision as of 21:37, 27 August 2022

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 $n=8,4,3,2,1$ (various ways of splitting $8x$ into two factors). However, for solutions $n=4,2,1$, $x$ isn't an integer, therefore leaving solutions $n=8$ and $n=3$.

$n=8$ 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.

Hence, the answer must be $n=3$, or $\boxed{\textbf{(E) }\text{none of these}}$

~JcCC

See also

1978 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1978_AHSME_Problems/Problem_30&oldid=177604"