Difference between revisions of "1978 AHSME Problems/Problem 30"
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− | ==Solution== | + | == Problem 30 == |
+ | |||
+ | In a tennis tournament, <math>n</math> women and <math>2n</math> 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 <math>7/5</math>, then <math>n</math> equals | ||
+ | |||
+ | <math>\textbf{(A) }2\qquad | ||
+ | \textbf{(B) }4\qquad | ||
+ | \textbf{(C) }6\qquad | ||
+ | \textbf{(D) }7\qquad | ||
+ | \textbf{(E) }\text{none of these} </math> | ||
+ | |||
+ | ==Solution 1== | ||
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. | ||
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</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 22:37, 27 August 2022
Contents
Problem 30
In a tennis tournament, women and
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
, then
equals
Solution 1
Since there are women, the number of matches between only women is
and similarly, there are
matches between only men. Since every woman plays every man exactly once, there are
matches which are between a man and a woman. Call these
matches co-ed matches, and let
be the number of co-ed matches won by women.
Then it follows that
which can be simplified to
The number of matches won by women must be less than the total number of matches, so we obtain the inequality
Rearranging and factoring gives
and the only integers which satisfy this inequality are
and
Clearly, there could not have been people in the tournament, so
If
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 back into the equation
and solving for
gives
Since
must be an integer,
cannot be
It follows that
so the answer is (E). (When
solving gives
)
Solution 2
Since there are women and
men, there are a total of
players. Hence, the number of total matches must be
. We also know that the ratio of the number of matches won by women to the number of matches won by men is
and that there were no draws, so the total number of matches must be
for some value
. This gives
So
It follows that
(various ways of splitting
into two factors). However, for solutions
,
isn't an integer, therefore leaving solutions
and
.
is invalid. This can be shown as when
,
The ratio of matches won by gender is
, so the number of matches won by women must be at least
. 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
. As the maximum number of matches won is smaller than
, the solution is extraneous/invalid.
Hence, the answer must be , or
~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 |
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