Difference between revisions of "2015 AMC 12A Problems/Problem 13"
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\qquad\textbf{(B)}\ \text{There must be an even number of even scores.}\\ | \qquad\textbf{(B)}\ \text{There must be an even number of even scores.}\\ | ||
\qquad\textbf{(C)}\ \text{There cannot be two scores of }0\text{.}\\ | \qquad\textbf{(C)}\ \text{There cannot be two scores of }0\text{.}\\ | ||
| − | \qquad\textbf{(D) | + | \qquad\textbf{(D)}\ \text{The sum of the scores must be at least }100\text{.}\\ |
\qquad\textbf{(E)}\ \text{The highest score must be at least }12\text{.}</math> | \qquad\textbf{(E)}\ \text{The highest score must be at least }12\text{.}</math> | ||
==Solution== | ==Solution== | ||
| − | We can eliminate answer choices <math>\textbf{(A)}</math> and <math>\textbf{(B)}</math> because there are an even number of scores, so if one is false, the other must be false too. | + | We can eliminate answer choices <math>\textbf{(A)}</math> and <math>\textbf{(B)}</math> because there are an even number of scores, so if one is false, the other must be false too. Answer choice <math>\textbf{(C)}</math> must be true since every team plays every other team, so it is impossible for two teams to lose every game. Answer choice <math>\textbf{(D)}</math> must be true since each game gives out a total of two points, and there are <math>\binom{12}{2} = 66</math> games, for a total of <math>132</math> points. Answer choice <math>\boxed{\textbf{(E)}}</math> is false. If everyone draws each of their 11 games, then every team will tie for first place with 11 points each.<li> |
| − | Answer choice <math>\textbf{(C)}</math> must be true since every team plays every other team, so it is impossible for two teams to lose every game. | + | Quick question, why does the answer key say 3? |
| − | Answer choice <math>\textbf{(D)}</math> must be true since each game gives out a total of two points, and there are <math>\ | ||
| − | Answer choice <math>\textbf{(E)}</math> is false | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2015|ab=A|num-b=12|num-a=14}} | {{AMC12 box|year=2015|ab=A|num-b=12|num-a=14}} | ||
Revision as of 23:43, 3 October 2021
Problem
A league with 12 teams holds a round-robin tournament, with each team playing every other team exactly once. Games either end with one team victorious or else end in a draw. A team scores 2 points for every game it wins and 1 point for every game it draws. Which of the following is NOT a true statement about the list of 12 scores?
Solution
We can eliminate answer choices 
and 
because there are an even number of scores, so if one is false, the other must be false too. Answer choice 
must be true since every team plays every other team, so it is impossible for two teams to lose every game. Answer choice 
must be true since each game gives out a total of two points, and there are 
games, for a total of 
points. Answer choice 
is false. If everyone draws each of their 11 games, then every team will tie for first place with 11 points each.
See Also
| 2015 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 12 |
Followed by Problem 14 |
| 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 | |
| All AMC 12 Problems and Solutions | |
