Difference between revisions of "2019 AMC 8 Problems/Problem 19"

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==Problem 19==
 
==Problem 19==
In a tournament there are six team taht play each other twice. A team earns <math>3</math> points for a win, <math>1</math> point for a draw, and <math>0</math> points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?
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In a tournament there are six team that play each other twice. A team earns <math>3</math> points for a win, <math>1</math> point for a draw, and <math>0</math> points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?
  
 
<math>\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30</math>
 
<math>\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30</math>

Revision as of 17:30, 20 November 2019

Problem 19

In a tournament there are six team that play each other twice. A team earns $3$ points for a win, $1$ point for a draw, and $0$ points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?

$\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30$

Solution 1

See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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 AJHSME/AMC 8 Problems and Solutions

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