Difference between revisions of "2019 AMC 8 Problems/Problem 19"
(Added Problem from pamphlet) |
(→Problem 19) |
||
Line 1: | Line 1: | ||
==Problem 19== | ==Problem 19== | ||
− | In a tournament there are six team | + | 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 16:30, 20 November 2019
Problem 19
In a tournament there are six team that play each other twice. A team earns points for a win, point for a draw, and 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?
Solution 1
See Also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.