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

(Problem 19)
(Solution 1)
Line 6: Line 6:
 
==Solution 1==
 
==Solution 1==
  
 +
After fully understanding the problem, we immediately know that the three top teams, say team A, team B, and team C, must beat the other three teams D,E,F. Therefore, A,B,C must each obtain <math>(3+3+3)=9</math> points. However, they play against each team twice, for a total of <math>18</math> points against D,E and,F.
  
 
==See Also==
 
==See Also==

Revision as of 20:01, 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

After fully understanding the problem, we immediately know that the three top teams, say team A, team B, and team C, must beat the other three teams D,E,F. Therefore, A,B,C must each obtain $(3+3+3)=9$ points. However, they play against each team twice, for a total of $18$ points against D,E and,F.

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

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

Invalid username
Login to AoPS