2006 AMC 8 Problems/Problem 20

Problem

A singles tournament had six players. Each player played every other player only once, with no ties. If Helen won 4 games, Ines won 3 games, Janet won 2 games, Kendra won 2 games and Lara won 2 games, how many games did Monica win?

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution

Since there are 6 players, a total of $\frac{6(6-1)}{2}=15$ games are played. So far, $4+3+2+2+2=13$ games finished (one person won from each game), so Monica needs to win $15-13 = \boxed{\textbf{(C)}\ 2}$ .

2006 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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All AJHSME/AMC 8 Problems and Solutions

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2006 AMC 8 Problems/Problem 20

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