Difference between revisions of "2006 AMC 8 Problems/Problem 20"

(Solution)
 
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==Solution==
 
==Solution==
Since there are 6 players, a total of <math>5+4+3+2+1=15</math> games are played. So far, <math>4+3+2+2+2=13</math> games finished (one person won from each game), so Monica needs to win <math>15-13 = \boxed{\textbf{(C)}\ 2}</math>.
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Since there are 6 players, a total of <math>\frac{6(6-1)}{2}=15</math> games are played. So far, <math>4+3+2+2+2=13</math> games finished (one person won from each game), so Monica needs to win <math>15-13 = \boxed{\textbf{(C)}\ 2}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 04:05, 12 November 2016

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 $15-13 = \boxed{\textbf{(C)}\ 2}$.

See Also

2006 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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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