Difference between revisions of "2006 AMC 8 Problems/Problem 20"
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==Solution== | ==Solution== | ||
| − | Since there are 6 players, a total of <math> | + | 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>. |
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| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/M4SEFkUCZ94 | ||
| + | |||
| + | ==See Also== | ||
{{AMC8 box|year=2006|n=II|num-b=19|num-a=21}} | {{AMC8 box|year=2006|n=II|num-b=19|num-a=21}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 17:09, 8 November 2024
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?
Solution
Since there are 6 players, a total of 
games are played. So far, 
games finished (one person won from each game), so Monica needs to win 
.
Video Solution by WhyMath
See Also
| 2006 AMC 8 (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
