Difference between revisions of "2006 AMC 8 Problems/Problem 20"
AlcumusGuy (talk | contribs) (Created page with "==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...") |
(→Solution) |
||
| (5 intermediate revisions by 5 users not shown) | |||
| Line 1: | Line 1: | ||
==Problem== | ==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? | 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? | ||
| + | |||
<math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4 </math> | <math> \textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4 </math> | ||
==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>. |
| + | |||
| + | ==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 03:05, 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?
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 
.
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. 
