2006 AMC 8 Problem 20

Below is the professionally curated solution for Problem 20 of the 2006 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 8 solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:combinations

Difficulty rating: 1260

20.

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?

$0$

$1$

$2$

$3$

$4$

Solution:

In every match, there was exactly one winner. There are $6 \cdot 5 \div 2 = 15$ games and therefore $15$ wins.

There are already $4 + 3 + 2 + 2 + 2 = 13$ wins accounted for, so Monica won $15 - 13 = 2$ games.

Thus, C is the correct answer.

← Problem 19 Full Exam Problem 21 →

Problem 20 in Other Years

1985 AMC 8 · 1986 AMC 8 · 1987 AMC 8 · 1988 AMC 8 · 1989 AMC 8 · 1990 AMC 8 · 1991 AMC 8 · 1992 AMC 8 · 1993 AMC 8 · 1994 AMC 8 · 1995 AMC 8 · 1996 AMC 8 · 1997 AMC 8 · 1998 AMC 8 · 1999 AMC 8 · 2000 AMC 8 · 2001 AMC 8 · 2002 AMC 8 · 2003 AMC 8 · 2004 AMC 8 · 2005 AMC 8 · 2007 AMC 8 · 2008 AMC 8 · 2009 AMC 8 · 2010 AMC 8 · 2011 AMC 8 · 2012 AMC 8 · 2013 AMC 8 · 2014 AMC 8 · 2015 AMC 8 · 2016 AMC 8 · 2017 AMC 8 · 2018 AMC 8 · 2019 AMC 8 · 2020 AMC 8 · 2022 AMC 8 · 2023 AMC 8 · 2024 AMC 8 · 2025 AMC 8 · 2026 AMC 8