2006 AMC 8 第 20 题

先试着解答 2006 AMC 8 第 20 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2006 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

20.

一场单打锦标赛有六名选手。每名选手与其他每名选手只比赛一次,没有平局。如果 Helen 赢了 44 场,Ines 赢了 33 场,Janet 赢了 22 场,Kendra 赢了 22 场,Lara 赢了 22 场,那么 Monica 赢了几场?
A singles tournament had six players. Each player played every other player only once, with no ties. If Helen won 44 games, Ines won 33 games, Janet won 22 games, Kendra won 22 games and Lara won 22 games, how many games did Monica win?
00
11
22
33
44
答案:C
知识点:组合
小提示:
数出六人循环赛共有多少场比赛。
Count how many games are played in a six-player round robin.
大提示:
每场比赛恰好有一个胜者,所以总胜场数等于总比赛数。
Every game has exactly one winner, so total wins equal total games.
解答:
每场比赛恰好有一个胜者。六名选手共有 6⋅5÷2=156 \cdot 5 \div 2 = 15 场比赛,因此共有 1515 个胜场。
已经计入的胜场数为 4+3+2+2+2=13 4 + 3 + 2 + 2 + 2 = 13 场,所以 Monica 赢了 15−13=215 - 13 = 2 场。
所以正确答案是 C。
In every match, there was exactly one winner. There are 6⋅5÷2=156 \cdot 5 \div 2 = 15 games and therefore 1515 wins.
There are already 4+3+2+2+2=13 4 + 3 + 2 + 2 + 2 = 13 wins accounted for, so Monica won 15−13=215 - 13 = 2 games.
Thus, C is the correct answer.
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