AMC8 2006

AMC8 2006 · Q20

AMC8 2006 · Q20. It mainly tests Basic counting (rules of product/sum), Casework.

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?

一个单打锦标赛有六名选手。每位选手只与其他选手各对战一次，没有平局。如果海伦赢了 4 场比赛，伊内斯赢了 3 场，珍妮特赢了 2 场，肯德拉赢了 2 场，拉拉赢了 2 场，那么莫妮卡赢了多少场比赛？

(A) 0 0

(B) 1 1

(D) 3 3

(E) 4 4

Answer

Correct choice: (C)

正确答案：（C）

Solution

(C) Each of the six players played 5 games, and each game involved two players. So there were $\frac{6\cdot 5}{2}=15$ games. Helen, Ines, Janet, Kendra and Lara won a total of $4+3+2+2+2=13$ games, so Monica won $15-13=2$ games.

（C）六名选手每人打了5场比赛，每场比赛有两名选手参与。因此总共有 $\frac{6\cdot 5}{2}=15$ 场比赛。Helen、Ines、Janet、Kendra 和 Lara 一共赢了 $4+3+2+2+2=13$ 场，所以 Monica 赢了 $15-13=2$ 场。

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