Question

彼得、艾玛和凯勒互相下棋。彼得赢了 $4$ 局，输了 $2$ 局。艾玛赢了 $3$ 局，输了 $3$ 局。如果凯勒输了 $3$ 局，他赢了多少局？

Peter, Emma, and Kyler played chess with each other. Peter won $4$ games and lost $2$ games. Emma won $3$ games and lost $3$ games. If Kyler lost $3$ games, how many games did he win?

$0$

$1$

$2$

$3$

$4$

答案：B

知识点：逻辑推理

难度评级：1020

小提示：

每一场胜利都对应另一个人的一场失败

Every recorded win contributes one recorded loss to another player.

大提示：

比较三人总胜场数和总负场数

Compare total wins and total losses across all three players.

视频讲解：

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文字解答：

把三个人的记录合在一起，每一场胜利都对应一场失败；和棋若有，则对胜负总数都没有贡献。总负场数为 $2+3+3=8$ ，所以凯勒赢了 $8-4-3=1$ 局。

所以正确答案是 B。

Across all players, every win is matched by one loss; drawn games, if any, contribute neither. The number of losses is $2+3+3=8,$ so the number of games Kyler won is $8-4-3=1.$

Thus, B is the correct answer.

2017 AMC 8 第 13 题