1993 AMC 8 第 23 题

先试着解答 1993 AMC 8 第 23 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 1993 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。

23.

五名赛跑者 PP、QQ、RR、SS、TT 比赛,且 PP 领先 QQ,PP 领先 RR,QQ 领先 SS,并且 TT 在 PP 之后、QQ 之前完成。谁不可能获得第三名?
Five runners, P,P, Q,Q, R,R, S,S, T,T, have a race, and PP beats Q,Q, PP beats R,R, QQ beats S,S, and TT finishes after PP and before Q.Q. Who could not have finished third in the race?
PP 和 QQ
PP and QQ
PP 和 RR
PP and RR
PP 和 SS
PP and SS
PP 和 TT
PP and TT
PP、SS 和 TT
P,P, SS and TT
答案:C
知识点:逻辑推理
小提示:
PP 领先 QQ、RR 和 TT,而 QQ 又领先 SS;据此判断 PP 的名次
PP finishes ahead of Q,Q, R,R, and T,T, while QQ finishes ahead of S;S; determine PP’s place
大提示:
如果某个选手前面必须有好几个人,那么他不可能排到第三
If several runners must all finish ahead of a given runner, that runner cannot be as high as third
解答:
因为 PP 领先 QQ 和 RR,也在 TT 之前,而 QQ 又领先 SS,所以 PP 第一,因此不可能第三。
条件还给出完赛先后链:PP、TT、QQ、SS 依次在前。所以 PP、TT 和 QQ 都在 SS 前,SS 最好也只能第四,不可能第三。
QQ、RR、TT 都可能第三。例如,PP、TT、QQ、RR、SS 使 QQ 第三;PP、RR、TT、QQ、SS 使 TT 第三;PP、TT、RR、QQ、SS 使 RR 第三。因此只有 PP 和 SS 不可能第三。
所以正确答案是 C。
Since PP beats QQ and R,R, finishes ahead of T,T, and QQ beats S,S, runner PP finishes first and so cannot be third.
The clues give the chain PP before TT before QQ before S.S. So P,P, T,T, and QQ all finish ahead of S,S, meaning SS is no better than fourth and cannot be third either.
Each of Q,Q, R,R, TT can finish third: for example P,P, T,T, Q,Q, R,R, SS puts QQ third; P,P, R,R, T,T, Q,Q, SS puts TT third; and P,P, T,T, R,R, Q,Q, SS puts RR third. So only PP and SS cannot be third.
Thus, the correct answer is C .
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