ChaseDream
搜索
返回列表 发新帖
查看: 1265|回复: 2
打印 上一主题 下一主题

大全 LSAT 1-2-19

[复制链接]
楼主
发表于 2009-8-23 12:59:00 | 只看该作者

大全 LSAT 1-2-19

19.   There are at least three people in the room. At most two people in the room recognize each other. At least one person in the room recognizes everybody else in the room.

Which one of the following is NOT consistent with the above?

(A) Four people are in the room.

(B) No two people in the room recognize each other.

(C) At most one person in the room recognizes everybody else in the room.

(D) Anyone in the room who recognizes any other person in the room is also recognized by that person.

(E) Two people in the room recognize every one else in the room.

答案是D。这道题完全是clueless啊。。唉 请教了

沙发
发表于 2009-8-24 02:09:00 | 只看该作者

借楼主沙发问个问题:C和E选项是互斥的,为什么不选其中一个呢?

板凳
发表于 2009-8-25 17:32:00 | 只看该作者
A 题目中说最少3人,那么4人不矛盾吧。
B 题目说最多2人彼此认识,那么没有两个人彼此认识也合理吧。
C 题目说最少一个人认识大家,这里说最多一个人,也合理吧。
D 选项说任何一个人如果认得别人,别人也认得他。题目说最少3人,最多2人彼此认识。这不是和题目矛盾了吗?
E 题目说至少一个人认识大家,这里说有两个人认识大家,不矛盾吧。

这种题目,LSAT里面好像有不少。一开始我也不明白。后来清楚了。这种题不是智力题,也不是数学题(不会让你算到底多少人,也算不出来)。只不过让你根据题目,分析选项是否可能。
您需要登录后才可以回帖 登录 | 立即注册

Mark一下! 看一下! 顶楼主! 感谢分享! 快速回复:

手机版|ChaseDream|GMT+8, 2024-11-27 02:29
京公网安备11010202008513号 京ICP证101109号 京ICP备12012021号

ChaseDream 论坛

© 2003-2023 ChaseDream.com. All Rights Reserved.

返回顶部