[求助]LSAT-1-4-19

ymxl4611

麻烦斑竹帮我看一下这个问题:

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.（D）

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

我知道D为什么是答案,可是我想请问为什么C可以满足呢?谢谢!

chelseayang

The key term here is "consistent", which means "compatible, being in agreement with". My experience told me that for this type of question, "being not in conflict with" is "consistent".

As to this question, "At most one person" in C is not in conflict with the "at least one person". For instance, it could be exactly "one person", which make both "at most one" and "at least one" valid.

My two cents. I hope it answers your question

BTW, welcom you to the newly opened LSAT board

ymxl4611

Thanks!

我想我明白了.

Bensontuo

ymxl4611 发表于 2004-11-25 20:02
麻烦斑竹帮我看一下这个问题:19.   There are at least three people in the room. At most two  ...

Spot the question type: Must be wrong

Core of the argument:

At least 3 p in the room

at most 2 know each other

at least 1 knows everyone in the room

Let see the answers.

A. Could be true

B. 2 p ---> no recognize each other, could be true, since if we have 4 people, then it must be true that it could be 2 out of 4 does not recognize each other

C. At least one means it could be either as 1 or more than 1, so, it could be true

D. It is not possible, at most 2 people recognize each other, but we do at least have 3 people in the room. so it must be wrong.

E. It could be true.