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. answer my choice:C
First of all, this is more of a logic game time of question in LSAT. You really do not need to pay attention to it as GMAT does not use this type of question.
D is not consistent. If there are at least three people in the room, say three: x,y,z. x recognizes y and z. With choice D, y and z should recognize x too. Then it violate "At most two people in the room recognize each other".
I think it is more important to figure out using a simple by special case. For example 3 people. use 代入法. At beginning, I do not know how to handle it. But with special case, it is easy. It is like GMAT algebra problem.
京公网安备11010202008513号 京ICP证101109号 京ICP备12012021号
发表于 2003-9-2 18:28:00
my choice:C
楼主
