LSAT逻辑题分析:第三类人
2011-09-06 16:39:00   来源：网络资源    双击单词自动翻译

　　14-9-9. Some people are Montagues and some people are Capulets. No Montague can be crossed in love All Capulets can be crossed in love .Therefore, Capulets are not Montague Anyone who is not a Montague is intemperate.

　　Assume that all of the statements in the passage are true, If it is also true that no Montague is intemperate, then which one of the following must be true?

　　(A) The only people who can be crossed in live are intemperate Capulets.

　　(B) Anyone who is not a Copulet is a Montagues

　　(C) All intemperate people can be crossed in love.

　　(D) All intemperate people are Capulets

　　(E) All Capulets are intemperate.

　　这种题目我看到就晕，能否再讲讲.答案是e 。

　　原题的信息: 1. ALL Montague are not crossed in love 2. ALL Capulets are crossed in love3. Anyone who is not a Montague is intermerate 4. Montague 和Capulets集合不重叠

　　A提到了crossed in love的信息,而且是crossed in love 的必要条件..原文1,2信息根本无提到.错.

　　(B) 表示二选一..原文没提到.

　　(C) All intemperate people = NOT Montague can be crossed in love. 原文没提到,可能还有第三类人.

　　(D) All intemperate people = NOT Montague are Capulets 错

　　(E) All Capulets are intemperate. 由 3,4可得

　　可以看到BCD的错误其实是一样的...都是ASSUME除了Montague和 Capulets外没有第三类人.

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