LSAT－10－2－23

23.(A)      A poor farmer was fond of telling his children: “In this world, you are either rich or poor, and you are either honest or dishonest. All poor farmers are honest. Therefore, all rich farmers are dishonest.”

The farmer’s conclusion is properly drawn if the argument assumes that

(A) every honest farmer is poor

(B) every honest person is a farmer

(C) everyone who is dishonest is a rich farmer

(D) everyone who is poor is honest（A）

(E) every poor person is a farmer

像这种题目我一见就害怕，不知道考场上大家是怎么应付的，要在草纸上推出过程吗，那不是很费时间吗？

BTW, anyone helps me to find the reasoning line. I am too dizzle to figure it out ,after spending all a day in reviewing those wrong questions. Thanks advance!

my approach to this questio is: look for the contrapositive of the conclusion

Conclusion is: rich----dishonest

Contrapositive: honest----poor

I am not sure if such method works for all questions of similar type.

I see ,thanks!!

斑竹的方法很好！

这种题我通常是这么考虑的（自认为也不错）：结论是“all rich farmers are dishonest”，为了使它成立，只需要排除它不成立的情况即可，它不成立的情况只有一种（如果情况多了我这也就不叫什么方法了）即“一部分rich farmers are honest”，A 表明honest的人都是poor，正好排除这种使它不成立的情况，所以是答案。反复想几遍，以后碰到包你5秒钟之内搞定。

其实这种题还有一种更简单的解决办法，我叫它 “ABCD”法。应用到本题A－poor，B-rich,C－honest，D－dishonest。AB之间、CD之间不用管，考察的是A与CD，B与CD之间的关系，要想让B与D结婚，只要表明A与C结婚了且没有婚外情就可以了。题干中All poor farmers are honest就是A只爱C一个，答案A则表明C也只爱A一个，那么B只有和D结婚了，蛮符合国情的！这种题你就找C也只爱A一个的答案就行了，三秒钟搞定！