急问JJ中一题答案及其出处，万分感谢

bbmandbbg

The percentage of households with an annual income of more than $40,000 is higher in M city than in any other city. However, the percentage of households with an annual income of  $60,000 or more is highest in S city.

If the above information is true, 下面那个可以推出？

A. No households in M city has an annual income of $60,000 or more.

B. Some households in M city have an annual income between $40,000 and $60,000

C. The number of households with an annual income of more than $40,000 is greater in M than in S city.

D. Average annual household income is higher in S than in M city.

E. The percentage of households with an annual income of $80,000 is higher in S than in M city.

[此贴子已经被作者于2004-9-25 17:29:38编辑过]

beibeikang

怎么没人呀，马上要考了，急呀

lawyer_1

不知处处，但答案为B。如果M没有人的收入在40K和60K之间，则都在60K以上，则S市的60K以上的收入的人数PERCENTAGE最大和M市的40K以上最大构成矛盾。故M市必有人收入在40K和60K之间。易混的是E，但S是可以没有人收入80K以上，都低于80K，而S市可以有人高于80K。故E错。其他易排除

wanze

这题我也选b，答案是e。不解中......

携隐

E明显错误。应该是B。wanze你的答案哪里来的？

wanze

这是管为东13套题中的一题，

lawyer_1

GWD的题答案错误的较多

sweetsweet

我认为E是对的。

题干中的条件the percentage of households with an annual income of  $60,000 or more is highest in S city　已经说明了The percentage of households with an annual income of $80,000 is higher in S than in M city。

B应该是无关选项吧

携隐

来推导一下。设S₁和S₂分别为S市大于$40,000和大于$60,000的，M₁和M₂分别为M市大于$40,000和大于$60,000的，S和M分别为S市和M市所有人口，则根据原文条件有：

S₁/S<M₁/M

M₂/M<S₂/S

以上两式相加为S₁/S+M₂/M<M₁/M+S₂/S

把相同字母的项移一移，就是：(S₁-S₂)/S<(M₁-M₂)/M，也就是说，收入位于4-6之间的人口比例，S市小于M市。

来假设一种极端的情况，就是S市没有收入在4-6之间的人，那么S₁-S₂=0，带入以上不等式，得出：

(M₁-M₂)/M>0。既然大于0，那肯定至少有一些人收入在4-6之间，所以B对啊。

E举反例就可以排除了。

jnlvo

同意楼上的！