[求助]请教一道prep的题目

xingxing23

Although the discount stores in Goreville's central shopping district are expected to close within five years as a result of competition from a SpendLess discount department store that just opened, those locations will not stay vacant for long. In the five years since the opening of Colson's, a nondiscount department store, a new store has opened at the location of every store in the shopping district that closed because it could not compete with Colson's.

Which of the following, if true, most seriously weakends the argument?

A. Many customers of Colson's are expected to do less shopping there than they did before the SpendLess store opened.

B. Increasingly, the stores that have opened in teh central shopping district since Colson's opened have been discount stores.

C. At present, the central shopping district has as many stores operating in it as it ever had.

D. Over the course of the next five years, it is expected that Goreville's population will grow at a faster rate than it has for the past several decades.

E. Many stores in the central shopping district sell types of merchandise that are not available at either SpendLess or Colson's.

糖棠8875

B?

redhead

我觉得的E

因为discount store竞争不过Spendless and nondiscount store竞争不过Colson。但是如果卖的东西跟它们无关的时候就不能构成竞争也就不会关门了

zhu_eagle

个人觉得选A。

关系：由于竞争不过SpendLess, the discount stores关。（这是the discount stores和SpendLess的关系）

前提：a new store会在所有竞争不过Colson's的商店的location开店。（这是a new store和Colson's的关系）

结论：由于a new store开店，the discount store的Location不会空很久。（这是a new store和the discount store的关系）

由前提和结论知，"所有竞争不过Colson's的商店的location”和"the discount store的Location”应该一致，即文中隐含了the discount store竞争不过Colson's(这是the discount store和Colson's的关系)。为什么竞争不过Colson's呢？因为竞争不过Spendless。于是知道，文章隐含竞争不过Spendless，就一定竞争不过Colson's(这是Spendless和Colson's的关系)。只要削弱这个关系即可。

A选项：顾客在Spendless开店后，买Colson's的东西变少了，即Colson's的竞争力下降了。这样虽然the discount store竞争不过Spendless,但不一定竞争不过Colson's。于是a new store不一定会在the discount store的location开店, 那个location可能空很久。削弱了结论。

可能以下这个更清晰。

T=the discount stores, S=SpendLess, C=Colson's, A=a new store

(A--C)&(A--T) --> (C--T), 已知(T--S), 需要(C--S)。于是削弱(C--S)

[此贴子已经被作者于2008-5-9 21:35:26编辑过]