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GMAT-6-22请教求解思路

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楼主
发表于 2005-11-11 13:53:00 | 只看该作者

GMAT-6-22请教求解思路



























The violent crime rate (number of violent crimes per 1,000 residents) in Meadowbrook is 60 percent higher now than it was four years ago.  The corresponding increase for Parkdale is only 10 percent.  These figures support the conclusion that residents of Meadowbrook are more likely to become victims of violent crime than are residents of Parkdale.

The argument above is flawed because it fails to take into account











changes in the population density of both Parkdale and Meadowbrook over the past four years





how the rate of population growth in Meadowbrook over the past four years compares to the corresponding rate for Parkdale





the ratio of violent to nonviolent crimes committed during the past four years in Meadowbrook and Parkdale





the violent crime rates in Meadowbrook and Parkdale four years ago





how Meadowbrook’s expenditures for crime prevention over the past four years compare to Parkdale’s expenditures
沙发
发表于 2005-11-11 14:27:00 | 只看该作者

C,


做个乘法。

板凳
发表于 2005-11-11 22:05:00 | 只看该作者

选D


只有知道4年前的各自暴力犯罪率,才能知道上升各自的百分比后,今年的犯罪率。


欢迎讨论

地板
发表于 2005-11-11 22:43:00 | 只看该作者

同意C。


问的是哪一个more likely,所以用比率。

5#
 楼主| 发表于 2005-11-12 10:52:00 | 只看该作者

明白了,谢谢。支持Deanjiang的解释,选D.


[此贴子已经被作者于2005-11-12 10:52:57编辑过]
6#
发表于 2005-12-17 20:36:00 | 只看该作者

原来1% 增加60% 现在是 1。6%


原来40%增加10%现在是44%

7#
发表于 2005-12-18 00:53:00 | 只看该作者

D is the best

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