T-4-Q20 天山-7-22

T-4-Q20
            天山-7-22

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

1. Changes in the population density of both Meadowbrook and Parkdale over the past four years.

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

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

4. The violent crime rates in Meadowbrook and Parkdale four years ago

5. How Meadowbrooks’ expenditures for crime prevention over the past four years compare to Parkdale’s expenditures.

答案：D。请教下这道题为什么是D不是A呢？

题干说M地过去4年暴力犯罪率上升60%，而P地只上升了10%。注意是上升了。基数没有说。

打个比方，M原来1%犯罪率，P30%的话，上升了60%，10%之后，分别是1.6%和33%，你说哪个高？？

谢谢楼上的解答，我明白D了。但是仍然对A存在疑问。A也解释的通啊。比如M地区和P地区人口原来都是1000人，原来犯罪率是10%，那么原来两地的犯罪人数都是100.后来，假设M地区人口密度下降了，只有625人，P地区的人口变成了909人，这样即使犯罪人数依然都还是原来的100人，M地区的犯罪率变成了16%，P地区的犯罪率变成了11%。符合原文的条件，但是否定了结论即M地区的居民更容易成为罪犯。这样理解看上去也没啥问题啊。请多多指教！

照你的说法，岂不是支持题干的结论了吗？题目要求你找FLAW，即反对结论的选项。

另外，本题都是围绕犯罪率的比较，不涉及人口基数。即便是变化后的犯罪率，也还是由犯罪分子数量比上变化后的人口基数算出来的。犯罪率是衡量居民是否更容易成为罪犯的唯一标尺。

很佩服你的推导，我比较懒，看了A直接PASS。不过我认为你想多了。