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The violent crime rate (number of violent crimes per 1,000

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The violent crime rate (number of violent crimes per 1,000 [#permalink] New post 16 Dec 2005, 00:10
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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

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

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

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

D. the violent crime rates in Meadowbrook and Parkdale four years ago

E. how Meadowbrook’s expenditures for crime prevention over the past four years compare to Parkdale’s expenditures
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 [#permalink] New post 16 Dec 2005, 00:33
A is my answer
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 [#permalink] New post 16 Dec 2005, 05:13
Clear D for me.
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 [#permalink] New post 16 Dec 2005, 05:17
i too take A
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 [#permalink] New post 16 Dec 2005, 07:55
D) !

A) the density doesnt matter in this case b/c the calc of the rate is fixed on 1000 residents.
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 [#permalink] New post 16 Dec 2005, 08:05
its D
unless u know what the crime rate is, how can u find out the number of people affected
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 [#permalink] New post 16 Dec 2005, 08:16
i agree , my first pick was D but got confused......
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 [#permalink] New post 16 Dec 2005, 14:09
it should be A.
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 [#permalink] New post 16 Dec 2005, 20:12
D

10 to 16 vs 60 to 66
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 [#permalink] New post 19 Dec 2005, 20:20
(D) is the best answer.
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  [#permalink] 19 Dec 2005, 20:20
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