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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]

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01 Sep 2007, 00:27
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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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01 Sep 2007, 00:40
E only.

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01 Sep 2007, 12:30
I'd also choose E. What's the OA?

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01 Sep 2007, 13:15
i'll go for A

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01 Sep 2007, 13:19
Hey Guys,
I have a question. In the arguments it states in parathesis " (number of violent crimes per 1,000 residents)" - Doesn't this refer to Meadowbrook only? Or to Parkdale as well?

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01 Sep 2007, 14:07
I will go with B. Only the rate of population growth is important here. If the population growth is Parkdale was much higher than Meadowbrook, it can explain why the crime rate is low.
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Regards

Subhen

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01 Sep 2007, 14:28
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solidcolor wrote:
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

Definitely D.
Comparing increase in crime rate for two cities cannot be possible without knowing previous rate.

Consider:
If Meadowbrook's previous rate was 1, 60% increase would be 1.6 now.
If Parkdale previous rate was 10, 10% increase would be 11 now.
If you change the previous rate around, you get different results.

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01 Sep 2007, 14:56
I was debating between A and B.
I'll pick B- The changes in the rate of population make more sense than changes in population density.

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01 Sep 2007, 16:55
bkk145 wrote:
Definitely D.
Comparing increase in crime rate for two cities cannot be possible without knowing previous rate.

Consider:
If Meadowbrook's previous rate was 1, 60% increase would be 1.6 now.
If Parkdale previous rate was 10, 10% increase would be 11 now.
If you change the previous rate around, you get different results.

PERFECT!

OA is D.

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Director
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01 Sep 2007, 19:55
Violent crime rate in Meadowbrook has increased quite a bit in comparison to the other town.

So, author saying so the probability that the citizens of Meadowbrook can be victim has also increase. We have to refute this very point.

Probability has not increased if the number of people moved in Meadowbrook is way higher than other town.

B provides us that

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Manager
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01 Sep 2007, 22:56
Asaf, I highly recommend you read this post by bkk145. His answer is perfect.

bkk145 wrote:
Definitely D.
Comparing increase in crime rate for two cities cannot be possible without knowing previous rate.

Consider:
If Meadowbrook's previous rate was 1, 60% increase would be 1.6 now.
If Parkdale previous rate was 10, 10% increase would be 11 now.
If you change the previous rate around, you get different results.

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Director
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02 Sep 2007, 02:23
solidcolor wrote:
Asaf, I highly recommend you read this post by bkk145. His answer is perfect.

bkk145 wrote:
Definitely D.
Comparing increase in crime rate for two cities cannot be possible without knowing previous rate.

Consider:
If Meadowbrook's previous rate was 1, 60% increase would be 1.6 now.
If Parkdale previous rate was 10, 10% increase would be 11 now.
If you change the previous rate around, you get different results.

Got it! Thanks bkk145 and solidcolor!

I think key is what is given in parentheses 'crimes per thousand people'.

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Re: A typical CR   [#permalink] 02 Sep 2007, 02:23
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# The violent crime rate (number of violent crimes per 1,000

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