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The violent crime rate (number of violent crimes per 1,000
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01 Sep 2007, 01:27
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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
Archived Topic
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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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Re: The violent crime rate (number of violent crimes per 1,000
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01 Sep 2007, 15:28
1
Kudos
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.
Re: The violent crime rate (number of violent crimes per 1,000
[#permalink]
01 Sep 2007, 14:19
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?
Re: The violent crime rate (number of violent crimes per 1,000
[#permalink]
01 Sep 2007, 15:07
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.
Re: The violent crime rate (number of violent crimes per 1,000
[#permalink]
01 Sep 2007, 17:55
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.
Re: The violent crime rate (number of violent crimes per 1,000
[#permalink]
01 Sep 2007, 23:56
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.
Re: The violent crime rate (number of violent crimes per 1,000
[#permalink]
02 Sep 2007, 03:23
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'.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
gmatclubot
Re: The violent crime rate (number of violent crimes per 1,000 [#permalink]