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

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Re: The violent crime rate (number of violent crimes per 1,000 [#permalink] New post 09 Aug 2012, 18:16
I went with answer choice A. but the OA is D.
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Re: The violent crime rate (number of violent crimes per 1,000 [#permalink] New post 09 Feb 2014, 18:27
x2suresh wrote:
e.g
ratio of crimes M: P = 100: 200

M -->1000 --> 100 (four years ago) --> 160 (now : 60% more)
P -->1000 --> 200 (four years ago) --> 220 (now : 10% more)

Who is more likely to become victims : P..

C is the best answer.


Lots of confidence, but perhaps the wrong answer?

I think D is the right answer? I like the logic though.
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Re: The violent crime rate (number of violent crimes per 1,000 [#permalink] New post 26 Mar 2015, 16:07
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Re: The violent crime rate (number of violent crimes per 1,000 [#permalink] New post 02 Apr 2015, 07:58
Hi folks

I initially thought of D as well. But isnt it stated in the argument that "The corresponding increase for Parkdale is only 10 percent."

What does corresponding mean here.... proportion?

Hence selected A. Population density i.e. the Denominator , if that increases or decreases , the answer fluctuates accordingly

Can someone pls clarify

Thanks
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Re: The violent crime rate (number of violent crimes per 1,000 [#permalink] New post 05 Apr 2015, 20:12
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the answer is D because what we are trying to do is find a reason why the conclusion may be flawed. The conclusion is:

These figures support the conclusion that residents of Meadowbrook are more likely to become victims of violent crime than are residents of Parkdale.

Even thought Meadowbrook has increased at a rate 6 times that of Parkdale over the past four years, what we don't know is their current rates.

For example, let's say four years ago that Meadowbrook had a rate of 100, and Parkdale had a rate of 1000.

Meadowbrook is now at 160, while Parkdale is now at 1100. Clearly the conclusion is now invalid.

Ans is D
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Re: The violent crime rate (number of violent crimes per 1,000 [#permalink] New post 25 Jun 2015, 20:04
janet1511 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



The problem with B is that we have already been provided crime RATE (number of violent crimes per 1,000 residents), so any increase in population will not impact the possibility of being a victim.
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Re: The violent crime rate (number of violent crimes per 1,000 [#permalink] New post 30 Jun 2015, 03:58
Quote:
I can see why D was selected, but can someone explain why A is incorrect?

If the population density if each city is different, would that not have an effect on the figures that are given in the stimulus?


Population density is irrelevant.Read the opening line --> violent crime rate (number of violent crimes per 1,000 residents) - it takes into account population density.

Here is an example for you -

Tom's salary is 80% higher than it was four years ago. Harry's is only 40% higher. Therefore, Tom is more likely than Harry to be doing well financially or rich. --> what is the flaw in this statement? Think and you will understand why the OA is correct!
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Re: The violent crime rate (number of violent crimes per 1,000 [#permalink] New post 01 Jul 2015, 06:19
restore wrote:
I can see why D was selected, but can someone explain why A is incorrect?

If the population density if each city is different, would that not have an effect on the figures that are given in the stimulus?


Hi restore ,

Question says:
The violent crime rate (number of violent crimes per 1,000 residents) that is Crime rate / 1000 population not Crime rate/ Entire population.

While comparing Crime rate / 1000 population both the denominator are same so over population density won't impact.

Hope it clears few of your doubt.
Re: The violent crime rate (number of violent crimes per 1,000   [#permalink] 01 Jul 2015, 06:19

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