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

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The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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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

Originally posted by janet1511 on 12 Mar 2009, 08:04.
Last edited by hazelnut on 06 Apr 2018, 01:38, edited 2 times in total.
Edited the question.
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Re: The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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New post 12 Mar 2009, 08:23
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Its B
let me explain

A. changes in the population density of both Parkdale and Meadowbrook over the
past four years ---density has no corelation here
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 --ratio does not matter
D. the violent crime rates in Meadowbrook and Parkdale four years ago --OOS
E. how Meadowbrook’s expenditures for crime prevention over the past four years
compare to Parkdale’s expenditures--totally OOS
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Re: The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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x2suresh

do you mean D? C is talking about the ratio of violent to nonviolent crimes


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.
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Re: The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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New post 12 Mar 2009, 08:39
Agreed D :P
I misread it completely
I meant that if there is high base % increase does not represent true picture
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Re: The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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New post 12 Mar 2009, 08:58
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janet1511 wrote:
x2suresh

do you mean D? C is talking about the ratio of violent to nonviolent crimes


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.



Sorry yeap.. I mean D.
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Re: The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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New post 12 Mar 2009, 09:24
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D, since the argument only shows a relative increase and not an absolute increase, e.g. a 10% rise from 100 crimes per 1,000 residents (+10) is more than a 60% rise from 10 crimes per 1,000 residents (+6).
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Re: The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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New post 28 Feb 2010, 18:10
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Agree with D.

4 yrs ago M has violent rate 100. 4 yrs after it is increased by 60%...it becomes 160

4 yrs ago P has violent rate 200. 4 yrs after it is increased by 10%...it becomes 220

This shows that residents of P are more victims than M. Argument fails to consider this. Therefore, IMO d
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Re: The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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New post 01 Mar 2010, 09:25
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First my choice was B. But after reading the reasons for D, I too feel D is the correct option.

What is OA?
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Re: The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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New post 13 Nov 2011, 00:06
I agree with D. The number and statistic in this question quite easy, but the thing we need is to read the right answer express correctly our idea.
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Re: The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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Re: The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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New post 17 Jan 2012, 10:30
Any time the argument talks about percentages, we need to look for an answer that talks about absolute numbers.

Hence, D!
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Re: The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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the best answer is D.But i feel even D is flawed.We need to know the absolute number of crimes in both the cities 4 years ago,the rates are not going to give a clear answer.
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Re: The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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New post 13 Apr 2012, 21:46
the argument fails to look at what the starting point is. and only D addresses this.
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Re: The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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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?
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Re: The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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New post 09 Aug 2012, 19:16
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 residents) [#permalink]

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New post 09 Feb 2014, 19:27
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 residents) [#permalink]

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New post 02 Apr 2015, 08:58
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 residents) [#permalink]

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New post 05 Apr 2015, 21:12
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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 residents) [#permalink]

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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 residents) [#permalink]

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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!
Re: The violent crime rate (number of violent crimes per 1,000 residents)   [#permalink] 30 Jun 2015, 04:58

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