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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]

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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 [#permalink]

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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..

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]

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26 Mar 2015, 17:07
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Re: The violent crime rate (number of violent crimes per 1,000 [#permalink]

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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 [#permalink]

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05 Apr 2015, 21:12
Expert's post
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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]

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25 Jun 2015, 21:04
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
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]

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30 Jun 2015, 04:58
1
KUDOS
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]

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01 Jul 2015, 07:19
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.
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Re: The violent crime rate (number of violent crimes per 1,000 [#permalink]

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23 Apr 2016, 06:50
I also thought the same. and made mistake. it was on my 6th question.

But what i forgot to notice was the the increase motioned here is with base to 4 years old number. so it can very well be that even if the question said in parkdale the rate decreased by 10%. the actual number would be bigger.

nitya34 wrote:
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 [#permalink]

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13 May 2016, 03:45
But the problem with B is that no matter how fast the population growth rate might be,we are calculating it per 1000 residents only.

So,if In city B-lets say 4 years back the violent crime per 1000 people was 500,4 years later it increased by 60%,it would become 800.
However In city P-what if the crime 4 years earlier was higher than city A,lets say 600 violent crime per 1000 people,then 40% rise would be around 840 people.

So this concludes that if we don't know the past crime rate data(crimes/1000 people),it's unlikely which city is going to become the victim of crime.
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Re: The violent crime rate (number of violent crimes per 1,000 [#permalink]

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16 May 2016, 06:33
Question is playing on the higher percent, higher value myth.
60% higher and 10% higher, however what is the baseline figure for these percentages? Only then can we make a valid conclusion regarding the two figures.
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Re: The violent crime rate (number of violent crimes per 1,000 [#permalink]

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08 Sep 2016, 04:44
Premise: Violent crime rate (No. of violent crimes per thousand residents) in meadowbrook is 60% higher now than it was four years ago. Corresponding increase in parkdale is only 10 percent.

To justify the conclusion we need to know the violent crime rate from 4 years ago.

St:1 Changes in population density have got nothing to with the crime rate.
St:2 Rate of population growth will have no bearing.
St:3 Out of scope
St:4. 'Bang on'. If the crime rate 4 years in MB was significantly lower, then 60% wouldn't be a great increase when compared to the rise in crime rates in Parkdale. [For instance, Population of MB=1000 & of PD=5000, crime rate in MB can be assumed as 5%.i.e, 5/1000. 60% increase would make it 8. If crime rate in Parkdale is 20%, then VCR would be 100 and a 10% increase would make it 110.]
St:5 Out of scope.
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Re: The violent crime rate (number of violent crimes per 1,000 [#permalink]

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08 Sep 2016, 04:55
If violent crime rate in M was 10/100 4 years ago and in P 80/100

Then 60% increase in M is 16 and 10% increase is 88 in P . By these numbers M is still safer city than P. So we need the number of violent crime rates 4 years ago to find actual crime rates now.

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

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08 Sep 2016, 04:56
urfrndniks wrote:
If violent crime rate in M was 10/100 4 years ago and in P 80/100

Then 60% increase in M is 16 and 10% increase is 88 in P . By these numbers M is still safer city than P. So we need the number of violent crime rates 4 years ago to find actual crime rates now.

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That's why right answer is D

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Re: The violent crime rate (number of violent crimes per 1,000   [#permalink] 08 Sep 2016, 04:56

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