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

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

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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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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 residents) [#permalink]

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30 Jun 2015, 04:58
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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!

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

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30 Jun 2017, 10:17
vishwaprakash wrote:
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.

Can anyone elaborate on this?
Taking option A into consideration - if population density of M is 10,000 and that of P is 20,000, then according to the premise the crime rate in M is 10 [per 1000] and in P is 20 [per 1000]. Hence crime rate in P>M. But if I reverse the numbers i.e. M = 20,000 and P = 10,000, the crime rate in M > P.

So how come considering population density is out? Can anyone please explain?

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Last edited by Sash143 on 01 Jul 2017, 01:47, edited 1 time in total.

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

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30 Jun 2017, 11:18
Sash143 wrote:
vishwaprakash wrote:
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.

Can anyone elaborate on this?
Taking option A into consideration - if population density of M is 10,000 and that of P is 20,000, then according to the premise the crime rate in M is 10 [per 1000] and in N is 20 [per 1000]. Hence crime rate in P>M. But if I reverse the numbers i.e. M = 20,000 and P = 10,000, the crime rate in M > P.

So how come considering population density is out? Can anyone please explain?

Hi Sash143,

if population density of M is 10,000 and that of P is 20,000, then according to the premise the crime rate in M is 10 [per 1000] and in N is 20 [per 1000]

Crime rate=No. of crimes reported for every 1000 people; Numerator is no. of crimes, in the example you cited, you have taken the population density as the numerator. So what you are doing is not a comparison of crime rates.

For crime rate now after the increase; we need to know the crime rate 4 years back. Author fails to consider the start value of the crime rate and concludes with that flaw in mind.
Say Crime rate in M 4 years back was 10 in 1000 residents; now it becomes 16 in 1000 residents
Say Crime rate in P 4 years back was 100 in 1000 residents; now it becomes 110 in 1000 residents; can we still say Meadowbrook residents are more likely to become victims? No. Thats the flaw in the argument.
Hope that helps
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Re: The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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01 Jul 2017, 02:15
Quote:
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.

Quote:
Can anyone elaborate on this?
Taking option A into consideration - if population density of M is 10,000 and that of P is 20,000, then according to the premise the crime rate in M is 10 [per 1000] and in N is 20 [per 1000]. Hence crime rate in P>M. But if I reverse the numbers i.e. M = 20,000 and P = 10,000, the crime rate in M > P.

So how come considering population density is out? Can anyone please explain?

Quote:
Hi Sash143,

if population density of M is 10,000 and that of P is 20,000, then according to the premise the crime rate in M is 10 [per 1000] and in N is 20 [per 1000]

Crime rate=No. of crimes reported for every 1000 people; Numerator is no. of crimes, in the example you cited, you have taken the population density as the numerator. So what you are doing is not a comparison of crime rates.

For crime rate now after the increase; we need to know the crime rate 4 years back. Author fails to consider the start value of the crime rate and concludes with that flaw in mind.
Say Crime rate in M 4 years back was 10 in 1000 residents; now it becomes 16 in 1000 residents
Say Crime rate in P 4 years back was 100 in 1000 residents; now it becomes 110 in 1000 residents; can we still say Meadowbrook residents are more likely to become victims? No. Thats the flaw in the argument.
Hope that helps

Hi origen87,

Thanks for that explanation. I agree that looking for the crime rate for M and P 4 years ago is apt and precise reasoning. Could you please cite a numerical example for the above text you have mentioned [green highlight], debunking option A ?

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

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03 Jul 2017, 09:54
Hi Sash143

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

The way I interpreted A - changes in population density - certain areas in P which were much dense 4 years ago are not dense now. Same with M. But certain other areas which were not dense 4 years ago can now be dense. Hence, A does not confirm a change in population and can be skipped.

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

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04 Jul 2017, 11:10
It is such an easy question only if we read carefully .
Here percentages is mentioned .
Rate of increase is 60 % for Meadowbrook
For Parkdale it is 10 %
But we do know the absolute number for both of the cities .
As Meadowbrook can have 100 violent crimes before 4 years and increased to 160 because of 60% increase
But Parkdale can have 1000 violent crimes before and increased to about 1100 after 4 years
Hence D is the answer .
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Re: The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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05 Jul 2017, 21:32
saikarthikreddy wrote:
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.

True..and SO I selected B.considering B gives clearer picture.

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

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12 Jul 2017, 19:10
Violent crime rate in Mtown increased more than Pdale by x%

Whenever we are given a % increase or decrease statistic, i look for the original value. If not, then move on to sample space. Here the population growth is a nice trap answer but answer D is stronger because if the total number was questionable then the argument falls apart even before we consider the population
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Re: The violent crime rate (number of violent crimes per 1,000 residents)   [#permalink] 12 Jul 2017, 19:10

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