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

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Joined: 24 Oct 2012
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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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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]

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12 Jul 2017, 23:15
When comparing the crime rates in two cities in a present context, the crime rates that occurred 4 yrs ago are to be considered.

only Option D does that and is therefore the answer.
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Re: The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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26 Aug 2017, 13:29
Hello Experts,

Although I marked D and got the answer correct, I was actually stuck between B and D.

Can you explain how did you eliminate option B?
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Re: The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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05 Nov 2017, 04:46
Hello Experts (e-GMAT/GMATNinja),

I agree that D is correct. But can you please explain why B is incorrect by taking an example.

I am confused why option B is wrong.

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

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09 Nov 2017, 16:47
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Expert's post
snjainpune wrote:
Hello Experts (e-GMAT/GMATNinja),

I agree that D is correct. But can you please explain why B is incorrect by taking an example.

I am confused why option B is wrong.

pikolo2510 wrote:
Hello Experts,

Although I marked D and got the answer correct, I was actually stuck between B and D.

Can you explain how did you eliminate option B?

Quote:
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 passage is concerned with the violent crime RATES of the two cities (the number of violent crimes per 1,000 residents). We do not need information about population growth to compare violent crime RATES.

For example, say that the populations of both cities have always grown at the same rate and that both cities start with the same violent crime rate. In that case, a 60% increase in the violent crime rate in Meadowbrook vs a 10% in Parkdale would obviously mean that there are now more TOTAL violent crimes in Meadowbrook.

Now let's say that we start with the same violent crime rates and the same populations and use the same percent changes to the violent crime rates. If Meadowbrook's population has not changed and Parkdale's population has skyrocketed, it is POSSIBLE that there are now more TOTAL violent crimes in Parkdale. But that wouldn't change the fact that Meadowbrook still has more violent crimes PER 1,000 RESIDENTS. Thus, the LIKELIHOOD of becoming victims of violent crime would still be higher in Meadowbrook, even if Parkdale has a higher TOTAL number of violent crimes.

For example, if the odds of winning the lottery in India are one in a million, India will have over 1,000 winners. If the odds of winning the lottery in Aruba are one in a thousand, Aruba will have about 100 winners. There are FEWER winners in Aruba, but people in Aruba have a higher chance of winning.

I hope that helps!
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The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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10 Nov 2017, 00:58
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GMATNinja wrote:
snjainpune wrote:
Hello Experts (e-GMAT/GMATNinja),

I agree that D is correct. But can you please explain why B is incorrect by taking an example.

I am confused why option B is wrong.

pikolo2510 wrote:
Hello Experts,

Although I marked D and got the answer correct, I was actually stuck between B and D.

Can you explain how did you eliminate option B?

Quote:
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 passage is concerned with the violent crime RATES of the two cities (the number of violent crimes per 1,000 residents). We do not need information about population growth to compare violent crime RATES.

For example, say that the populations of both cities have always grown at the same rate and that both cities start with the same violent crime rate. In that case, a 60% increase in the violent crime rate in Meadowbrook vs a 10% in Parkdale would obviously mean that there are now more TOTAL violent crimes in Meadowbrook.

Now let's say that we start with the same violent crime rates and the same populations and use the same percent changes to the violent crime rates. If Meadowbrook's population has not changed and Parkdale's population has skyrocketed, it is POSSIBLE that there are now more TOTAL violent crimes in Parkdale. But that wouldn't change the fact that Meadowbrook still has more violent crimes PER 1,000 RESIDENTS. Thus, the LIKELIHOOD of becoming victims of violent crime would still be higher in Meadowbrook, even if Parkdale has a higher TOTAL number of violent crimes.

For example, if the odds of winning the lottery in India are one in a million, India will have over 1,000 winners. If the odds of winning the lottery in Aruba are one in a thousand, Aruba will have about 100 winners. There are FEWER winners in Aruba, but people in Aruba have a higher chance of winning.

I hope that helps!

Now I got it. Thanks for the great explanation, GMATNinja!!
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Re: The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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03 Dec 2017, 14:53
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 statement tells us that violent crime in Meadowbrook has increased by 60% in the past four years and that violent crime in Parkdale has increased by 10% in that same timeframe. Based on this the author concludes that Meadowbrook residents are more likely to be victims of violent crimes than Parkdale residents.
We are looking for the flaw/assumption being made in the argument. The clear flaw is that we don't have a starting figure in order to compare the 60% increase in Meadowbrook to the 10% increase in Parkdale.

(A) changes in the population density of both Parkdale and Meadowbrook over the past four years
This is out of scope because we are looking at violent crime rate, not comparing the population density

(B) how the rate of population growth in Meadowbrook over the past four years compares to the corresponding rate for Parkdale
We are looking to compare the violent crime rates, not compare the population growth rate.

(C) the ratio of violent to nonviolent crimes committed during the past four years in Meadowbrook and Parkdale
We are specifically looking for the ratio of violent crime in Meadowbrook to the ratio of violent crime in Parkdale, not the ratio of violent to nonviolent crime.

(D) the violent crime rates in Meadowbrook and Parkdale four years ago
This is the correct answer! We're looking for the starting point to be able to compare the stats above.

(E) how Meadowbrook’s expenditures for crime prevention over the past four years compare to Parkdale’s expenditures
Knowing how much each town has spent on crime prevention is out of scope for this question
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Re: The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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23 Dec 2017, 08:31
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?

D is the correct answer. Since the question already takes about the crime rate, so any references to population will not be the answers.
Look at why other answers including A are incorrect.

(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
-----irrelevant.. we are only talking about the violent crimes

(D) the violent crime rates in Meadowbrook and Parkdale four years ago
-----correct. What if the crime rate of Meadowbrook was more than that of Parkdale, 4yrs ago?

(E) how Meadowbrook’s expenditures for crime prevention over the past four years compare to Parkdale’s expenditures
-----irrelevant.. expenditures will not help in any inference related to crime rate.

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Re: The violent crime rate (number of violent crimes per 1,000 residents)   [#permalink] 23 Dec 2017, 08:31

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