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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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New post Updated on: 11 Sep 2018, 20:41
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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


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Originally posted by janet1511 on 12 Mar 2009, 08:04.
Last edited by Bunuel on 11 Sep 2018, 20:41, edited 3 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 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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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 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 21 Apr 2012, 11:49
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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 30 Jun 2015, 04:58
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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?


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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New post 05 Nov 2017, 04:46
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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.

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

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New post 09 Nov 2017, 16:47
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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.

Thanks in advance!


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

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New post 13 Mar 2018, 14:22
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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 information in the passage AND the conclusion should sound somewhat "fishy"
All we're told is that Meadowbrook's and Parkdale's crime rate have increased 60% and 10% respectively in the past 4 years. HOWEVER, we know nothing about the crime rate statistics from 4 years ago.

Consider this possible scenario.
4 years ago: Meadowbrook's crime rate was 10 violent crimes per 1,000 residents
4 years ago: Parkdale's crime rate was 900 violent crimes per 1,000 residents

So, given the increases in crime, we can conclude that:
Present: Meadowbrook's crime rate is 16 violent crimes per 1,000 residents (going from 10 to 16 represents a 60% increase)
Present: Parkdale's crime rate was 990 violent crimes per 1,000 residents (going from 900 to 990 represents a 10% increase)

Can we conclude that Meadowbrook's residents are more likely to become victims of violent crime than are residents of Parkdale?
Absolutely not!

So, we should be looking for an answer choice that explains why we need to know the crime rates from 4 years ago.

Only answer choice D does this.

Answer: D

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

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New post 20 Mar 2018, 23:11
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


This is a trap question regarding percentages and it forces us to believe to take absolute numbers for percentages.
Also we given violent crime per 1000 .
Imo D

If we know the crime rates in the two cities before four years we can reach some conclusion.If they were more in one country and less in some country they only we can reach conclusion.
A This is not required and it is irrelevant as we already have rates that have increased.
B Again this not required as we are already given rates.
C Not required as we just concerned with violent crimes .
D correct
E 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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New post 11 Sep 2018, 20:30
Ron's explanation:

* Say the number of accidents in Chicago is higher than the number of accidents in San Francisco. Does that mean you're more likely to get in a crash in Chicago? (Not necessarily; it could just mean that Chicago has more people on the road.)
* Say the accident rate -- i.e., number of accidents per driver or per mile driven -- is higher for Chicago than for San Francisco. Does that mean you're more likely to get in a crash in Chicago? Yes (unless there are other factors -- weather, etc. -- that the study failed to take into account.)

The entire purpose of per-capita statistics, such as crime "rates", is to reflect the likelihood of an event. So, there you go.
Sue's salary is 60% higher than it was four years ago. Tom's is only 10% higher. Therefore, Sue is more likely than Tom to be doing well financially.
--> In this case, I'm betting it's pretty obvious why we would need to know what their salaries actually were four years ago (i.e., choice D).
As I'm sure you'll also agree, this reasoning doesn't depend on quibbling over exactly how "doing well financially" is defined. (So, at the end of the day, the question in the original post doesn't matter much.)
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Re: The violent crime rate (number of violent crimes per 1,000 residents) &nbs [#permalink] 11 Sep 2018, 20:30
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