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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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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

Source : GMATPrep Default Exam Pack

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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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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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: CR THE VOILENT CRIME RATE  [#permalink]

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24 Nov 2016, 05:44
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It is better to tabulate data given for such questions -

2012 2016
Crime rate in Meadowbrook x 1.6x
Crime rate in Parkdale y 1.1y

the conclusion states that -
1.6 x > 1.1 y OR (x/y) > (1.1/1.6) ;
We cannot assume this. We need the values of x and y to determine whether this is true. Which answer states this ? D.

A - population density does not matter. No matter what the population density is the violent crime rate remains the same.

B - does not matter. population growth has no effect on violent crime rate. Violent crime rate would still be (No. of violent crimes/ 1000 people )

C - we are concerned only with violent crimes.

D - Correct answer. Gives information about the values of x and y.

E - Not relevant. We are only concerned with violent crime rates, not expenditures.
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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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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.

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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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: CR THE VOILENT CRIME RATE  [#permalink]

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13 Oct 2016, 11:37
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$$Population \ density = \frac{Population}{Land \ Area}$$

Option (A) talks about increase in population Density , not increase in Population.

Now, Population Density can Increase only in the following Circumstances -

1. Increase in Numerator = Increase in Population ( Considering there is no change in Land Area)
2. Decrease in Denominator = Decrease in Land Areas ( Considering there is no change in Land Area)

Certainly this is a far fetched conclusion and the exact reason for increase in Population Density can not be found, nor can we comment on the victims of violent crime.

Whereas option (D) states -

The argument doesn't consider the violent crime rates 4 years ago...

Might be 4 years ago the situation was such that the percentage of Violent Crime to total Crimes was higher in Parkdale tha in Meadowbrook

Further the data compares data in Percentage terms not the actual number of Victims of Violent Crimes..

Errors in the options -

Quote:
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

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

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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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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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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.

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

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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.

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

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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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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: CR THE VOILENT CRIME RATE  [#permalink]

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30 Oct 2018, 01:51
Bunuel,

Exact Similar Question is available here with more explanations
https://gmatclub.com/forum/the-violent- ... 76503.html

Can this be merged to the other one

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

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30 Oct 2018, 02:03
Probus wrote:
Bunuel,

Exact Similar Question is available here with more explanations
https://gmatclub.com/forum/the-violent- ... 76503.html

Can this be merged to the other one

Probus

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

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29 Jun 2019, 11:19
if a statistics-based CR problem isn't immediately intuitive, try writing an analogy for the problem, using statistics and concepts that are easier for you to think about.

E.g., here's an analogy for this problem:
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).

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

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12 Jul 2019, 23:56
Say 4 years b4, violent crime rate in M was x and in P was y
Currently, violent crime rate in M is 1.6x and in P is 1.1y

Now, what if x=10% and y=50%, then in that case currently violent crime rate in M is 16% while in P is 55%

Hence D is the right answer
Re: The violent crime rate (number of violent crimes per 1,000 residents)   [#permalink] 12 Jul 2019, 23:56
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