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Crime rate

by VIGNESHWARR » Wed Jun 26, 2013 3:06 am
The violent crime rate(number of violent crimes per 1000 residents) in Meadowbrook is 60 perct 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 arguement above is flawed because it fails to take into account

1.changes in the population density of boyh Parkdale and Meadowbrook over the past four years.
2.how the rate of the population density of both Parkdale over the past four years compares to the corresponding rate for Parkdale.
3.the ratio of nonviolent crimes commited during the past four years compares in Meadowbrook and Parkdale.
4.the violent crime rate in Meadowbrook and Parkdale four ago
5.how Meadowbrook's expenditure for crime prevention over the past four years compare to Parkdale's expenditure
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by Ugo2602 » Thu Jun 27, 2013 1:12 am
IMO: A

What is the OA please?
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by VIGNESHWARR » Thu Jun 27, 2013 9:22 am
answer in GMAT Prep is 4
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by GMATGuruNY » Thu Jun 27, 2013 5:21 pm
VIGNESHWARR wrote:The violent crime rate(number of violent crimes per 1000 residents) in Meadowbrook is 60 perct 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 arguement above is flawed because it fails to take into account

1.changes in the population density of boyh Parkdale and Meadowbrook over the past four years.
2.how the rate of the population density of both Parkdale over the past four years compares to the corresponding rate for Parkdale.
3.the ratio of nonviolent crimes commited during the past four years compares in Meadowbrook and Parkdale.
4.the violent crime rate in Meadowbrook and Parkdale four ago
5.how Meadowbrook's expenditure for crime prevention over the past four years compare to Parkdale's expenditure
Case 1:
Meadowbrook's violent crime rate 4 years ago = 1000, so its violent crime rate now = 1600.
Parkdale's violent crime rate 4 years ago = 10, so its violent crime rate now = 11.
Here, Meadowbrook's current rate is far HIGHER than Parkdale's, SUPPORTING the conclusion that residents of Meadowbrook are more likely to become victims of violent crime than are residents of Parkdale.

Case 2:
Meadowbrook's violent crime rate 4 years ago = 10, so its violent crime rate now = 16.
Parkdale's violent crime rate 4 years ago = 1000, so its violent crime rate now = 1100.
Here, Meadowbrook's current rate is far LOWER than Parkdale's, WEAKENING the conclusion that residents of Meadowbrook are more likely to become victims of violent crime than are residents of Parkdale.

As the cases above illustrate, since we don't know the rates in Meadowbrook and Parkdale 4 years ago, we cannot determine which city has the higher rate now.
Answer choice D describes this flaw in the reasoning:
The argument above is flawed because it fails to take into account the violent crime rate in Meadowbrook and Parkdale four years ago.

The correct answer is D.

Answer choice A: The argument above is flawed because it fails to take into account changes in the population density of both Parkdale and Meadowbrook over the past four years.
Who cares?
If Meadowbrook has a higher violent crime rate -- if the ratio of VIOLENT CRIMES TO RESIDENTS in Meadowbrook is higher -- then the likelihood that a Meadowbrook resident will fall victim to a violent crime is higher, whether there are 1000 people in Meadowbrook or 1,000,000.
Eliminate A.
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by vishugogo » Fri Jun 28, 2013 5:34 am
Dear Mitch,

Kindly explain in detail as why A is incorrect.
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by Gaurav 2013-fall » Wed Jul 03, 2013 1:09 am
+1 for D
Let me tell you something you already know. The world ain't all sunshine and rainbows. It is a very mean and nasty place and it will beat you to your knees and keep you there permanently if you let it. You, me, or nobody is gonna hit as hard as life. But it ain't how hard you hit; it's about how hard you can get hit, and keep moving forward. How much you can take, and keep moving forward. That's how winning is done. Now, if you know what you're worth, then go out and get what you're worth. But you gotta be willing to take the hit, and not pointing fingers saying you ain't where you are because of him, or her, or anybody. Cowards do that and that ain't you. You're better than that! (Rocky VI)
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by ramitagrawal » Wed Jul 03, 2013 3:57 am
vishugogo wrote:Dear Mitch,

Kindly explain in detail as why A is incorrect.
Hi,

Option A states: 1.changes in the population density of boyh Parkdale and Meadowbrook over the past four years.

population density = number of people/area.

Even if population density changes either due to change in number of people or due to change in area, it doesn't change the data in question. The argument is talking about:

number of crimes/1000 people. So even if the population changes it doesn't change this value.

Hope it helps.
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