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Violent Crime Rate

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Violent Crime Rate

by vineetbatra » Wed Oct 21, 2009 7:34 am
I will post the OA later.

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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by pandeyvineet24 » Wed Oct 21, 2009 10:52 am
D for me.
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by vineetbatra » Wed Oct 21, 2009 12:54 pm
Can you please explain why D?
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by Ludacrispat26 » Wed Oct 21, 2009 1:56 pm
IMO 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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by Testluv » Wed Oct 21, 2009 9:42 pm
Ludacrispat26's solution is very good, and this is another very good example of how GMAT likes to test quant skills in the verbal (GMAT also likes to test verbal skills in quant).
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by delhiboy1979 » Thu Oct 22, 2009 12:15 am
Is the OA D? I thought it should be B
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by vineetbatra » Thu Oct 22, 2009 8:26 am
Ludacrispat26 wrote: 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.

Ans is D
That is good explanation and OA is D.
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by uptowngirl92 » Fri Oct 30, 2009 5:46 am
Hey guys..agreed D is the answer..can somebody please give me examples of A,B and C?I was a bit confused.....
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by 2010gmat » Fri Oct 30, 2009 10:13 am
great explanation above....

how to solve such questions faster??

i was stuck between B and D and used numbers to get rid off B....but i know that during the exam under time pressure i would screw up...

I tried to support B because i thought that P might have had a population blast and that resulted in crime rate growth nos lower than those of town M....

but after pluggin in the nos. i realised that the final conclusion will depend on the base nos...
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by ssgmatter » Thu May 20, 2010 7:48 am
although D for sure makes sense here but can any one touch on the option B why it is wrong here?....

Thanks!
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by ThiMBA » Thu May 20, 2010 10:15 am
How I approached it

Meadowbrook 2006

10 crimes/1000 residents

Meadownbrook 2010

16 crimes/1000 residends

Parkdale 2006

10 crimes/1000 residents

Parkdale 2010

11 crimes/1000 residents

In this instance, you can see that the crime rate in Parkdale is clearly lower now in comparison to Meadowbrook which supports the conclusion

Since we do not the crime rate figures of 4 years ago, we can also assume that Parkdale's 2006 figures could've been something like 100/1,000. IN that case, its 2010 crime rate would be 160/1,000 which would indicate the opposite of the conclusion - hence D is the correct answer
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by kevincanspain » Thu May 20, 2010 2:17 pm
vineetbatra wrote:I will post the OA later.

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
B is not the flaw, since the argument uses as evidence the number of violent crimes per 1,000 residents, not the total number of crimes
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