GMAT Prep question

[anksm22]

OA D

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Hi anksm22,

Since this is a CR prompt, you should have posted it in THAT Sub-Forum. I'll still be happy to walk you through it though.

This is a FLAW prompt. You typically see 1 of these on Test Day and it's always written around some type of flaw in the logic.

One of the common Flaws that the GMAT typically tests is the concept of Percents vs. Values. For example, is a 10% pay raise at your job necessarily a lot more money? In that same way, would a $20,000 raise be considered a big raise? In both situations, the real question is "what did you start with?" Without that information, there's no way to know if a percent or value increase/decrease is significant or not.

This prompt is all about percentage data. A 60% increase in violent crime might seem like a big increase, but it's not necessarily a big increase if we're basing it off of a really small number of violent crimes. The correct answer matches the issue above: what were the violent crime numbers that we started with? The author does NOT consider that point, which is why his argument is flawed.

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

[GMATGuruNY]

The violent crime rate (number of violent crimes per 1000 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 is 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

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.