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 inrease 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
Will post answer soon, I need someone to guide me. Not sure why the answer is what it is. Thanks in advance.
Violent Crime Rate
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Data: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 inrease 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
The violent crime rate in Meadowbrook is 60 percent higher now than it was four years ago.
The corresponding increase for Parkdale is only 10 percent.
Conclusion:
Residents of Meadowbrook are more likely to become victims of violent crime than are residents of Parkdale.
Consider the following cases:
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.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
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