Solutions

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Question

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


 

Option A:

changes in the population density of both Parkdale and Meadowbrook over the past four years

Option B:

how the rate of population growth in Meadowbrook over the past four years compares to the corresponding rate for Parkdale

Option C:

the ratio of violent to nonviolent crimes committed during the past four years in Meadowbrook and Parkdale

Option D:

the violent crime rates in Meadowbrook and Parkdale four years ago

Option E:

how Meadowbrook’s expenditures for crime prevention over the past four years compare to Parkdale’s expenditures.

Difficulty Level

Medium

Solution

Option D is the correct answer.


Option Analysis

Question type: Flaw 

Summary of the argument: Meadowbrook’s crime rate is increasing at a higher rate than is for Parkdale. Hence residents of Meadowbrook are more likely to become criminals that residents of Parkdale are.  

 

A) Population density is the number of people living per unit area. This doesn’t affect the crime rate nor does it explain the increase or the projection.  

B) Population growth doesn’t correspond to the increase in violent crime rate.  

C) The ration doesn’t attribute to violent crime rates.  

D) Correct Answer 

E) We are not worried about the expenditure.


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