Answered

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

Answer :

estryzy

Answer:

(D) the violent crime rates in Meadowbrook and Parkdale four years ago

Explanation:

Let us not forget the conclusion of the excerpt: "These figures support the conclusion that residents of Meadowbrook are more likely to become victims of violent crime than are residents of Parkdale. "

In this light, we are trying to look for a reason why there is a flaw in the above conclusion.

Although the first statement tells us that the crime rate in Meadowbrook has increased 6 times that of Parkdale over the last four years, it fails to tell us the current crime rates of these cities. With this, we can judge and come to the conclusion above.

Other Questions