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# The violent crime rate (number of violent crimes per 1,000 residents)

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Joined: 12 Sep 2015
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Re: The violent crime rate (number of violent crimes per 1,000 residents)  [#permalink]

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13 Mar 2018, 14:22
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janet1511 wrote:
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

The information in the passage AND the conclusion should sound somewhat "fishy"
All we're told is that Meadowbrook's and Parkdale's crime rate have increased 60% and 10% respectively in the past 4 years. HOWEVER, we know nothing about the crime rate statistics from 4 years ago.

Consider this possible scenario.
4 years ago: Meadowbrook's crime rate was 10 violent crimes per 1,000 residents
4 years ago: Parkdale's crime rate was 900 violent crimes per 1,000 residents

So, given the increases in crime, we can conclude that:
Present: Meadowbrook's crime rate is 16 violent crimes per 1,000 residents (going from 10 to 16 represents a 60% increase)
Present: Parkdale's crime rate was 990 violent crimes per 1,000 residents (going from 900 to 990 represents a 10% increase)

Can we conclude that Meadowbrook's residents are more likely to become victims of violent crime than are residents of Parkdale?
Absolutely not!

So, we should be looking for an answer choice that explains why we need to know the crime rates from 4 years ago.

Only answer choice D does this.

Cheers,
Brent
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Re: The violent crime rate (number of violent crimes per 1,000 residents)  [#permalink]

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20 Mar 2018, 23:11
janet1511 wrote:
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

This is a trap question regarding percentages and it forces us to believe to take absolute numbers for percentages.
Also we given violent crime per 1000 .
Imo D

If we know the crime rates in the two cities before four years we can reach some conclusion.If they were more in one country and less in some country they only we can reach conclusion.
A This is not required and it is irrelevant as we already have rates that have increased.
B Again this not required as we are already given rates.
C Not required as we just concerned with violent crimes .
D correct
E Out of scope .
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Re: The violent crime rate (number of violent crimes per 1,000 residents)  [#permalink]

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22 Mar 2018, 08:28
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.

what if 4 years ago
Maedowbrook crime rate was 100 -> after 60% increase its 160. diff- 60
Parkdale crime rate was 1000-> after 10% increase its 1100 diff -100

Hence knowing the crime rate prior to 4 years is crucial. ANs is D
Re: The violent crime rate (number of violent crimes per 1,000 residents) &nbs [#permalink] 22 Mar 2018, 08:28

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