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

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Manager
Joined: 05 Nov 2014
Posts: 112

Kudos [?]: 14 [0], given: 87

Location: India
Concentration: Strategy, Operations
GMAT 1: 580 Q49 V21
GPA: 3.75
Re: The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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12 Jul 2017, 22:15
When comparing the crime rates in two cities in a present context, the crime rates that occurred 4 yrs ago are to be considered.

only Option D does that and is therefore the answer.

Kudos [?]: 14 [0], given: 87

Manager
Joined: 05 Jul 2017
Posts: 183

Kudos [?]: 47 [0], given: 179

Location: India
Concentration: Entrepreneurship, Technology
GMAT 1: 700 Q49 V36
GPA: 4
Re: The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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26 Aug 2017, 12:29
Hello Experts,

Although I marked D and got the answer correct, I was actually stuck between B and D.

Can you explain how did you eliminate option B?

Kudos [?]: 47 [0], given: 179

Intern
Joined: 09 Nov 2016
Posts: 41

Kudos [?]: 3 [0], given: 7

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

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05 Nov 2017, 03:46
Hello Experts (e-GMAT/GMATNinja),

I agree that D is correct. But can you please explain why B is incorrect by taking an example.

I am confused why option B is wrong.

Kudos [?]: 3 [0], given: 7

GMAT Club Verbal Expert
Status: GMAT and GRE tutor
Joined: 13 Aug 2009
Posts: 1223

Kudos [?]: 2032 [0], given: 462

Location: United States
GMAT 1: 780 Q51 V46
GMAT 2: 800 Q51 V51
GRE 1: 340 Q170 V170
Re: The violent crime rate (number of violent crimes per 1,000 residents) [#permalink]

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09 Nov 2017, 15:47
snjainpune wrote:
Hello Experts (e-GMAT/GMATNinja),

I agree that D is correct. But can you please explain why B is incorrect by taking an example.

I am confused why option B is wrong.

pikolo2510 wrote:
Hello Experts,

Although I marked D and got the answer correct, I was actually stuck between B and D.

Can you explain how did you eliminate option B?

Quote:
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 passage is concerned with the violent crime RATES of the two cities (the number of violent crimes per 1,000 residents). We do not need information about population growth to compare violent crime RATES.

For example, say that the populations of both cities have always grown at the same rate and that both cities start with the same violent crime rate. In that case, a 60% increase in the violent crime rate in Meadowbrook vs a 10% in Parkdale would obviously mean that there are now more TOTAL violent crimes in Meadowbrook.

Now let's say that we start with the same violent crime rates and the same populations and use the same percent changes to the violent crime rates. If Meadowbrook's population has not changed and Parkdale's population has skyrocketed, it is POSSIBLE that there are now more TOTAL violent crimes in Parkdale. But that wouldn't change the fact that Meadowbrook still has more violent crimes PER 1,000 RESIDENTS. Thus, the LIKELIHOOD of becoming victims of violent crime would still be higher in Meadowbrook, even if Parkdale has a higher TOTAL number of violent crimes.

For example, if the odds of winning the lottery in India are one in a million, India will have over 1,000 winners. If the odds of winning the lottery in Aruba are one in a thousand, Aruba will have about 100 winners. There are FEWER winners in Aruba, but people in Aruba have a higher chance of winning.

I hope that helps!
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Kudos [?]: 2032 [0], given: 462

Intern
Joined: 09 Nov 2016
Posts: 41

Kudos [?]: 3 [1], given: 7

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

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09 Nov 2017, 23:58
1
KUDOS
GMATNinja wrote:
snjainpune wrote:
Hello Experts (e-GMAT/GMATNinja),

I agree that D is correct. But can you please explain why B is incorrect by taking an example.

I am confused why option B is wrong.

pikolo2510 wrote:
Hello Experts,

Although I marked D and got the answer correct, I was actually stuck between B and D.

Can you explain how did you eliminate option B?

Quote:
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 passage is concerned with the violent crime RATES of the two cities (the number of violent crimes per 1,000 residents). We do not need information about population growth to compare violent crime RATES.

For example, say that the populations of both cities have always grown at the same rate and that both cities start with the same violent crime rate. In that case, a 60% increase in the violent crime rate in Meadowbrook vs a 10% in Parkdale would obviously mean that there are now more TOTAL violent crimes in Meadowbrook.

Now let's say that we start with the same violent crime rates and the same populations and use the same percent changes to the violent crime rates. If Meadowbrook's population has not changed and Parkdale's population has skyrocketed, it is POSSIBLE that there are now more TOTAL violent crimes in Parkdale. But that wouldn't change the fact that Meadowbrook still has more violent crimes PER 1,000 RESIDENTS. Thus, the LIKELIHOOD of becoming victims of violent crime would still be higher in Meadowbrook, even if Parkdale has a higher TOTAL number of violent crimes.

For example, if the odds of winning the lottery in India are one in a million, India will have over 1,000 winners. If the odds of winning the lottery in Aruba are one in a thousand, Aruba will have about 100 winners. There are FEWER winners in Aruba, but people in Aruba have a higher chance of winning.

I hope that helps!

Now I got it. Thanks for the great explanation, GMATNinja!!

Kudos [?]: 3 [1], given: 7

Intern
Joined: 18 Jan 2017
Posts: 37

Kudos [?]: 7 [0], given: 158

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

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03 Dec 2017, 13:53
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 statement tells us that violent crime in Meadowbrook has increased by 60% in the past four years and that violent crime in Parkdale has increased by 10% in that same timeframe. Based on this the author concludes that Meadowbrook residents are more likely to be victims of violent crimes than Parkdale residents.
We are looking for the flaw/assumption being made in the argument. The clear flaw is that we don't have a starting figure in order to compare the 60% increase in Meadowbrook to the 10% increase in Parkdale.

(A) changes in the population density of both Parkdale and Meadowbrook over the past four years
This is out of scope because we are looking at violent crime rate, not comparing the population density

(B) how the rate of population growth in Meadowbrook over the past four years compares to the corresponding rate for Parkdale
We are looking to compare the violent crime rates, not compare the population growth rate.

(C) the ratio of violent to nonviolent crimes committed during the past four years in Meadowbrook and Parkdale
We are specifically looking for the ratio of violent crime in Meadowbrook to the ratio of violent crime in Parkdale, not the ratio of violent to nonviolent crime.

(D) the violent crime rates in Meadowbrook and Parkdale four years ago
This is the correct answer! We're looking for the starting point to be able to compare the stats above.

(E) how Meadowbrook’s expenditures for crime prevention over the past four years compare to Parkdale’s expenditures
Knowing how much each town has spent on crime prevention is out of scope for this question

Kudos [?]: 7 [0], given: 158

Re: The violent crime rate (number of violent crimes per 1,000 residents)   [#permalink] 03 Dec 2017, 13:53

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