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

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Manager
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Joined: 05 Nov 2014
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Kudos [?]: 13 [0], given: 85

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

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New post 12 Jul 2017, 23:15
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 [?]: 13 [0], given: 85

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

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New post 26 Aug 2017, 13:29
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 [?]: 45 [0], given: 176

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

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New post 05 Nov 2017, 04:46
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.

Thanks in advance!

Kudos [?]: 2 [0], given: 7

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

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New post 09 Nov 2017, 16:47
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.

Thanks in advance!

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 [?]: 1863 [0], given: 443

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

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New post 10 Nov 2017, 00:58
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.

Thanks in advance!

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 [?]: 2 [0], given: 7

The violent crime rate (number of violent crimes per 1,000 residents)   [#permalink] 10 Nov 2017, 00:58

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