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

Population density is irrelevant.Read the opening line --> violent crime rate (number of violent crimes per 1,000 residents) - it takes into account population density.

Here is an example for you -

Tom's salary is 80% higher than it was four years ago. Harry's is only 40% higher. Therefore, Tom is more likely than Harry to be doing well financially or rich. --> what is the flaw in this statement? Think and you will understand why the OA is correct!

D, since the argument only shows a relative increase and not an absolute increase, e.g. a 10% rise from 100 crimes per 1,000 residents (+10) is more than a 60% rise from 10 crimes per 1,000 residents (+6).

I can see why D was selected, but can someone explain why A is incorrect?

If the population density if each city is different, would that not have an effect on the figures that are given in the stimulus?

\(Population \ density = \frac{Population}{Land \ Area}\)

Option (A) talks about increase in population Density , not increase in Population.

Now, Population Density can Increase only in the following Circumstances -

1. Increase in Numerator = Increase in Population ( Considering there is no change in Land Area)
2. Decrease in Denominator = Decrease in Land Areas ( Considering there is no change in Land Area)

Certainly this is a far fetched conclusion and the exact reason for increase in Population Density can not be found, nor can we comment on the victims of violent crime.


Whereas option (D) states -

The argument doesn't consider the violent crime rates 4 years ago...

Might be 4 years ago the situation was such that the percentage of Violent Crime to total Crimes was higher in Parkdale tha in Meadowbrook

Further the data compares data in Percentage terms not the actual number of Victims of Violent Crimes..

Errors in the options -



Hope this helps..

It is better to tabulate data given for such questions -

2012 2016
Crime rate in Meadowbrook x 1.6x
Crime rate in Parkdale y 1.1y

the conclusion states that -
1.6 x > 1.1 y OR (x/y) > (1.1/1.6) ;
We cannot assume this. We need the values of x and y to determine whether this is true. Which answer states this ? D.

A - population density does not matter. No matter what the population density is the violent crime rate remains the same.

B - does not matter. population growth has no effect on violent crime rate. Violent crime rate would still be (No. of violent crimes/ 1000 people )

C - we are concerned only with violent crimes.

D - Correct answer. Gives information about the values of x and y.

E - Not relevant. We are only concerned with violent crime rates, not expenditures.

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!

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!

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.

Answer: D

Cheers,
Brent

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 .

Ron's explanation:

* Say the number of accidents in Chicago is higher than the number of accidents in San Francisco. Does that mean you're more likely to get in a crash in Chicago? (Not necessarily; it could just mean that Chicago has more people on the road.)
* Say the accident rate -- i.e., number of accidents per driver or per mile driven -- is higher for Chicago than for San Francisco. Does that mean you're more likely to get in a crash in Chicago? Yes (unless there are other factors -- weather, etc. -- that the study failed to take into account.)

The entire purpose of per-capita statistics, such as crime "rates", is to reflect the likelihood of an event. So, there you go.
Sue's salary is 60% higher than it was four years ago. Tom's is only 10% higher. Therefore, Sue is more likely than Tom to be doing well financially.
--> In this case, I'm betting it's pretty obvious why we would need to know what their salaries actually were four years ago (i.e., choice D).
As I'm sure you'll also agree, this reasoning doesn't depend on quibbling over exactly how "doing well financially" is defined. (So, at the end of the day, the question in the original post doesn't matter much.)

Bunuel,

Exact Similar Question is available here with more explanations
https://gmatclub.com/forum/the-violent- ... 76503.html

Can this be merged to the other one

Probus

__________________
Merged. Thank you.

if a statistics-based CR problem isn't immediately intuitive, try writing an analogy for the problem, using statistics and concepts that are easier for you to think about.

E.g., here's an analogy for this problem:
Sue's salary is 60% higher than it was four years ago. Tom's is only 10% higher. Therefore, Sue is more likely than Tom to be doing well financially.
--> In this case, I'm betting it's pretty obvious why we would need to know what their salaries actually were four years ago (i.e., choice D).

Posted from my mobile device

Say 4 years b4, violent crime rate in M was x and in P was y
Currently, violent crime rate in M is 1.6x and in P is 1.1y

Now, what if x=10% and y=50%, then in that case currently violent crime rate in M is 16% while in P is 55%

Hence D is the right answer

nehasomani33 - This one, also?

Dear Friends,

Here is the detailed explanation to this question-




Mind-map: Meadowbrook’s violent crime rate % increase > Parkdale’s → Meadowbrook residents’ chances of facing violent crime > Parkdale’s

Missing link: Between greater % increase in Meadowbrook’s crime rate than in Parkdale’s and conclusion that Meadowbrook’s citizens are more likely to face violent crime.

Expectation from the correct answer choice: To undermine conclusion that Meadowbrook’s greater percentage violent crime rate increase means its citizens are more likely to face said crimes.

Choice A: This answer choice only provides for different change rates of the cities’ respective population densities, which would not affect the likelihood of citizens to face violent crime; population density measures the number of people located in a specific region of the city relative to other regions, while the likelihood of citizens to face violent crime takes into account the ratio of violent crimes to the population of the city as a whole; this answer choice does not address the link between the greater percentage increase in Meadowbrook’s violent crime rate and the corresponding greater likelihood of its citizens to face violent crime; therefore, it is an incorrect answer choice.
Choice B: This answer choice provides for differential changes in the cities’ respective population growth rates, but this would have no effect on the violent crime rate, as violent crime rates are calculated per 1000 citizens, and the population growth rate only measures changes in the absolute number of citizens; as it does not undermine the conclusion that Meadowbrook’s greater percentage violent crime rate increase means its citizens are more likely to face said crimes, it is an incorrect answer choice.
Choice C: This answer choice only provides for the relative difference between violent and non-violent crime occurrences in the given cities, thereby only allowing us insight into the likelihood of citizens to face violent crime as opposed to non-violent crime within either city; it has no bearing on the comparative likelihood of citizens of Meadowbrook and Parkdale to face violent crime; as it thereby fails to undermine the conclusion that Meadowbrook’s greater percentage violent crime rate increase means its citizens are more likely to face said crimes, it is an incorrect answer choice.
Choice D: This answer choice would allow one to look at older absolute values for violent crimes and contextualize the current percentage increases in absolute numerical terms as well; for example, if the respective violent crime rates in Meadowbrook and Parkdale four years ago stood at 200 and 2000 respectively, current violent crime rates at the respective percentage increases of 60% and 10% would be 320 and 2200, contradicting the argument in the passage; thereby, it undermines the conclusion that Meadowbrook’s greater percentage violent crime rate increase means its citizens are more likely to face said crimes and is the correct answer choice.
Choice E: This answer choice is irrelevant as it only provides for changes in crime prevention expenditure, and does not signify to what extent this expenditure has affected the violent crime rate in each city; it has no bearing on the current absolute violent crime rates in the cities and consequently, does not address the conclusion that Meadowbrook’s greater percentage violent crime rate increase means its citizens are more likely to face said crimes, making it an incorrect answer choice.

Hence, D is the best answer choice.

To understand the concept of “Characteristics of an Evaluation Statement on GMAT Critical Reasoning,” you may want to watch the following video (~3 minutes):



All the best!
Experts' Global Team

Remember!!

In strengthen/weaken questions, when there is percentage, we say there are two types

1. percentage in total --> based on the total number

2. percentage of growth --> based on the previous year's/month's/any time period's number

In this problem, we have 60 percent and 10 percent as growth of violent crime rate. Therefore, we want the violent crime rate four years ago.

If we want A and B to work, then we need to change the stimulus to:

There are 60% of people who commit violent crimes / who were victims of violent crimes, and Parkdale 14%...

Since the new stimulus talks about percentage in total, we look for total number change. Therefore, in this case, A and B would work.