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08 Dec 2020, 15:10
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We're told about violent crime rate increases in Meadowbrook and Parkdale. Over the last four years, Meadowbrook increased by 60% and Parkdale increased by 10%.
We can't conclude that residents of Meadowbrook are more likely to become victims of violent crime than are residents of Parkdale UNLESS we know the violent crime rates in each city four years ago. We only know the increase over the span of four years.
D is the correct answer.
We can't conclude that residents of Meadowbrook are more likely to become victims of violent crime than are residents of Parkdale UNLESS we know the violent crime rates in each city four years ago. We only know the increase over the span of four years.
D is the correct answer.
18 Aug 2022, 22:49
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The Story
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. – Violent Crime Rate is defined as the number of violent crimes per 1000 residents.
The VCR has gone up by 60% in the last four years in a particular town (Town 1). What’s gone up by 60%? The crime rate. So, say there used to be 10 violent crimes per 1,000 residents four years ago, now there are 16 crimes per 1,000 residents.
The corresponding increase for Parkdale is only 10 percent. – In another town (Town 2), the VCR increased by 10%.
(I’m thinking: Since we don’t know the actual rate for either town, we can’t say which town saw a bigger increase in the number of violent crimes.)
These figures support the conclusion that residents of Meadowbrook are more likely to become victims of violent crime than are residents of Parkdale. – The author concludes that the residents of the first town have a higher chance of encountering a violent crime than do residents of the second town.
Gist:
Residents of town 1 are more likely to encounter violent crimes than are residents of town 2 (main point), because town 1 has a higher percentage increase in VCR (60%) than town 2 (10%) (basis).
Gap(s) in logic:
The passage gives us only the percentage increases in VCR’s for the two towns. The argument does not tell us about the actual rates. To conclude that residents of the first town are more likely to become victims of violent crime, we need to know the actual violent crime rates for the two towns.
For example:
Town 1: Say the VCR goes from 10 to 16. That’s a 60% increase. So, the town now has 16 violent crimes per 1,000 residents.
Town 2: Say the VCR goes from 50 to 55. That’s a 10% increase. So, the town now has 55 violent crimes per 1,000 residents.
Do you see that in this scenario Town 2 residents would be more likely to encounter violent crimes?
Question Stem
The argument above is flawed because it fails to take into account
We have come up with one gap above. There could, of course, be more gaps (flaws) in the argument too. Let’s go through the answer choices.
Answer choice analysis
(A) changes in the population density of both Parkdale and Meadowbrook over the past four years
Incorrect. Population density: Total Population/ Total Area
The areas of the two towns are unlikely to change. So, for argument’s sake, let’s take ‘changes in population density’ to imply ‘changes in population’. Basically, the argument fails to take into account changes in the population
Is this a flaw in the argument?
The argument talks about violent crime rate i.e. number of violent crimes per 1,000 residents. In this case, the changes in population will have no impact on the argument. Let’s try to understand this.
Let’s say that the population of Meadrowbrook increased by 100% from 1000 to 2000 in the last 4 years. Let’s say the violent crime rate four years ago was 10 per 1000 residents.
Given that the violent crime rate increased by 60%, the violent crime rate now is 16 per 1000 residents.
The number of violent crimes now is 161000*2000 i.e., 16*2 = 32.
As we can see from the above calculations, if we increase the population of Meadrowbrook, the number of violent crimes will also increase. Thus, in this case, a change in the population of Meadrowbrook will have no impact on the likelihood that residents become victims of violent crime (the likelihood of residents becoming victims of violent crimes = the number of violent crimes/ the population). Since the argument talks about violent crime rate, any change in population will have an equivalent change in the number of violent crimes. Thus, a change in population will not have an impact on the likelihood of residents becoming victims of violent crime.
As the populations of the towns increase, the number of violent crimes would also increase. However, I would not expect the violent crime rates to increase. Increase in violent crime rates is not linked to changes in population or the rate of population change.
I have explained a scenario in ‘additional notes’ in which option A is correct. You may want to go through that to better understand why option A is wrong in the original question.
(B) how the rate of population growth in Meadowbrook over the past four years compares to the corresponding rate for Parkdale
Incorrect. I talked about this already in the explanation for answer choice A.
Say the population of a certain town increases. Would that directly translate into an increase in the violent crime rate?
No.
An increase in population will relate with an increase in the number of violent crimes, however, I would not expect an increase in the rate of violent crimes. It is not that the argument fails to consider the rate of population growth in the two towns. Irrelevant.
I have explained a scenario in ‘additional notes’ in which options A and B are correct. You may want to go through that to better understand why these answer choices are wrong in the original question.
(C) the ratio of violent to nonviolent crimes committed during the past four years in Meadowbrook and Parkdale
Incorrect. The entire argument, including the conclusion, is about violent crimes. The argument need not take into account the ratio of violent to non-violent crimes.
Irrelevant.
(D) the violent crime rates in Meadowbrook and Parkdale four years ago
Correct. Bam! The argument needs to include VCR’s from four years ago to be logical.
This one is in-line with our initial gap analysis. To understand that residents of Meadowbrook are actually more likely to become victims of violent crime than are residents of Parkdale, we need to know the actual violent crime rates in the two towns. Just learning about the percentage increase in the VCR is not enough. If we have the VCR’s from four years ago. And we club that information with the percentage increases mentioned in the passage. We can get the VCR’s for the two towns today. Then that information would be able to tell us residents of which town are more likely to become victims of violent crimes.
(E) how Meadowbrook’s expenditures for crime prevention over the past four years compare to Parkdale’s expenditures
Incorrect. Expenditures? And that too expenditures for crime prevention? Nope.
How the expenditures for crime prevention compare is irrelevant to the argument. This is not a flaw in the argument.
Additional Notes
Answer choices A and B seem attractive to many GMAT candidates. At the time of writing this solution, 34% students mark A or B as their answer on GMAT Club. That is over one-third of the people who attempt this question - a significant chunk.
I have tried to explain a scenario below in which answer choices A and B would be correct. It is fairly long, and I discuss options A and B all the way in the end. But that’s the logical path I could think of to reach there.
Here goes.
To see clearly why the two answer choices are wrong, we need to be clear that the argument is not dealing with the number of violent crimes but with violent crime rates.
What if the passage did deal with an increase in the number of violent crimes?
I.
The number of violent crimes in Town 1 is 60 percent higher now than it was four years ago. The corresponding increase for Town 2 is only 10 percent. These figures support the conclusion that residents of Town 1 are more likely to become victims of violent crime than are residents of Town 2.
Does this argument make sense? Do you notice any gaps here?
The argument is not logical. Let’s start with some new towns - X and Y - to understand why the argument isn’t logical.
Say,
- 1. Town X had 10 murders last year.
2. Town Y had 100 murders last year.
Does that mean that residents of Town Y are more likely to get murdered than are residents of Town X?
Don’t we need to know about the populations of the two towns? We do, right?
10 murders in a town of population 100 means there’s a 10% likelihood of someone getting murdered.
100 murders in a town of population 10,000 means there’s a 1% likelihood of someone getting murdered.
So, just knowing the number of murders on its own is not enough to conclude about the likelihood of getting murdered.
Now let’s go to the variation of the argument I have created above.
Say,
- 1. Town 1:
- Number of violent crimes: 100 → 160 (60% increase)
- a. Number of violent crimes: 50 → 55 (10% increase)
b. Number of violent crimes: 200 → 220 (10% increase)
(I’ve made two cases - one in which the final number of violent crimes in this town is lower and one in which it is higher.)
Based on 160 v/s 220, can we say that residents of Town 2 are more likely to become victims of violent crime than are residents of Town 1?
The answer to both these questions is NO. We need to know the populations of the two towns (the base or the denominator) to be able to make such claims.
II.
Now, what if we knew the populations of the two towns also? In fact, let’s say the populations of the two towns were equal four years ago. And then we’re given that the number of violent crimes increased by 60% and 10% respectively.
Can we then conclude that residents of Town 1 are more likely to become victims of violent crime than are residents of Town 2?
I think we can’t. Because we still need to know the initial numbers of violent crimes in the two towns.
Let me explain this with numbers:
Population of each town four years ago: 10,000
Number of violent crimes in Town 1: 50 → 80
Number of violent crimes in Town 2: 300 → 330
It is still not logical to conclude that residents of Town 1 are more likely to become victims of violent crime than are residents of Town 2.
III.
This is the flaw highlighted by options A and B.
Ok, now let’s say we were also given the initial number of violent crimes for each town too. In fact, let’s say, the initial numbers were equal.
Four years ago:
Population of Town 1: 10,000
Population of Town 2: 10,000
Number of violent crimes in Town 1: 100
Number of violent crimes in Town 2: 100
Current situation:
Number of violent crimes in Town 1: 100 → 160
Number of violent crimes in Town 2: 100 → 110
Can we then conclude that residents of Town 1 are more likely to become victims of violent crime than are residents of Town 2?
We still can’t. We need to know how the populations of the two towns changed in the last four years.
E.g. If the population of Town 1 increased by 60% (10,000 → 16,000) and the population of Town 2 increased by 10% (10,000 → 11,000), the likelihood of encountering a violent crime would be the same in both the towns.
To summarise,
- 1. We need to know the initial population of the each of the two towns
2. We need to know the initial violent crime rates for the two towns
3. We need to know the rate at which the population of each town has increased
Answer choices A and B play on the third gap mentioned above. So, answer choices A and B would point out a flaw in this modified argument about the number of violent crimes.
If the argument were about the number of violent crimes instead of violent crime rates, both answer choices A and B would be correct.
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22 Aug 2022, 05:42
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in the year 2018
M = 10
P= 10
in the year 2022
M= 16
P=12
Conclusion holds
in the year 2018
M = 10
P= 20
in the year 2022
M= 16
P=22
Conclusion breaks
The conclusion is not true in every case. therefore the flaw in the argument is non-availability of 2018 data.
M = 10
P= 10
in the year 2022
M= 16
P=12
Conclusion holds
in the year 2018
M = 10
P= 20
in the year 2022
M= 16
P=22
Conclusion breaks
The conclusion is not true in every case. therefore the flaw in the argument is non-availability of 2018 data.
