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For similar homes and comparable residents, home insurance for theft

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For similar homes and comparable residents, home insurance for theft  [#permalink]

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New post Updated on: 13 Aug 2019, 22:42
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For similar homes and comparable residents, home insurance for theft has always cost more in Springfield than in Shelbyville. Police studies, however, show that homes owned by Springfield residents are, on average, slightly less likely to be robbed than homes in Shelbyville. Clearly, therefore, insurance companies are making a greater profit on home theft insurance in Springfield than in Shelbyville.

In evaluating the argument, it would be most useful to compare


(A) the population density of Springfield with the population density of Shelbyville

(B) the cost of compensating theft losses in Springfield with the cost of compensating theft losses in Shelbyville

(C) the rates Springfield residents pay for auto insurance with the rates paid for auto insurance by residents of Shelbyville

(D) the condition of Springfield's roads and streets with the condition of Shelbyville's roads and streets

(E) the cost of home theft insurance in Springfield and Shelbyville with that in other cities

Originally posted by amulya619 on 14 Oct 2016, 23:59.
Last edited by Bunuel on 13 Aug 2019, 22:42, edited 3 times in total.
Edited the question.
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Re: For similar homes and comparable residents, home insurance for theft  [#permalink]

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New post 15 Oct 2016, 01:25
If the cost of compensating the home theft losses in Springfield is more than the cost of compensating the home theft losses in Shelbyville, even when the homes owned by Springfield residents are slightly less likely to be robbed than the homes owned by Shelbyville residents, we cannot assuredly say if the home insurance companies are making a greater profit in Springfield. Taking this option to two extremes strengthen as well as weaken the argument!

Therefore, pick D.
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New post 01 Aug 2017, 22:18
D.the condition of Springfield's roads and streets with the condition of Shelbyville's roads and streets-Out of scope

Ans is B
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Re: For similar homes and comparable residents, home insurance for theft  [#permalink]

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New post 20 Apr 2019, 19:14
amulya619 wrote:
For similar homes and comparable residents, home insurance for theft has always cost more in Springfield than in Shelbyville. Police studies, however, show that homes owned by Springfield residents are, on average, slightly less likely to be robbed than homes in Shelbyville. Clearly, therefore, insurance companies are making a greater profit on home theft insurance in Springfield than in Shelbyville.

In evaluating the argument, it would be most useful to compare


A. the population density of Springfield with the population density of Shelbyville

B. the cost of compensating theft losses in Springfield with the cost of compensating theft losses in Shelbyville

C. the rates Springfield residents pay for auto insurance with the rates paid for auto insurance by residents of Shelbyville

D. the condition of Springfield's roads and streets with the condition of Shelbyville's roads and streets

E. the cost of home theft insurance in Springfield and Shelbyville with that in other cities


Use variance technique...

Conclusion :- insurance companies are making a greater profit on home theft insurance in Springfield than in Shelbyville.

Option B -->

YES..the cost of compensating theft losses in Springfield is greater than the cost of compensating theft losses in Shelbyville.

The conclusion gets HURT.

We cant conclude that "insurance companies are making a greater profit on home theft insurance in Springfield than in Shelbyville."

No...the cost of compensating theft losses in Springfield is not greater than the cost of compensating theft losses in Shelbyville.

But the home insurance for theft has always cost more in Springfield than in Shelbyville. So We can conclude that "insurance companies are making a greater profit on home theft insurance in Springfield than in Shelbyville."

The conclusion gets HELPED.

So option B is answer.

Please give me kudo s if you liked my explanation.
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Re: For similar homes and comparable residents, home insurance for theft  [#permalink]

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New post 20 Aug 2019, 00:03
amulya619 wrote:
For similar homes and comparable residents, home insurance for theft has always cost more in Springfield than in Shelbyville. Police studies, however, show that homes owned by Springfield residents are, on average, slightly less likely to be robbed than homes in Shelbyville. Clearly, therefore, insurance companies are making a greater profit on home theft insurance in Springfield than in Shelbyville.

In evaluating the argument, it would be most useful to compare


(A) the population density of Springfield with the population density of Shelbyville

(B) the cost of compensating theft losses in Springfield with the cost of compensating theft losses in Shelbyville

(C) the rates Springfield residents pay for auto insurance with the rates paid for auto insurance by residents of Shelbyville

(D) the condition of Springfield's roads and streets with the condition of Shelbyville's roads and streets

(E) the cost of home theft insurance in Springfield and Shelbyville with that in other cities


OFFICIAL EXPLANATION:



Reading the question: we can use opinion-charged words to identify the pieces of the argument here. "Clearly" introduces the conclusion. The word "however" also is a clue: it tells us that the first sentence is a data point, not just filler. The logical structure is: A is true, but B is true, therefore C is true. A is the difference in cost in home insurance between these two cities, B is the rate of theft. The argument is not too strong.

Creating a filter: as the page summarizing the Critical Reasoning Strategy mentions, a prediction of the correct answer, even a vague or unrealistic prediction, is most powerful filter to evaluate answer choices. Predicting isn't always easy, but on this question, there are many reasons why insurance companies might have to pay more for losses in Springfield than in Shelbyville. Maybe the thieves in Springfield are more skilled and they manage to steal more per theft than in Shelbyville. That possibility is unlikely to be an answer choice, but we can still use it: "thieves in Springfield are more skilled and steal more."

Applying the filter, we evaluate the answer choices. Choice (B) is actually pretty close to our prediction. Choice (C) involves auto rates, which wouldn't shed light on this question without further information. (D) and (E) also involve comparisons with other things that we know nothing about, so they cause problems rather than solve problems. Back to (A), we can see it doesn't directly concern whether companies profit more from fewer thefts. Notice that our prediction was quite different from choice (B), but it was similar enough to help us spot (B) quickly.

Logical proof: we can use analysis by extreme cases to establish that choice (B) is correct. If the losses per theft were identical in Shelbyville and in Springfield, the conclusion would be true and the argument would stand; if they were wildly different, the conclusion could be false.

The correct answer is (B).
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Re: For similar homes and comparable residents, home insurance for theft   [#permalink] 20 Aug 2019, 00:03
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