PT151.S2.Q15

SECTION 1 · rc

PASSAGE 1
Question 1 Question 2 Question 3 Question 4 Question 5 Question 6

PASSAGE 2
Question 7 Question 8 Question 9 Question 10 Question 11 Question 12 Question 13 Question 14

PASSAGE 3
Question 15 Question 16 Question 17 Question 18 Question 19 Question 20 Question 21

PASSAGE 4
Question 22 Question 23 Question 24 Question 25 Question 26 Question 27

SECTION 2 · lr

Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 10 Question 11 Question 12 Question 13 Question 14 Question 15 Question 16 Question 17 Question 18 Question 19 Question 20 Question 21 Question 22 Question 23 Question 24 Question 25 Question 26

SECTION 3 · lr

Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 10 Question 11 Question 12 Question 13 Question 14 Question 15 Question 16 Question 17 Question 18 Question 19 Question 20 Question 21 Question 22 Question 23 Question 24 Question 25

SECTION 4 · lr

Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 10 Question 11 Question 12 Question 13 Question 14 Question 15 Question 16 Question 17 Question 18 Question 19 Question 20 Question 21 Question 22 Question 23 Question 24 Question 25

PrepTest 151 - Section 2 - Question 15

Question

Explanation

Analytics

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Support All the apartments on 20th Avenue are in old houses. ████████ █████ ███ █████ ██ ████ ██████████ ██ ████ ██████ ██ █████ ███ ███ ███████ ██████████ ████ ███ ██████ ██ ████ ██████ ███████ ████ ████ ███ ██████████

Summarize Argument

The author concludes that most old houses on 20th Avenue contain more than one apartment. He supports this by saying that all the apartments are in old houses, and there are twice as many apartments as old houses.

Identify and Describe Flaw

The author concludes that most— i.e. more than half— of the old houses have more than one apartment, just because there are twice as many apartments as old houses. But let’s say there are 10 old houses and 20 apartments. What if one old house contains all 20 apartments? What if 4 old houses contain 5 apartments?

It’s not necessarily true that most of the old houses contain more than one apartment. In fact, it’s possible that most of the old houses don’t contain any apartments at all!

Analysis by EleanorRoberts

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15.

The reasoning in the argument █████ ██ ████ ██████████ ██ █████████ ██ ███ ███████ ████ ███ ████████

overlooks the possibility ████ ████ ██ ███ █████████ ██ ████ ██████ ███ ███ ███ ██████

The author doesn't overlook this possibility. He says that there are twice as many apartments as old houses and that all the apartments are in old houses. But there might still be other buildings that are not old houses.

draws a conclusion ████ ██████ ████████ ███ ██ ███ ████████ ███████ ██ ███████ ██ ███ ██████████

This is the cookie-cutter flaw of circular reasoning, where the conclusion is simply a restatement of one of the premises. The author doesn’t make this mistake. His premises may not support his conclusion well, but they are distinct from his conclusion.

fails to consider ███ ███████████ ████ ████ █████████ ██ ████ ██████ ███ █████ █████ ██ ██████ █████████████ █████ ████ ██████████

The author doesn’t overlook this possibility. His argument is only about the old houses and apartments on 20th Ave. There might be some buildings that are hotels or rental houses, but this doesn’t affect his argument.

confuses a condition █████ ████████ █████ ██ ██████████ ██ ██████ ███ █████ ██ ███ ██████████ ██████████ ████ █ █████████ █████ ████████ ██ ████████ ██ █████ ███ ███ ██████████ ██ ██ ████

This is the cookie-cutter flaw of confusing necessary and sufficient conditions. The author doesn’t make this mistake; he never presents a condition sufficient to ensure the truth of the conclusion in the first place.

fails to address ███ ███████████ ████ █ ███████████ ██████ ██ ███ ██████ ██ ████ ██████ ███████ █████ ██ ████ ██████████

If a significant number of old houses have 3 or more apartments, the author can't conclude that most houses have at least one. For example, if there are 10 old houses and 20 apartments, it's possible that just 4 houses have 5 apartments, meaning most houses have no apartments.

Difficulty

Question

56% of people who answer get this correct

This is a difficult question.

CURVE

Score of students with a 50% chance of getting this right

25% 151

159

75% 166

Analysis

Question Stem Tags

Flaw or descriptive weakening

Stimulus Tags

Conditional reasoning

Math

Quantifier

Answer Popularity

Popularity Avg. score

154

151

155

29%

157

56%

164

Question history

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