Logical reasoning PrepTest 151 · Section 2 · Question 15
Question prompt
Why the credited answer is right
Credited answer: E
The notes below walk through why it fits the stem and how to eliminate the rest.
Question Type
Answer choices
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Aoverlooks the possibility that Remaining source text redacted.
Why choice A is not credited
Incorrect. This argument leaves open that possibility. However, it notes that all apartments are in old houses, and then concludes something about the number of apartments per old house. The newer houses are irrelevant to the consideration. -
Bdraws a conclusion that Remaining source text redacted.
Why choice B is not credited
Incorrect. This answer reflects a circular reasoning flaw, but the premises are both distinct from the conclusion. -
Cfails to consider the Remaining source text redacted.
Why choice C is not credited
Incorrect. Since the argument limits itself to discussing old houses and apartments, this answer is out of scope. If the conclusion were about the number of renters, then this answer would need to be considered. -
Dconfuses a condition whose Remaining source text redacted.
Why choice D is not credited
Incorrect. While the argument does use conditional language, it doesn't commit a conditional logic flaw. No statement given provides a sufficient condition for the conclusion. This answer choice would be correct in an argument that confused the first statement to mean that anything in an old house on 20th Ave. was an apartment. -
Efails to address the Remaining source text redacted.
Why choice E matches the stem
Correct. Argument or Facts:
Argument
Valid or Flawed:
Flawed
Question Type:
Errors in Reasoning
Stimulus Summary:
Apt. on 20th → In an old house
Twice as many apartments as old houses
Therefore, most old houses on 20th have multiple apartment
Answer Anticipation:
This answer jumps between a statement about the number of an element (twice as many as old houses) to a statement about the percent of an element (over 50% of old houses). That's problematic.
To highlight this, think about a street with 3 old houses, one of which has 6 apartments, and the other two of which have none. There would be twice as many apartments as old houses, but only 1 out of 3 has multiple apartments.
Answer Explanation:
If the old houses have 3 or more apartments in them, then it can't be concluded that the old houses have the apartments distributed evenly enough to guarantee that most of them have multiple apartments.
Key Takeaway:
Be careful in arguments that treat the average and distribution among a group as interchangeable. Here, while the "average" would be two apartments per old house, there's no guarantee that the apartments are distributed evenly.
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Flaw 1 reply
Started by @chris_va