Q15 - All the apartments on 20th Avenue are in

[ohthatpatrick]

Question Type:
Flaw

Stimulus Breakdown:
Conclusion: Most old houses on 20th Ave contain more than one apartment.
Evidence: All the apartments on 20th are in old houses, and there are twice as many apartments as old houses.

Answer Anticipation:
This math flaw is interpreting average too literally. If there are twice as many apartments as old houses (say, 20 apartments vs. 10 houses), then there is an average of 2 apartments per old house. That's where the author is deriving her conclusion.

However, if we were trying to contradict her conclusion, and say that only a minority of old houses had more than one apartment, we could just spread those 20 apartments out in lumpy fashion, not evenly distributed fashion.
You could have 4 old houses that each contain 5 apartments (getting us our 20 apartments) and 6 old houses that contain 0 apartments. This fits all the premise facts, but it contradicts the conclusion. In this scenaro, most of the old houses (6 out of 10) do not contain more than one apartment; in fact, we made it so that most old houses contain ZERO apartments.

The flaw is thinking that these 20 apartments are smoothly distributed among the 10 old houses, when a "lumpy" example could disprove the conclusion.

Correct Answer:
E

Answer Choice Analysis:
(A) Would this weaken? No. The author wasn't saying all houses are old. It wouldn't bother her to admit that some are not old.

(B) Was this a circular argument? No. The premises say different things from the conclusion.

(C) Would this weaken? No. The author wouldn't be troubled by other rental accommodations.

(D) Was this a Conditional Logic flaw? No. The first sentence IS conditional logic (if we want it to be). But a conditional logic flaw stemming from that conditional rule would sound something like this: "All the apartments on 20th Avenue are in old houses. Ben lives in an old house. Thus, there is an apartment in the building where Ben lives." If you were trying to make (D) work by relating it to the conclusion, you might think, "The condition whose presence is required in order for the conclusion to be true is the idea that [the apartments are distributed in a roughly even fashion across the pool of old houses]." There's no condition in the argument whose presence would ensure that [the apartments are distributed in a roughly even fashion].

(E) Would this weaken? YES. This sounds like our "lumpy" objection.

Takeaway/Pattern: The "all", the "twice", and the "most" warn us that this will be testing mathematical thinking. Our goal on Flaw is to be able to argue the Anti-Conclusion, in this case that "50% or less of the old houses contain more than 1 apartment".

If you struggle to see the math flaw, it's worth jotting down some small numbers to test the concept. If you're dealing with the concept of "most", you should probably pick an odd number. If there are 5 houses, then most = 3 or more. We need to argue that 3 of them have only one apartment or no apartments. 5 houses would mean 10 apartments. Can you allocate 10 apartments into 5 houses, while making sure that 3 of them have only 1 or 0 in side of them? Sure. We could just do 10, 0, 0, 0, 0. We could do 4, 4, 1, 1, 0. There are lots of ways to avoid the conclusion, and seeing one or two of them will help us to understand why an "IMBALANCED distribution" (like what E talks about) is crucial to objecting to the author's math.

#officialexplanation

[SharonT609]

Does "a significant number of" in E means many and therefore equals to some? I ruled out E because I thought the significant number meant "most". But now I think even nearly a half could count as "a significant number" if the situation is really rare. Is that right?

[EricaL584]

Let's say there are 100 old houses and 200 apartments on the avenue. If 40 among the 100 old houses have averagely 5 apartments, it will meet the goal of 200 apartments, and the rest 60 old houses can each have only one apartment safely.

I am little confused about D though. I thought is was saying that "most old houses on 20th avenue contain more than one apartment" is sufficient to make sure that "there are twice as many apartments on 20th avenue as there are old houses" , but it is not a necessary condition for that.
......

5 minutes after my post, I realized why D can't be selected. Come on.."most old houses on 20th avenue contain more than one apartment" is NOT AT ALL sufficient to make sure that "there are twice as many apartments on 20th avenue as there are old houses" . 90*2 + 10=190, it is not twice of 100.