PT151 S2 Q15 ExplanationAll the apartments on 20th

Answer choices, explained

1. A

   Not an Objection7% picked this

   overlooks the possibility that some of the buildings on 20th Avenue are

   It doesn't affect this argument if some of the houses are not old. We only care about the old houses because that's where all the apartments are found and that's what the conclusion is about.

   7%
2. B

   Conclusion ? Premise4% picked this

   draws a conclusion that simply restates one of the premises offered in support

   The conclusion says that "most old houses have more than one apartment". Neither premise says that. This answer describes the famous flaw Circular Reasoning, which is almost always wrong.

   4%
3. C

   Not an Objection6% picked this

   fails to consider the possibility that some buildings on 20th Avenue may offer types of rental

   Just like (A), we don't care about the fact that there are also non-apartments on 20th avenue. We're just looking at apartments, which are all found in old houses, and analyzing whether there is more than one apartment in most of those old houses.

   6%
4. D

   Bad Premise / Conclusion Match25% picked this

   confuses a condition whose presence would be sufficient to ensure the truth of the argument’s conclusion with a condition whose presence is required in

   Is there a condition whose presence would be sufficient to ensure that "most old houses on 20th have more than one apartment"? No, there's no conditional logic in the argument that would deliver us the idea in the conclusion. This answer describes the famous flaw of Necessary vs. Sufficient. There is one conditional statement in the argument: "apartment on 20th ? in an old house" If the author had said, "All the apartments on 20th Avenue are in old houses. However, Bill does not live in an apartment on 20th Avenue. Thus Bill does not live in an old house", that would have been committing the error that (D) describes.

   25%
5. E

   Correct57% picked this

   fails to address the possibility that a significant number of old houses on 20th Avenue contain

   Why this is right

   Yes, this presents the mathematical objection we considered. We know that old houses are averaging two apartments per house, but if a significant number of those old houses have 3 or more apartments, then it's still possible that the majority of old houses has 1 or 0 apartments. Our hypothetical example from before is getting at this idea. 10 apartments, 5 old houses: House 1: 1 apartment House 2: 1 apartment House 3: 1 apartment House 4: 3 apartments House 5: 4 apartments A significant number (2 out of 5) have 3 or more apartments, which allowed for the average to still be 2 apartments per house even though the conclusion was wrong: most of the old houses don't have more than one house.

   Skill tested: Flaw · how this choice captures the argument's function is the move to repeat next time.

   57%