Logical Reasoning

PT151 · S2 · Q15 All the apartments on 20th

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

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 there are old houses.

Evaluation

This is very math-y, so using some numbers can maybe help us make sense of it. Let's start with "twice as many apartments as old houses". We'll say there are 10 apartments and 5 old houses.

All 10 apartments are in those 5 old houses. That's an average of 2 apartments per house, so the author is reasoning that "most of these old houses" have more than one apartment.

But does an average value have to be true of most data points within a group?

Can a group of houses have an average of 2 apartments, even if most of those houses have less than 2 apartments?

Sure. If you want the average value to be unlike most data points in the set, just make a couple data points be total outliers.

Take any ten people off the street: their average wealth is probably about $50,000. Now add Jeff Bezos and Bill Gates to that group and you have a dozen people with an average wealth of about $10 billion / person. Is it true to say that most of the people in that group have more than $1 billion in wealth? No, only 2 of the 12 do. It's just that their influence skews the average very high.

Similarly, we could allocate our 10 apartments over five old houses very unevenly: House 1: 1 apartment House 2: 1 apartment House 3: 1 apartment House 4: 3 apartments House 5: 4 apartments

It would be true that they have an average of 2 apartments / house, even though most of the houses do not have more than one apartment (only 2 out of 5 have more than one).

Goal

Look for an answer that calls out this mismatch between "an average" for a group and . Or look for an answer raising the specific objection that the apartments might be allocated to the old houses in a very uneven way, such that there are a bunch of apartments found in a minority of the houses, with only 1 or 0 apartments found in the majority of the houses.

15.

The reasoning in the argument above is most vulnerable to criticism on the grounds that the argument

  1. overlooks the possibility that some

    Not an Objection

    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.

    8% picked this

  2. draws a conclusion that simply

    Conclusion ? Premise

    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% picked this

  3. fails to consider the possibility

    Not an Objection

    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% picked this

  4. confuses a condition whose presence

    Bad Premise / Conclusion Match

    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.

    26% picked this

  5. Correct

    fails to address the possibility

    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.

    56% picked this

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