September 2014 LSAT
Section 3
Question 10

September 2014 LSAT
Section 3
Question 10

Reply

Shunhe on August 11, 2020

Hi @Mwins,Thanks for the question! So we’re told here that X’s speech is at 3PM, so we should see what we can infer from that. Well, remember that M has to before L, and on the same day. So since there’s only one speech at 3PM, and X is giving it, that means that M has to be at 1PM, and L has to be at 2PM, and they have to be on the same day.

So there’s two possibilities for where ML can speak. Let’s first say they both speak in the Gold Room. Well, then based on the last rule, we know that X and Z are both in the Rose Room. And now we have to place Y in the Rose Room too. Because there’s only one room with 3 speeches, and it’s the room with X. So Z, Y, and X are all in that room. Specifically, it goes ZYX, since X goes at 3PM, and Y has to go after Z.

Now let’s say that ML both go in the Rose Room. If X is in the same room as them, then Z and Y go in the other room (obviously). And if X is in the other room, that’s the room with 3 people, so Z and Y both still have to go in the Gold Room with X! So In all these scenarios, Z and Y are together, and so L is never with Y. Thus, (A) can never happen.

Hope this helps! Feel free to ask any other questions that you might have.

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