October 1991 LSAT
Section 3
Question 10

October 1991 LSAT
Section 3
Question 10

Reply

Ben on May 11, 2020

Hi Tkj97, thanks for the question.This question involves making some further inferences after placing the KP duo as you did.

Here is what we should have after your inference.

5 __ __

4 M N

3 K __

2 P __

1 __ __

and

5 __ __

4 M N

3 __ __

2 K __

1 P __

What we have left over in the first diagram is L on its own. Meaning L can only be placed in 1 or 5. And we also have J, O, Q. Since O and Q cannot go beside each other. One of them must go with J, while the other goes with K. P is not an option because floor 2 is always solitary.

In the second diagram. We have L on its own which can go in 3 or 5. We again have J, O, and Q. Since O and Q cannot go beside each other, one of them must go with J and the other with P. K is not an option because floor 2 is always solitary. Since Q cannot be on the first two floors, then O and P must be together, while J/Q and L switch between 3 and 5.

As you see, based not the correct answer, we didn't need all of these deductions. Simply knowing that L is 1/5 or 3/5 would have been enough. But I hope this helps you better understand the cascade of inferences.

Please let me know if you have any other questions.

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