October 2008 LSAT
Section 4
Question 11

October 2008 LSAT
Section 4
Question 11

Reply

Emil on July 9, 2022

Hi Tyler808,For this question, we are given the constraints that we have six messages, and 5 is L. Since we know there cannot be more than two of each, it must be H H J J L L.

_ _ _ _ L _

We also know that L is not 1, 1=6, there is 1 HJ block, and that J's messages have one in the first 3, and one in the last three.

I would then make two scenarios based on the J rule, since we know that J must be either 4 or 6

_ _ _ J L _

_ _ _ _ L J

We know that 6=1, so if J is last then it is also one, and if J is not last, then H must be, since L cannot be either. So

H _ _ J L H

J _ _ _ L J

We can then place our HJ blocks- remembering that we must have one J in the first three.

H J _ J L H

J _ _ _ L J is impossible, since we cannot place an HJ.

So, it must be the case that H J L J L H- so J and L are not interchangeable at 3/4, since we would then have two Js in the first three.

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