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#10 - Local

LSAT Novice
Posts: 1
Joined: Thu Jul 12, 2018 9:59 pm
Points: 1

Hi Dave,

I'm reviewing this test, and I'm struggling to understand why answer choice E isn't correct for this question.

I would really appreciate it if you could walk me through how to figure this out.

Jennifer Janowsky
PowerScore Staff
PowerScore Staff
Posts: 91
Joined: Sun Aug 20, 2017 11:58 pm
Points: 91

Hi! This game is a little tricky, but let's see if we can walk through the setup first.

Basically, there are 6 messages, in order. Here's the spots and the distributions for that:


1 2 3 4 5 6

Up to three of those have been made by one person, making 3 possible distributions: 3-1-1-1, 2-1-1-1-1, 1-1-1-1-1-1. This means that at least 4 people must have left messages.

When you string all of the rules together from the problem, you get something like this:

G --> F --> P-T
P --> H-L

The two strands can be connected at P to make a powerful super-strand, which can be followed backwards for its contrapositive. This is where you get an interesting inference--if P is not included, you cannot have G, T or F. That is too many people to eliminate (there must be at least 4). Therefore, P must be included, AND T, H, and L must follow from P.

For question 10, we are given a situation in which P must be 5th and asked what must be true.

Looking at the rules, if P then T, H, and L. T must be after P, so we already have 2 placed:

__.__.__.__ P T
1 2 3 4 5 6

H must be before L, so H would have to be in spaces 1, 2 or 3. But if H is in 1, then P must be in 6, which is already occupied by T. Therefore, H must be in 2 or 3, followed by L in either 3 or 4.

__ H/ H/L L/ P T
1 2 3 4 5 6

Looking at the answer choices, the only one that could be true is that Liam left exactly two messages--one could be on both 3 and 4. Answer (E), that Fleure left the third and fourth messages, is not possible because one of those must be L.

I hope that clears everything up for you! :-D