#10 - Local, Could Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?f=296&t=2115)

The correct answer choice is (C)

If P leaves exactly one message, then any of the three distributions are still possible. If P’s message is the fifth, then from the contrapositive of the third rule H cannot leave the first message. This information eliminates answer choice (A).

From the sixth rule, when P leaves a message, then H and L also leave messages, with H > L. Because H cannot leave the first message, the earliest message H could leave is the second, and thus the earliest message that L could leave is the third. Thus, answer choice (D) can be eliminated.

The remaining three answers all concern individuals leaving two messages.

Answer choice (B): if T leaves exactly two messages, under the 2-1-1-1-1-0 distribution, F, P, T, H, and L each leave a message, with P > T and H > L. But, with P leaving the fifth message, and the fifth rule stipulating that all of P’s messages precede all of T’s messages, there is not sufficient room to accommodate both of T’s messages and still conform to the fifth rule. Thus, this answer choice is incorrect.

Answer choice (E): if F leaves the third and fourth messages, under the 2-1-1-1-1-0 distribution, F, P, T, H, and L each leave a message, with P > T and H > L. But, with P leaving the fifth message, T must leave the last message, forcing H to leave the first message and L to leave the second message. Because this violates the third rule, this answer cannot be correct.

Thus, answer choice (C) is correct. L could leave the third and fourth messages.

[2018LSATReppin]

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.

Thanks!

[Jennifer Janowsky]

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:

FGHLPT(6)

__.__.__.__.__.__
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!

[gab1234]

When re-working this problem, I correctly chose C and understood the reason A was incorrect. Could you elaborate to share the concrete reason C is correct?

[Jeremy Press]

Hi gab!

In a general sense, with P on 5 (and only 5), the last rule triggers, and H/L must be included with all H's ahead of all L's. That means L cannot be first, and H cannot be last. In addition, H cannot be first because the only P is not on space 6 (triggering the contrapositive of the third rule). We can't immediately take the diagram of this local scenario much further than that.

To see how answer choice C is a possible solution, consider what happens if Liam leaves exactly two messages. That means, because of the first rule, exactly 4 other people must leave messages. We know who those four are. They can't include G, because when G leaves a message, it eventually triggers the inclusion of every other variable (with each variable then being used only once). With the other 5 included (F, H, L, P, T), the F rule forces T to be in the 6th position (after P). F would then have to be in the first position, since no other variable would be available to go first (we already knew from the above paragraph that H, L and P cannot be first, and now T cannot be first because, in light of the F rule, all of its occurrences must be after P). There would be one remaining H, which, because of the P rule, would have to come before both L's. But that works, with H on 2, and L's on 3 and 4, validating answer choice C! See diagram below:

Screen Shot 2020-09-28 at 11.00.25 AM.png

I hope this helps!

[kenlars5]

Hello,

So if F is not there and therefore only HLPT, does the rule that all P precede T still apply? I guess I'm just confused because I thought that F is what made that rule come into play, but without F such order wouldn't matter? I see why H--L always applies because both H and L are ALWAYS required, but how do you know if F is sufficient to bring about this condition or if this is a rule that applies at all times?

I'm just asking because one of the previous explanations said that since P is 5th we know T is 6th. Regardless you could still get to the right answer because the other ones did not work, but it's always helpful to know exactly what is going on because I didn't initially place T right after P which would've more quickly enabled me to see that F was needed because H can't go first and H--L; rather than just going straight to all the answers and seeing which could be true. If F is sufficient to bring about the rule of P--T than I would assume I had the proper approach? If someone could just clarify for me the best approach that would be great!

Thanks!

[Jeremy Press]

Hi ken,

You're correct that if you did have a group of only HLPT leaving messages, then the T's would not have to follow the P's, because the sufficient condition of the 5th rule with F wouldn't trigger. But, and here's the key, that scenario (which in general could be possible under the condition given in the question, e.g., with a solution like T-H-L-L-P-L) doesn't work with any of the given answer choices. So you need a group of variables that includes more than just HLPT to find a workable answer. The group will need to be either FHLPT (which works with answer choice C) or all six of FGHLPT (which, again, doesn't work with any answer choice).

Your best bet on this question is to think first about the three workable groups (HLPT, FHLPT, or GFHLPT), then think about whether each answer choice would work within the general parameters of any of those workable groups. It's definitely a tricky question!

I hope this helps!

[kenlars5]

Thanks for the clarification and explanation Jeremy! That helped a lot.

[frk215]

Hey folks! I wanted to ask about the correct strategy for this question. I spent about 2 minutes and 40 seconds on it and I was wondering how I could lower that time. Here was my attack:

1. noticed that a + d would have the same connotation in terms of placing h and violating the h1 rule - crossed em out
2. noticed that with two ts we would have a grouping of t, t, p, h, l with one more space to fill with either g or f. g will make f join and doesn't allow for repeats so it has to be f— but with f the rule: p —t kicks with not enough room after p to place both ts — crossed out b
3. made a hypothetical to map out c and see if it works: F H L L P T —> works great, choose and move on

Is there anything else I could've done to move faster?

[Adam Tyson]

This sounds like a solid approach to me, frk215, and if you had not taken the time in the initial setup to do the numeric distributions you would probably have enough time to do this work. With practice, you'll get there faster! But doing the distribution up front would sure help here, because both of those answers about people leaving exactly 2 messages would have triggered the 2-1-1-1-1-0 distribution where G is out and everyone else, including F, is in. That allows you to more quickly prove that there cannot be two T's, as you did.