June 2017 LSAT Section 4 Question 23

# Which one of the following, if substituted for the condition that Fan and Haley cannot visit the same city as each ot...

2 Replies

Jacob on December 1, 2018

Hi @shylahmarieg@gmail.comIâ€™m happy to help. I will start by saying that this type of logic game problem is about as hard as they come â€” substitution equivalence. So donâ€™t be worried if it is causing you to struggle â€” and also donâ€™t be afraid to skip over it on the real exam and come back to it, rather than spend a ton of time on it. Every question is worth the same amount of points!

That said, letâ€™s dive in.

We have F G H and I visiting

M S and T.

Each manager goes to at least one, and each city will be visited by EXACTLY two.

That already tells us a bunch: we can draw our game as

M S T

1 _ _ _

2 _ _ _

(We know therefore that some of the managers will repeat: we have FGHI, and only 6 slots â€” so there will be two repeats.)

Back to the rules.

I x 2 (he visits exactly two cities)

(We also now know that the sixth slot will either be F/G/H)

No F/H above H/F

If MG, then TH

(If not TH, then not MG)

No SG

Notice how restrictive this final rule is. It gives us three game possibilities for where G can go: Either G visits M, or visits T, or visits both. That is it!

Letâ€™s work those possibilities through.

G visits M:

M S T

1 G _ _

2 _ _ _

G visits T:

M S T

1 _ _ G

2 _ _ _

G visits both:

M S T

1 G _ G

2 _ _ _

We know from a rule if MG, then TH. That impacts our first and third possibilities, because in both of those MG is triggered.

G visits M:

M S T

1 G _ H

2 _ _ _

G visits T:

M S T

1 _ _ G

2 _ _ _

G visits both:

M S T

1 G _ G

2 _ _ H

We also know that I visits exactly two cities, and that F has to visit one city, so possibility 3 is completely solved:

G visits M:

M S T

1 G _ H

2 _ _ _

G visits T:

M S T

1 _ _ G

2 _ _ _

G visits both:

M S T

1 G F G

2 I I H

So letâ€™s think about the first two. In the first possibility, we know we canâ€™t put F and H together, and that it doesnâ€™t make sense to have two Hs on the same visit, so we must have I visiting T. Where does the second I go? If It visits M, then we would have to have F in S, and that wouldnâ€™t work because then we would have either another F (which doesnâ€™t make sense) or F and H (which canâ€™t go together. So our first possibility ends up looking like:

G visits M:

M S T

1 G I H

2 _ _ I

(With one remaining F and one remaining F/H)

G visits T:

M S T

1 _ _ G

2 _ _ _

G visits both:

M S T

1 G F G

2 I I H

Finally, letâ€™s look at the second possibility. Again, no F with H, and we need to place two Is. You should notice that I canâ€™t go to T, because then F and H would be forced together in either M or S, which canâ€™t happen. So the Is must go to M and S.

G visits M:

M S T

1 G I H

2 _ _ I

(With one remaining F and one remaining F/H)

G visits T:

M S T

1 I I G

2 _ _ _

(With F, H, and F/H remaining)

G visits both:

M S T

1 G F G

2 I I H

WHEW! That was brutal. Now letâ€™s look at the question. We are now substituting out our F/H rule, but we need a rule that will get us to these same answer possibilities (â€œwould have the same effect in determining the assignment of managers to cities.)

What does that mean? The wrong answer will EITHER knock out an existing game possibility or it will create a game possibility that we didnâ€™t have before.

A says G and I canâ€™t be together, but we know from our possibility 3 that G and I visit M. So A is wrong, because it knocks out that possibility.

B says that if F visits S, then H has to visit T. But in our possibility two, F could visit F, and F could visit T, and H would then visit M. So that possibility was ok according to our old rules, and according to B it would not be ok. So B is wrong.

D says that any visit that F does not visit, must be visited by H. Letâ€™s check: Possibility 3 allowed us to have a city (M) that was not visited by F, but it wasnâ€™t visited by H. So answer D is wrong.

C is the hardest wrong answer, so I am not surprised it is the one that tripped you up. C says that T must be visited by either F or H, but not both. C seems like it could be the right answer, because it exists with all 3 of our game possibilities. But remember, we also canâ€™t have an answer that gives us a new possibility â€” the rule needs to have the â€œSAME EFFECT.â€

Here is how you can prove that C does not have the same effect. Letâ€™s draw a possibility that is fine under C, but would be a violation of the old rules.

M S T

1 G F H

2 I H I

Notice that we have satisfied the new rule: T is visited by either F or H, but not both. And all of the other old rules are also satisfied, except for the clear violation of the old rule we subbed out (F and H cannot be together.) While that old rule is gone, we need to have a new rule that would have the â€œsame effectâ€ in determining the assignment of managers. And because it violates the old rule, it doesnâ€™t have the same effect. That is, under this new rule, we have a answer possibility that could not have existed under the old rules, so it cannot have the same effect in determining the assignments.

If you made it to here, bravo! This was an incredibly tough question. Rest assured that most logic game questions are not this hard, and do not take nearly as long to work through.

on November 21 at 11:08PM

What is the difference in A and E?