December 2017 LSAT Section 4 Question 1

# Which one of the following could be the tour's schedule, with the four cities included in the tour listed in the orde...

1 Reply

Irina on May 23 at 08:54PM

Hi @hfatima1,This is a linear game that requires us to determine the order of visits to four out of six cities - H J M O S T. Each city is visited only once. We have the following rules:

(1) H and T must be included in the tour but they cannot be visited consequently.

This rule tells us that H and T must appear on our schedule, and both appear exactly once but not next to each other, there must be at least one or two other cities between them.

___ ___ ___ ___ H T

(2) if O is included in the tour, S cannot be.

Conversely, we can infer that if S is included in the tour then O is not. These two visits are mutually exclusive, meaning only one of them could be visited or neither, which leaves us with two possible scenarios for a complete list of visited cities:

H T S/O J/M

H T J M

O -> ~S

S -> ~O

(3) If J is visited, it must be third.

J -> 3d

(4) If both J & M are included they must be visited consecutively.

We can see that the only way both J and M could be included is if both S & O are excluded, meaning the complete list of cities visited in that scenario is:

H T J M

Since JM must be consecutive and H T cannot be consecutive, we can infer that H and T must be in positions 1 or 4 and MJ in positions 2 and 3 respectively (remember J must always be 3d)

H/T M J H/T

Putting all the rules together we have the following setup

(1) H/T M J H/T

(2) ___ ___ __ __ H T S/O M/J

~HT (not consecutive)

J = 3d

The question asks us which of the following could be the tour's schedule:

(A) J T S H

Incorrect. J must always be third.

(B) M T J H

Incorrect. MJ must be consecutive when they are included

(C) O H S T

Incorrect. O and S are mutually exclusive.

(D) S T M H

Correct. This order complies with all the rules.

(E) T M J S

Incorrect. H and T must always be included.

Question 2 is asking if S is visited 4th which of the following must be true?

If S is visited 4th we know that we are in scenario 2, and H/T must be visited 1st and 3, thus J cannot be visited (it must always be 3d), and the only remaining city - M must be visited 2d.

H/T M H/T S

Let me know if this makes sense and if you have any other questions.