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December 2017 LSAT
Which one of the following could be the tour's schedule, with the four cities included in the tour listed in the orde...
on May 22 at 01:22AM
Can someone please show me the set up for this game?
on May 23 at 08:54PM
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.
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