#5 - Local, Must Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?f=1250&t=16333)

The correct answer choice is (C).


If O is visited second, then from the second rule S cannot be visited. Thus, the group of cities visited is limited to H, T, S, and then either J or M. Because H and T cannot be consecutive, one of H and T must then be first (and the other is third or fourth):

• _H/T_ _O_ ___ ___
    1           2    3    4

The third and fourth spaces must be occupied in some order by a choice of the remainder of H and T (T/H), and one of J or M (J/M). So, at first glance, it might seem like the answer here is four. However, don't forget that J can only be visited third, and so J can be removed from the list. Thus, the answer is three: H, T, and M.


Answer choice (C): This is the correct answer choice. Only H, T, and M could be visited fourth under the condition in the question stem.

[cmnoury1221]

Hello,
I want to be sure I had the right set ups for this question:
I initially chose (B) - 2 , but realized it is (C).
When O is second in one diagram H or T can be fourth and in a second diagram, when O is second M can also be fourth.
Is that the right way to get the answer here?

Thank you!
Carolyn

[Claire Horan]

Hi Cmnoury1221,

There are multiple ways to get to the right answer, and yours is one of them. Another way to think about it: there are 5 cities besides O. S can never be visited when O is visited. J can never go fourth. That leaves us with three cities. Since we can find at least one arrangement with O second and H, M, and T fourth, the answer is three, or C. It is a good idea to go from the highest number down, so that you are eliminating cities that cannot go fourth, rather than trying to generate combinations where each city goes fourth.

Thanks for the question!