#17 - Local, Must Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?f=362&t=3362)

The correct answer choice is (B)

If V and W are assigned to the same boat as each other, then in order to comply with the third rule they must be assigned to boat 2. This leaves room for only one adult on boat 2, and so two adults must be assigned to boat 1:

G3-Q17-d1.png

Because the second rule stipulates that when F is assigned to boat 2 then G is assigned to boat 2, F cannot be in boat 2 because there is not enough room for G as well. Thus, F must be assigned to boat 1. The remaining two adults—G and H—cannot be determined and they form a dual option. Y, the only unassigned person, then fills in the final spot on boat 1:

G3-Q17-d2.png

F and Y must be assigned to the same boat, and thus answer choice (B) is correct.

[ellenb]

Dear Powerscore,

For # 17 why did we chose to have F in in the first boat? and H/G dual option?

Thanks!

[Adam Tyson]

We know from the second rule that if V is in 1, W is in boat 2, and vice versa, so that if V and W are in the same boat as each other it could only be in boat 2. Let's start there:

1. __ __ __ __
2. V W __ __

Now we need to keep X and Z apart - those two don't get along:

1. X/Z __ __ __
2. Z/X V W __

Rule 2 kicks in here - if F is in 2, so is G. We don't have enough seats in boat 2 for both of them, though, so F has to go to boat 1, and the last seat in boat 2 must be filled with one of the other adults, G or H. That gives us this:

1. X/Z F G/H Y
2. Z/X VW H/G

See it now?

Hope that helped!

Adam M. Tyson
PowerScore SAT and LSAT Instructor