#6 - Local, Could Be True, Except

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=16406)

The correct answer choice is (D)

If L is in group 2, then from the contrapositive of the third rule H must be in group 2. Due to space restrictions, F and J must be in group 1, leaving K and N to form another dual-option:

D94_Game_#1_#6_diagram 1.png

The problem is still tricky because the dual-options are not completely free to rotate. The fourth rule affects both G and N, and assigning K and M—the remainder in each dual-option—to group 1 forces G and N into group 2 together, a violation of the fourth rule. Hence, answer choice (D) is correct.

Note that answer choice (B) is possible because G and N can be in group 1 together; the fourth rule just prohibits them from being in group 2 together, as discussed in the setup.

[rameday]

So the explanation in the back of the book still doesnt make sense to me.

I understand why FJ is in G1 and HL in G2 and i see the G/M and K/N dual option that is present in both groups but from there i have no idea how to get to the answer.

[Lucas Moreau]

Hello, rameday,

You're almost there! The trick is the contrapositive of the third rule. The third rule states that if H is in group 1, L must be in group 1.

H1 L1

So the contrapositive, if L is in group 2, means that H must be in group 2.

L2 H2

So we're looking, as you indicated, at a situation like this:

F J G/M K/N
H L M/G N/K

We know that K and M cannot be together, because that would cause it to look like this:

F J K M
H L G N

And if N is in group 2, G must be in group 1, which it isn't, so D is the correct answer choice.

Hope that helps,
Lucas Moreau

[apimlott]

Hello, I am confused why B is the right answer. I am using Powerscore's data and analytics interface, and question 6 says, "If L is in group 2, then each of the following is a pair of pair who could be together in group #1 EXCEPT."

so, correct answer would = Cannot be in group 1 together

but G and N can indeed be in group 1 together (as Dave says above), but not in group 2

is there a typo on the interface?

[Adam Tyson]

The correct answer to this question is D, not B, apimlott, and you are absolutely correct that G and N CAN be in Group 1 together (which is why B is not the correct answer). Our explanation at the top of this thread correctly identifies D as the correct response.