#2 - Local, Must Be True

[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 (B)

If K and N are in a group together, they create a block. F and J are also a block, and the two blocks cannot be in the same group due to lack of space. Thus, the two blocks are in separate groups, and we can reserve two more spaces in each group, even though we cannot determine exactly which variables are in each group:

D94_Game_#1_#2_diagram 1.png

With three spaces in each group reserved, we can Hurdle the Uncertainty and determine the placement of the remaining two variables, H and L. Because from the third rule if H is in group 1 then L must be in group 1, and there is insufficient room in group 1 for two more variables, we can infer that H cannot be in group 1 and therefore must be in group 2. Accordingly, answer choice (B) is correct.

To complete the diagram, as there is no more space in group 2, L must be in group 1.

[jpincus]

I see why B is the correct answer because it would make the groups consist of:
1: F,G,J,L
2: H,M,N,K

However, I am not seeing why choice E is wrong. If m is in group 1 couldn't it be diagrammed as:
1: N,L,H,M
2: K,F,J,G

Thank you!

[Dave Killoran]

Hi J,

Thanks for the question! First, your hypothetical has an issue since the question stem stipulates that "K is in the same group as N," and yours separates the two.

Second but equally importantly, keep in mind that this is a Must Be True question, so it's not whether M could be in Group 1, it's whether M must be in group 1. And, as it turns out, M can be in Group 2, so while (E) is possible, it is not something that has to occur.

Here's a hypothetical showing M in Group 2:

1: G - F - J - L
2: M - K - N - H

Please let me know if that helps. Thanks!