#9 - Local, Must Be True

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?f=371&t=15040)

The correct answer choice is (E).

This question also hinges on Hurdling the Uncertainty. With G assigned to two committees, G must be assigned to either the Quality or Sales committee, we can infer that L must be assigned to the committee that G is not on from the action of the fourth rule. Thus, although we cannot be certain which committee G or L is assigned to, we can jump over this uncertainty and prove that because M must be assigned to one of the committees, and there is only space available on the Quality Committee, M must be assigned to the Quality committee.

G2-Q9-d1.png

Accordingly, answer choice (E) is correct.

[CKwederis11@gmail.com]

From my understanding it would be that:

F G H / F G M / F K L

or

F G H / F L M / F K G

Because F is all 3 that limits Q and S. In addition, G must be in 2 committees so he is in either Quality or sales as well. Because of the GL rule, that limits the options.

Therefore answer E "M is assigned to the quality committee" Must Be True

[Rachael Wilkenfeld]

Hi CKwederis11,

Exactly. Once you place the 3Fs and the information from the game rules and scenario, you only have three slots left: two in Q, and 1 in S. We know those three slots need to be filled by G, L, and M. G because of the local rules, and L and M because the rule that all employees must be on at least one committee. Once we have that knowledge, we know that we need to have L and G on different committees. We put one on Q and one on S, leaving one spot for M in Q.

Great work!