#5 - Global, Must Be True, Minimum

[Dave Killoran]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?f=169&p=88159#p88159)

The correct answer choice is (C).

From the first two rules we established that either J or L must be on stage at all times. Thus, at least one dancer is always on stage, and answer choice (A) can be eliminated.

If J is on stage, then from the third rule F must be on stage, and from the fourth rule G must be on stage. Thus, with J on stage the minimum number of dancers on stage is three, and answer choice (E) can be eliminated.

However, instead of J on stage, L could be on stage. If L is on stage, then from the fourth rule G must be on stage. No other rules apply, and so L and G can be the only two dancers on the stage, eliminating answer choice (D) from consideration.

At this point, we have examined what occurs with both J and L. With J there must be a minimum of three dancers on stage, and with L there must be a minimum of two dancers on stage. Thus, because either J or L is always on stage, there is no option where there are fewer than two dancers on stage, and so answer choice (B) can be eliminated. Answer choice (C) is therefore correct.

[pwfquestions]

Is it correct that rule 1 does not conclude that either J or L must be on stage? You also need rule two to determine that, right?

[Adam Tyson]

That's correct, pwfquestions! The first rule means they can never both be onstage at the same time, but allows them to both be offstage. The second rule means they can never both be offstage at the same time, but would allow them to both be onstage. It's only when we combine those two rules that we can determine that there is always one onstage and one offstage!