LSAT Trainer LG Question pg 180

JustDoIt

So on page 180 (the irony) of the trainer, the last question of the second scenario brings up an interesting point. The scenario is sequencing with in out and the last question of the drill says "If N is performed before M, P will be performed before Q." Where only 5 elements are to be selected, does this mean that N-M guarantees that both P and Q are in? Or if one of P or Q are out, is the necessary still satisfied?

I know this is completely hypothetical and is supposed to be a drill. But I'm not sure if something similar will come up on an actual exam. Sorry if this is a silly question!

BinghamtonDave

Not a silly question at all. As written, it would be my understanding that the rule would imply that if N is performed before M then P and Q have to be both in. Being "in" in this instance here would seem to be a necessary condition for a P-Q ordering.

Alejandro

Yes! If N-M then you absolutely have to have P and Q in on a P-Q relationship. That only gives space for 1 more element in the group of 5. Also, if either P or Q are out then absolutely N cannot be before M. This holds whether M-N or whether either M and/or N are out. Hope that helps!

Alejandro

and yes, this is a crucial exercise. You can definitely expect this type of logic tested on the exam!

Cant Get Right

@JustDoIt said:
on page 180 (the irony)

Love this!

And yeah, @David3389 and @Alejandro have you covered.

JustDoIt

Got it. Thanks everyone!