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Hi guys, I am taking the december LSAT, and I just tried practising this game. @"J.Y. Ping" doesn't explain this part of the game. Please somebody help.
Why does P HAVE to precede T?
From Rule 5 we know P precedes T, if F has left any message.
If F has not left any message the rule should fail, Thus P doesn’t have to precede T.
Thus in Q2 option E should also be right and we will be left with 2 right answers.
P.S. Sorry for not posting the entire game here, too much to type and not sure if it may constitute a violation of LSAT's rules.
https://classic.7sage.com/lsat_explanations/lsat-30-section-1-game-2
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4 comments
@steve898 said it perfectly, what is happening here is a violation of the first rule on the list. I should add that something I have learned from a painstaking climb up the LG ladder is that if you somehow have 2 answer choices that are correct, you have more than likely not applied a rule somewhere.
Often looking smart about something, as I may appear to above, is a result of earlier feeling quite stupid about it.
Thanks. I feel so stupid now! Never get games wrong. Guess I was too stressed while doing it :p .
If (E) is correct, then T left the 1st and 6th message. Then we have four slots, 2 through 5, to fill. But at most one person, T in this case, can leave more than one message. In other words, if T leaves two, then we need at least four persons remaining from among F,G,H,L,P. If T leaves the first, then we cannot have F, because of rule 5: F—>(P & P—T). But if we don't have F, then we also don't have G, per rule 4's contrapositive. That leaves us with at most H,L,P to fill the four slots with no duplicates allowed, so (E) cannot be correct.