LSAT 90, Logic Game 4, Question 21, LSATHacks

This is a local question. If H is performed on day 2, then it is also performed on day 7 (rule 2). There’s only one spot open before H, and rule 5 says J has to go before H, so J goes first. This in turn requires J to go fifth (rule 1).

Remember, since J and H go twice, all other variables go only once. G, L and O are left to place, and a single M after one of the H’s (rule 4):

The only rule remaining is that O can’t go beside M (rule 3). So, if you place M third, then O can’t go fourth. Apart from that, there are no restrictions.

From this diagram, we can see that L can perform on day 8, so C is CORRECT.

I just imagine L sliding to day 8 in my mind. But you can also draw it out. If L went on day 8, then M would go on day 3, O on day 6 (to be away from M) and then G on day 4:

A is incorrect since J must be played on Day 1, so G can’t go there

B is wrong: it would force J to go three times. J is already playing on days 1 and 5, so if J went third it would go three times in total. Each variable can go twice at most: only those on days 1/5 and 2/7 can repeat.

D is wrong. Since we need an HM bloc, M has to play on either Day 3 or Day 8. And M can only go once, since J and H are the plays which go twice. So M cannot play on Day 4.

E is wrong. J goes first in order to be in front of H (rule 5). Therefore J also goes fifth (rule 1), and so there’s no space for O to go fifth.

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