June 2007 LSAT, Logic Game 2, Question 8

This question wants us to place the maximum number of films. But once Greed is shown, we can’t place Limelight. Here’s the main setup diagram for reference:

We need to first figure out who to place last on each day. We should place G last on the final day, in order to maximize the number of times we can place L:

H has to go last on the first day. And since we’re not placing G, we must place L last on the second day.

Next, we want to put as many H’s and L’s as we can. The little letters under each day indicate who is left that can be placed there. We can see those that we can place L thursday, H friday and L saturday:

That’s a total of 6. D is CORRECT.

You might wonder if there’s another way. What if we tried to maximize G instead? We’d get the same result. Above, we drew two H’s, three L’s, and one G. If we instead placed as many G’s as we could, we’d have three G’s, and one L. So it works out the same way. (We’d still have two H’s, because G blocks H on Saturday). Here’s what it looks like:

You don’t need to draw that diagram in timed conditions, as long as you can think through the logic. And it makes sense that putting G last is likely to get the max value, since it lets us fit in three L’s. Basically we have a tradeoff between putting 3G/1L or 3L/1G.