June 2007 LSAT, Game 2, #8

Let's continue through Game 2 of the June 2007 LSAT.  Last week, I created a setup for the game and answered questions six and seven. Today we'll tackle number eight, which says "If Limelight is never shown again during the festival once Greed is shown, then which one of the following is the maximum number of film showings that could occur during the festival?" To me, that's a pretty nasty question, because it involves a new rule and a lot of potential different scenarios where we'll have to try to maximize the number of films shown. I'm going to use a bit of intuition here, because I don't want to waste too much time on any one question. The problem with a question like this is that there's no very clean and certain way to test it. (Contrast this with question number six, which could be answered with an idiotproof process of elimination.) I'm going to try to squeeze in as many movies as possible, but in doing so I will have to make some guesses. Every time I take a guess, there's a path that I didn't take that might have worked out better.

I might start with the answers here, and work backward. I know, because seven is the maximum number of movies listed in any answer choice (E), that that's the absolute limit for the number of movies. So if I can figure out a way to make seven work, then E is the answer. If I can't make seven work, but I can make six work, then I'll choose D. And so on.

Can I make seven work? Let's see:

If I'm trying to maximize the movies shown, why wouldn't I go ahead and show all three films on the first day?

Thursday:  L  G  H

The problem with this approach is that, once I show G, I now can't show L anywhere in the remainder of the schedule. So Friday I could show H, then G:

Friday:  H  G

And Saturday, I could show just ONE of G and H:

Saturday:  H/G (doesn't matter which one).

Hmm. That's a total of only six movies. I was hoping to get to seven. So I'm going to try one more scenario:

On Thursday, if I want to avoid the problem of showing G and being unable to show L ever again, then I've got to avoid showing G. So I'll start my Thursday schedule with just L and H:

Thursday:  L  H

On Friday, I'll again want to avoid showing G early so I'll show just H and L:

Friday:  H  L

On Saturday, I'll now show L with G:

Saturday:  L  G

And once again, I can only get to six.

As I said earlier, this is a supernasty question, because I still haven't explored all the scenarios. What if I showed just L and H on Thursday, for example, but then showed all three films on Friday? Would I be able to get to seven then? (The answer is no. If I showed all three films on Friday then I'd only be able to show one film on Saturday, so once again I'd be stuck at six.)

Rather than exhaust all possibilities here, once I sensed that I was getting stuck at six I'd be happy going ahead and picking D. I might miss a question like this once in a while, but I'll 1) usually get it right and 2) never waste five minutes trying to be sure. I have bigger fish to fry in the remaining of the section, so I'll make an educated guess here and move on. Our answer is D.