LSAT 31, Game 4 Setup, Work Crew, LSATHacks

This is an explanation of the fourth logic game from Section I of LSAT Preptest 31, the June 2000 LSAT.

A work crew of up to five workers will be chosen from a group of seven. The seven workers are: George, Helena, Inga, Kelly, Leandra, Maricita, and Olaf (G, H, I, K, L, M, O). The crew will install a partition within three days, at most. They will complete five tasks in this order: framing, wallboarding, taping, sanding, and priming (F, W, T, S, P).

Game Setup

This is an unusual game, I’ve only seen one other like it. If you don’t set it up well (as I didn’t, the first time), it’s a very, very difficult game. But with the right diagram, things become simpler.

The key is arranging the diagram by tasks. The tasks are the central point. Different people can do each task, and the tasks are done on different days.

Those are the five tasks. The next step is to write each person who can do the task underneath.

For instance, G does T, so you can write G underneath T. H can do S and P, so you write T under those letters:

After making this kind of drawing, do yourself a favor and double check that you drew everything right. A mistake on this sort of list could be fatal.

The setup says that a crew of “up to five” workers will do the tasks. There are seven workers. So not every worker will be part of the crew. 

Three sets of workers are very important

There are three pairs of important workers however:

1. Only I and K can do F
2. Only L and O can do W
3. Only G and L can do T

So at least one person from each of those pairs must always be in the game. This knowledge answers questions 19 and 21.

Be sure to understand the effect of rules 1, 3 and 4. They make something simple seem complex. Here is what they mean:

“Tasks are done by separate workers, and finished in a day. There are no days without tasks.”

The tasks can be done in two days or three days

The number of days is important. The setup says the work takes “at most” three days. 

That means the tasks could be done faster. But not too fast – rule 2 says that T and P are on different days. So two days is the minimum. I’ve drawn the days over the diagram, for a two day job:

In this two day setup, T is on day one, and P is on day two. S could be done either on day one or day two.

A three day setup is also possible. There are more possibilities and it looks a bit messier, so bear with me:

F is on the first day, and P is on the third. T can be on days one or two; it just can’t be the same day as P.

S can go on day two or three. S can’t be done on day one, because then no task would be done on day two.

(Remember, the above setup is for a three day job)

I drew the possibilities for three day and two day installations over my main diagram:

These aren’t the main part of the game though. If you only know three things about the number of days, they should be:

1. T and P aren’t on the same day.
2. The tasks can be done in two or three days.
3. Each day has to have at least one task.

And remember, the other main point is that one person from each of these three pairs has to work:

1. IK
2. LO
3. GL

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