This is an explanation of the first logic game from Section III of LSAT Preptest 66, the June 2012 LSAT.
Six lab sessions for a chemistry class will be held over a span of three days: Wednesday, Thursday and Friday (W, T, F). One session will be in the morning and one in the afternoon. There will be six lab assistants, one for each session. They are Julio, Kevin, Lan, Nessa, Olivia and Rebecca (J, K, L, N, O, R). You must assign the lab assistants to labs based on the rules.
Game Setup
This game is a mix of linear and grouping. It tests your ability to remember the rules. If you have them in your head and can combine them, then this game is easy. If you have to constantly check the rules, then this game is hard.
I set up the game horizontally. If you set it up vertically, that’s fine too. It’s a matter of preference.
Next, KR have to be on the same day, in either order:
The line indicates that K and R can be reversed. Think of it like a suitcase handle.
L and O can’t be on the same day:
N can only go in the afternoon. I drew this on the diagram itself. I used a ‘not’ rule to show that it can’t be in morning.
That’s what we know for certain, it’s best to stick to what is definite.
Finally, J comes on an earlier day than O.
This diagram is slightly ambiguous. It just shows that J comes earlier. Morning is earlier than afternoon, so you have to remember that it means J comes on a day earlier than O does.
There aren’t any obvious deductions, but you can make some by thinking about how the rules fit together. If you suspect a game is restricted, it helps to make a sample scenario to see how things fit together.
KR take up an entire day. This is very restrictive. L/O are split between the other two days, and J-O are also split between the other two days.
In fact, JL always have to go together, and ON always go together. I’ll show this with an example.
By the way, I didn’t figure this out explicitly the first time I did the game, though I knew it intuitively well enough and could draw fast diagrams. You don’t need to make a perfect diagram to do well on a game.
Ok, time to prove the deduction. Let’s see what happens if you put KR on Wednesday. Remember that they’re reversible:
J goes before O (rule 4), so J is on Thursday and O is on Friday. Since L and O can’t go together, then L must also go on Thursday with J.
I haven’t placed O yet. I’ll get to that. First, I’ll explain what I’ve drawn.
KR have a box around them as a reminder that this is based on a rule.
JL have a line between them to indicate that they’re reversible, but there’s no box because no rule says they must be together. It just so happens that the other rules force them together.
This is an arbitrary style I made up for this game, because it’s unique. You could easily draw a box around J/L and do fine.
Ok, now let’s place O. We only have O and N left to place. And N can’t go in the morning. So this is the only possible diagram when KR go Wednesday:
If KR go Thursday or Friday then we get slightly different diagrams, but the principle is the same:
So there are only three possibilities for this game, and they’re based on the same idea. JL (in either order) are before ON (in that order). KR (in either order) fill one of three days.
Important. For the rest of the game, I’m going to say things like JL must come before ON. You must reread the above explanations if you don’t see why JL comes before ON. Otherwise this game won’t make sense.
The deduction is based on a combination of rules, which force JL to always go together. J goes before O, and L can’t go with O. KR splits JL and ON into two groups.
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Ben says
If you don’t make the JL and ON deduction, how does that change how you do the game? All of your answers to the questions seem to presuppose this deduction.
TutorRosalie (LSATHacks) says
Since the K/R block is determined, you can break down the game into three scenarios on which day K/R happens. You’ll realize as you place the other factors onto the board, that JL and ON have to be the case. You know that J has to come before O, and that L and O can’t go together. Since K/R are taken, then O can only go with N.