LSAT: Questions

Start out by SEALing the game. The Situation is doctoral candidates defending their theses. The Entities are A, B, C, D, E, F, G, H, and I. The Action is distribution, as we have nine candidates defending on three days, and our task is to figure out who is defending when. Limitations? 3 candidates per day.

Our sketch is pretty simple: 3 days, 3 per slots per day.

M: _ _ _ T: _ _ _ W: _ _ _

Go on to the rules.

The first rule states that B defends his thesis on Tuesday. That is concrete, and can be built directly into the sketch, giving us:

M: _ _ _ T: B _ _ W: _ _ _

The second and third rules are similar in that they dictate two pairs that must always defend their theses on the same day. This is essential. It means that wherever you place A, you must place E as well, and that wherever you C goes, F must follow. You can abbreviate these rules as ALWAYS AE, and ALWAYS CF.

The same principle applies to rules 4, 5, and 6. They tell us about entities that can NEVER be placed together. As such, whenever, you deal with G, always remember that H cannot be placed on the same day, and vice versa. The same goes for B and I and for C and I. You can put these rules in shorthand as NEVER GH, NEVER BI, and NEVER CI, respectively.

It does not seem like we have a whole lot of concrete information to work with - only one concrete rule that could be built into the sketch and no if-then rules to contrapose. But there are still a couple of really important things that we can deduce.

We know that A and E must be together, and that C and F must be together. So can A and C ever be together? No. A and C both carry along baggage (E and F, respectively), and there are only three slots per day, so A and C can never defend their theses on the same day. By the same token, A and F, E and C, and E and F are also all unacceptable.

Also, we know from the rules that C and I cannot be together. We also know that C must always be with F. Therefore, F and I can also never be together.

Update our never list - the list of pairs that can never be together:

NEVER:

G and H
B and I
C and I
F and I
A and C
A and F
E and C
E and F

Several questions hinge on these essential deductions, so if you were able to deduce them before getting into the questions that should have been a huge help. If not, you could still have gotten to the same conclusions indirectly through trial and error.

This question is an acceptability question. Use the Kaplan Method of using the rules to knock out the violators.

(A) is no good because A and E must be together.
(B) is out because G and H are together on Wednesday, which violates the fourth rule.
(C) can be eliminated because C and F, according to rule 3 must be together, but in this choice they are not.
(D) is also wrong because I and C are together, which, according to rule 6, they should not be.
That only leaves us with (E), which is the correct answer.