December 1991 LSAT
Section 1
Question 18

December 1991 LSAT
Section 1
Question 18

Reply

Jay-Etter on January 23, 2022

This question is rather unusual. Let's start with our diagram1: F/S, N/U, P/R

2: F/S, N/U, P/R

3: F/S, N/U, P/R

(there are multiple ways to diagram this one but I found this diagram to work).

After we put our rules in we get

1: F/S, N, P/R

2: F/S, N/U, P/R

3: F/S, U, P

(based on 1 being N, 3 being U, and U not with R).

The question is saying if new cars in the exhibition are research (aka. RN always paired), then which must be true:

We can pair the R and the N we already have in 1 to get

1: F, N, R (Not R because RS cannot go together)

2: F/S, N/U, P/R

3: F/S, U, P

Now, 2 is still pretty undetermined, so lets do two scenarios. 2 could either have New or Used, so lets draw those both out.

When 2 is new:

1: F, N, R

2: F, N, R

3: F/S, U, P.

When 2 is used:

1: F, N, R

2: F/S, U, P

3: F/S, U, P

Now we can look at our options.

A: Doesn't have to be true because we could have F and U in 3.

B: Doesn't have to be true because we could have FP in 3

C: Doesn't have to be true because we have FN in 1

D: Has to be true. Notice everywhere we have N we have F!

E: Doesn't have to be true, we could have PS in 3.

Hope this helps!

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