December 1991 LSAT Section 1 Question 18

If all the new cars in the exhibition are research models, then which one of the following statements must be true?

idonije on October 15, 2021

please clarify

If R-->N and if R-->F then how could A be incorrect while D is correct? The rule is basically stating R-->N & F. So, how can we conclude R-->N-->F when no explicit order is stated?

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Jay-Etter on January 23, 2022

This question is rather unusual. Let's start with our diagram
1: 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!