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