September 2016 LSAT
Section 3
Question 14

September 2016 LSAT
Section 3
Question 14

Reply

Skylar on December 21, 2019

@Ryan-Mahabir, happy to help!We are looking for the answer choice that CANNOT work.

(D) states that V and Z are the complete assignments to L. Rule #4 tells us that Y and Z must both be assigned to P if either is, and since Z is already assigned to a different class, we know that we can't put Y in P. Therefore, we must put it in M. Since there are exactly two spots in M, we must fill the second spot with R. S and T, which must go together, are both assigned to P. This gives us the following valid scenario:

L: VZ

M: YR

P: ST

(B) states that R and Z are the complete assignments to L. As in answer choice (D), we must remember that #4 says Y and Z must both be assigned to P if either is, and since Z is already assigned to a different class, we know that we can't put Y in P. Therefore, we must put it in M. There is exactly one spot left in M (because Rule #1 says M has exactly two assignments), so we cannot put S and T there because Rule #2 says they must go together. Therefore, S and T are assigned to P and the only remaining option to fill the second spot in M is V. This gives us:

L: RZ

M: YV

P: ST

However, this is an invalid scenario because it violates Rule #3, which states that V and Y can not be assigned to the same course. Therefore, (B) is the correct answer for a scenario that CANNOT work.

Does that make sense? Please reach out with any other questions!

GET $100

GET $200