PT42.S1.Q21 (G4) - for the school paper

Mike Baldwin

I do not understand how E can be the correct answer when it violates the rule stating that two students must have the same reviews. If O is given an S, then M must have a T and a U, but cannot have an S due to J having an S. Thus you would get J=S, K=T, L=U, M=U,T, and O=S,U,T. Can someone please help explain why E is the correct answer?

https://7sage.com/lsat_explanations/lsat-42-section-1-game-4/

akistotle

https://7sage.com/lsat_explanations/lsat-42-section-1-game-4/

@"Mike Baldwin" said:
If O is given an S, then M must have a T and a U, but cannot have an S due to J having an S. Thus you would get J=S, K=T, L=U, M=U,T, and O=S,U,T.

That conditional relationship isn't true. O can review Sunset.

The situation below satisfies all the rules and exactly three students are reviewing Undulation.

J: Tamerlane
K: Tamerlane
L: Undulation
M: Sunset and Undulation
O: Sunset, Tamerlane, and Undulation

Mike Baldwin

I see what I did now, thank you