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

Mike BaldwinMike Baldwin Free Trial Member
edited September 2017 in Logic Games 4 karma

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/

Comments

  • akistotleakistotle Member ⭐ 🍌🍌
    edited September 2017 9382 karma

    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 BaldwinMike Baldwin Free Trial Member
    4 karma

    I see what I did now, thank you

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