#11 - Global, Cannot Be True, List

[Administrator]

Complete Question Explanation
(The complete setup for this game can be found here: lsat/viewtopic.php?t=13972)

The correct answer choice is (B)

This is the fourth List question in this game. This question is identical to question #10, but this time the focus is on R, not J.

As shown in the discussion of the second rule, when L is addressed to R, the consequence is that F or S or both must be addressed to G. If both L and S are addressed to R, then F must be addressed to G, but there is no other piece of mail available to be addressed to G, creating a violation of the third rule (the other two variables are P and M; P must be addressed to J when L is addressed to R, and M can never be addressed to G according to the first rule). Therefore, answer choice (B) cannot occur and is correct.

[frank.dawson]

Hi,
would you please explain if L and S are pieces of R and M and P are pieces of J then why F can not be piece of G and if that is correct why B is the correct answer choice ? Answer choice E also besides B appears to have only S as a piece of G. I am still not clear why answer choice B is the correct one. Would you please explain in detail ?
Thank you

[Adam Tyson]

Hey there Frank! Take a look at the last rule of the game - F cannot be the only piece of mail addressed to anyone. If L and S go to R and M and P go to J, then F goes to G and is all alone. Whoops, can't let that happen!

The problem with B is that if L goes to R, P must go to J per the 2nd rule. What's left to go to G? M, S, and F. But M cannot go to G per the first rule, so that leaves only F and S. Since F cannot be there ALONE, S must go to G, and that means it cannot go to R. The second rule ends up telling us that when L goes to R, P goes to J, and when P goes to J, S MUST go to G.

The issue is not that F cannot go to G, but that F cannot go to G ALONE. It needs a partner, and in this case S has to be that partner.

I hope that helps!