June 2013 LSAT
Section 2
Question 20

June 2013 LSAT
Section 2
Question 20

Reply

Skylar on December 27, 2019

@nivensdc, maybe I can help.(B) and (C) are eliminated because if we have either K or L as the lone assignment to R, that leaves 4 variables to place. We know exactly 2 must go to T, which leaves exactly 2 to go in S. Under these conditions, it would be impossible to have either F or G as the sole assignment to a case, because there is no alone spot left. This violates Rule #2.

(D) and (E) are eliminated because they also violate Rule #2, which states that either F or G (but not both) must be the sole assignment to a case. We know that there are exactly 2 assignments to T, and both (C) and (D) assign exactly 2 variables to R. Therefore, neither F nor G can be assigned to T or R as a sole paralegal. This leaves S. However, Rule #3 tells us that H is assigned to S, so it would be impossible for F or G to be the sole assignment there either.

(A) is correct. Here is one valid scenario where G is the complete assignment to R and T has exactly two assignments.

R: G

S: HK

T: FL

Does that make sense? Let us know if you have any other questions or would like a more in-depth explanation of the setup of this game.

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