December 2010 LSAT
Section 4
Question 17

December 2010 LSAT
Section 4
Question 17

Reply

Irina on October 8, 2019

@Samir-Ghani,That's correct, three of the talks must have two employees. The question tells us that Q is the only employee to attend L, meaning we have four more talks - F G H I that are each attended not more than twice by Q R S T, thus we do have 2-2-2-1-1 scenario as you correctly identified.

If Q attends L, we can infer that R must attend F and I since R cannot attend G, H, or L per the rules.

If R attends F, we can infer that S must attend F as well per the rules.

Now, we know that Q attends the first talk that T attends per the rules and Q attends neither F nor H, we can infer that Q must attend G since Q&T must go together, and we only have one spot open in I.

Since S and T cannot attend the same talks, we can infer that one of them must attend H and one must attend I in either order. Thus, we can conclude that the only answer that could be false is (D) S could attend I instead, and T attend H in that scenario.

F: R S

G: Q T

H: /S/T

I: R /S/T

L: Q

Let me know if you have any further questions.

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