December 2010 LSAT Section 4 Question 16

Which one of the following is a complete and accurate list of the talks any one of which Rivera and Spivey could atte...

Meredith on October 18, 2019

Approaching this problem

How is it you can be comfortable with listing out what can and cannot be options for R and checking that with our deductions about S to arrive at your answer? I felt I had to double check I and L were options even though there were not restrictions for R or S for those talks since I didn't know if other rules would force them not to be options. How'd you know you didn't have to double check by working them out?

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Irina on October 18, 2019

@Meredith,

That's a great question. I think we do have to quickly check these scenarios, it should only take 30-45 seconds though. We know that R could only attend F I L. Let's try a scenario where they attend both talks together first:

F: R S
G: Q T
H: T
I : R S
L: Q

This scenario complies with all the rules so we only have to check L. We know that it is impossible for R S to attend I and L together because it would leave no room for Q since Q can only attend G I L. Therefore, we only need to try a scenario where RS attend F & L together.

F: R S
G: Q T
H: T
I : Q
L: R S

This scenario also complies with all the rules, so we can conclude that (A) is the correct answer.

joedonnelly840 on February 27, 2020

On question 16, how is it possible that L could be an option Spivey and Rivera? If each person needs to show up twice, and Spivey and Rivera show up together on Rivera's first talk, and L is the final talk, then there would be no second possible talk for Rivera to attend if he shows up with Spivey at L for his first talk. How is this possible?

Dro1215 on June 4, 2021

Yes it could, they could be First and Last. That one got me too