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 at 05:49PM

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?

2 Replies

Irina on October 18 at 09:39PM


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.

on February 27 at 04:30PM

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?