Approaching this problem

Started by Meredith · started 2019-10-18 17:49 · last activity 2021-06-04 09:48 · 3 replies

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?

Replies

  1. Irina · 2019-10-18 21:39

    @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.
  2. joedonnelly840 · 2020-02-27 16:30

    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?
  3. Dro1215 · 2021-06-04 09:48

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

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