Question #1740

Grouping Games Problem 9, question 23

Started by rockinrobinpa · started 2016-09-14 08:21 · last activity 2022-06-07 23:35 · 7 replies

As I work through the problem, I arrive at the correct answer (Answer choice E). However, the correct answer (that Nance & Palermo can't both be at Randsborough) seems to conflict with Rule 3 which establishes that if Longtree is at Souderton, then Nance and Palermo are at Randsborough. Why is it that when combined, Rules 4 and 5 turn Rule 3 into what seems like a "false" rule?? I realize you need the existence of the sufficient condition (Longtree at Souderton) in order to place N and R together at Randsborough for Rule 3 to apply, but it seems like Rule 3 could be made to conflict with Rules 4 and 5 in a problem. How would you know which rule (or combination of rules) would be deemed "True" while the others are deemed "False". Thanks.

Replies

  1. Mehran · 2016-09-14 19:06

    @rockinrobinpa thanks for your message. (E) cannot be true based on the combination of rules 4 and 5: Nr ==> Or ==> Ps Pr ==> Os ==> Ns In terms of how this affects rule 3, it means that L cannot be in S (i.e. L must be in R), because L in S leads to a situation that conflicts with rules 4 and 5 (i.e. Nr and Pr). Hope this helps! Please let us know if you have any other questions.

  2. rockinrobinpa · 2016-09-27 10:24

    Thank you for your reply. However, my question was meant to be more of a hypothetical one. Basically, I was wanting to know, if they were to ask a question similar to question 24, only wording it like: "If Longtree is at Souderton, which of the following must be true?"......, what would be the proper deductions you could make?? I would assume that the following would be true: Ls ====> Nr and Pr (Rule 3 must be true) But what (if anything) would be true with respect to the following? Nr====>Or ====>Ps (Rule 4 & contra positive of Rule 5) (this conflicts with Rule 3 and my hypothetical question)) Pr ====>Ks and Os (Rule 5). (Would any portion of this rule be "False"?? Thank you

  3. Mehran · 2016-10-04 02:08

    @rockinrobinpa this question would never appear because Longtree cannot be in Souderton. As I stated above, having Longtree at Souderton results in a situation that conflicts with our rules. Therefore, Longtree must be in Randsborough. Hope this helps! Please let us know if you have any other questions.

  4. kuvimec · 2020-07-29 15:16

    Can we use the contrapositive of rule 1 to rule out A?

  5. shunhe · 2020-08-04 18:39

    Hi @kuvimec, Thanks for the question! Unfortunately, we can’t use the contrapositive of rule 1 to eliminate (A). Let’s take a look at what that contrapositive is. The original rule tells us J at S —> K at R And so the contrapositive would be K at S —> J at R Now we’re asked here who can’t be a pair of doctors at Randsborough. So let’s say we put K or J in Randsborough. Well, neither of the sufficient cases are triggered, since they only tell us what happens if J or K are at S! So using the contrapositive to get rid of (A) would basically be a mistaken negation. It’s theoretically possible, based on this rule alone, that both J and K are at R; the rule basically tells us that both can’t be at S. Hope this helps! Feel free to ask any other questions that you might have.

  6. studying0 · 2022-06-03 02:15

    So is it okay to eliminate A through both necessary conditions that K and J are in R?

  7. Emil-Kunkin · 2022-06-07 23:35

    Hi Savannah0, I think the way to eliminate A would be to see that we could present a valid scenario for it with J and K in R. We know that if K is not at S, then P is not at R, so P is in S. This means that L cannot be in S. However, it looks like we could place N and O anywhere that is not prohibited. so, the following scenario is ok. R: J, K, L, N, O S: P For this sort of question type, I would actually try to see if any of the answer choices raise red flags, and prioritize attempting those before just going in order A-E.

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