Which one of the following CANNOT be a pair of the doctors at Randsborough?

rockinrobinpa September 14, 2016

Grouping Games Problem 9, question 23

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.

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Mehran September 14, 2016

@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.

rockinrobinpa September 27, 2016

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

Mehran October 4, 2016

@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.

kuvimec July 29, 2020

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

shunhe August 4, 2020

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.

studying0 June 3, 2022

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

Emil-Kunkin June 7, 2022

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.