Question #1717

R is assigned to the cage because the question asks for it

Started by randy-nodarse · started 2019-05-16 16:36 · last activity 2022-02-07 03:07 · 8 replies

Can I have another explanation on how to answer this question? 15 states "if R is assigned to a cage" so wouldn't that question not count as a valid scenario?

Replies

  1. randy-nodarse · 2019-05-16 16:52

    Never mind, I realized the rules would not allow for S and W to be exhibited at the same time. However, I would still like to know if I can use the scenario in #15 as a reference to answer this question because 15 asked for R to be in the cage? Thank you

  2. Ravi · 2019-05-16 19:03

    @randy-nodarse, Happy to help. If I were you, I wouldn't do that because you'd be confusing the sufficient for necessary conditions. In question 15, we know that R is in a cage, so R being in a cage is the sufficient condition. It's best to just work from what question 18 gives us. We're told that S is exhibited. What must be true? Based on rule 2, we know that any two birds that are both of the same sex and of the same kind as each other can't be caged together. Q, R, and S are all male parakeets. If S is exhibited, this means that Q and R must both be in cages since none of these birds can be caged together. This leads us to (E), which says R is assigned to a cage. Does this make sense? Let us know if you have any more questions!

  3. elawrencehenderson · 2020-06-17 05:51

    So, when the prompt refers to birds of the same sex and breed being unable to be caged together that also extends to being unable to be exhibited together? Is exhibition not different than being caged?

  4. rachelsilver · 2020-07-11 15:06

    I am also wondering why S and R can't be exhibited together (in different pairs)? In the answer explanation video, it refers to question 15 to show R not caged but that is only a could be true scenario. I can't find any restrictions that indicate R must be in a cage?

  5. rachelsilver · 2020-07-11 15:07

    *to Q15 to show R caged

  6. NULL · 2020-07-28 04:14

    R must be caged, because exhibitions are always 1m & 1f. PM: Q,R,S & PF: T,W. One of the rules states that if J or W are exhibited then S cannot. Contra positive is If S is exhibited then J & W cannot. S is being exhibited, and T must be its partner. Therefore, R must be caged, because the only other PF is W, which cannot be exhibited along side S. Hope that simplifies it!

  7. Dieg) · 2021-12-18 05:18

    This helps, thank you, Erica. I suspect they just missed this in the video bc it seems important to point out

  8. Ravi · 2022-02-07 03:07

    Erica's comment is on the money. @Dieg), I'm happy you were able to sort things out!

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