Group Games Questions - - Question 18

If S is one of the birds exhibited, it must be true that

randy-nodarse May 16, 2019

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

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?

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randy-nodarse May 16, 2019

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

Ravi May 16, 2019

@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!

elawrencehenderson June 17, 2020

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?

rachelsilver July 11, 2020

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?

rachelsilver July 11, 2020

*to Q15 to show R caged

Erica July 28, 2020

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!

Dieg) December 18, 2021

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

Ravi February 7, 2022

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