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All batangos are crentons , some franters are volns, and some kijuxes are not batangos. If no kijuxes are franters, which of the following CANNOT be true, given each group was observed to have no more than 2 members.

a) All volns are crentons, as are all franters
b) those kijuxes that are volns are not crentons
c)all kijuxes are volns and some are batangos
d) some batangos are franters and some are kijuxes

I have a word solution, and I can post that if requested, but I was wondering if there was a way to construct venn diagrams that encapsulate the information and are easy to read off.

The challenge is to solve this in 1 min and 30 secs so I imagine a venn diagram would be much faster than trying to run through each of the cases. So the challenge is to not only find a solution, but find the fastest way to solve this.

PiGuy314
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stackdsewew
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    Does the title belong to the problem statement? If so, please move it to the body of the question and add a real title. – Magdiragdag Jul 13 '16 at 04:23
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    It took me quite a bit more than $1$ minute and $30$ seconds and I didn't use Venn diagrams, so I'll just post this as a comment: I solved it by noting that the original restrictions on kijuxes are quite weak; they could be anything as long as they're not batangos or franters. Then I went through from the top and noted that a) and b) also don't place significant restrictions on the kijuxes, with b) not restricting them as long as they're not volns; and it's not hard to see that, ignoring the kijuxes, a) and b) don't generate any contradictions. – joriki Jul 13 '16 at 06:01
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    c), on the other hand, does draw a connection between the kijuxes and the other groups, so there's at least a chance for a contradiction here. And indeed c) forces there to be two kijuxes (one a batango and one not), they both have to be volns and both mustn't be a franter, so there are two volns that aren't franters, yet some franters are volns, which would make more than two volns, a contradiction. – joriki Jul 13 '16 at 06:03

1 Answers1

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First of all to make things easier, assign a simpler title to each category.
Consider batangos to be $A$.
Consider crentons to be $B$.
Consider franters to be $C$.
Consider volns to be $D$.
Consider kijuxes to be $E$.

You can rewrite the logic statements with the new names to get:
All $A$ are $B$.
Some $C$ are $D$.
Some $E$ are not $A$.
No $E$ are $C$.

This allows for an easier digestion of the important information in the question. The logic to proceed from here has been wonderfully covered in a comment by @joriki.

PiGuy314
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