October 1991 LSAT
Section 3
Question 19
Which one of the following statements CANNOT be true?
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Replies
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shunhe on August 17, 2020
Hi @ankita96,Thanks for the question! So remember, we know that S is a labrador, exactly one labrador wins a ribbon. We also know that S places ahead of Q and T, and since S finishes ahead of two dogs and only two dogs don’t get ribbons, S has to get a ribbon. So S is a labrador who has to get a ribbon. And exactly one female dog gets a ribbon, so S is the female labrador that gets a ribbon. U is a labrador, and since only one labrador gets a ribbon, U doesn’t get a ribbon. So now U, Q, and T are behind S, and P and R are ahead of S. Well, that means that S is in third place, and P and R are in first and second place, though we don’t know which is which. But that means first and second are both greyhounds. So there can’t be a labrador in second place. And that’s how we can eliminate (B).
Hope this helps! Feel free to ask any other questions that you might have.
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ankita96 on August 24, 2020
Yes that makes sense, thanks!!![](https://lsatmaxadmin.s3.us-west-1.amazonaws.com/556c5d9cab.jpg)
akinyiwilliams on December 24, 2020
So there is no scenario with Q and T in, where either P or R is out. Both P and T have to in the 4 with ribbons in any scenario, right?