October 1991 LSAT - Section 3 - Question 19

Which one of the following statements CANNOT be true?

ankita96 August 15, 2020

Explanation of the answer

Hi, Can I get an explanation for this answer. I am unable to eliminate B. Thanks in advance!

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shunhe 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.

ankita96 August 24, 2020

Yes that makes sense, thanks!!

akinyiwilliams 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?