1985 AJHSME Problems/Problem 25
Problem
Five cards are lying on a table as shown.
![\[\begin{matrix} & \qquad & \boxed{\tt{P}} & \qquad & \boxed{\tt{Q}} \\ \\ \boxed{\tt{3}} & \qquad & \boxed{\tt{4}} & \qquad & \boxed{\tt{6}} \end{matrix}\]](http://latex.artofproblemsolving.com/f/a/e/fae700ade12943277f8147039b058865f3fff819.png)
Each card has a letter on one side and a whole number on the other side. Jane said, "If a vowel is on one side of any card, then an even number is on the other side." Mary showed Jane was wrong by turning over one card. Which card did Mary turn over? (Each card number is the one with the number on it. For example card 4 is the one with 4 on it, not the fourth card from the left/right)
Solution 1
Logically, Jane's statement is equivalent to its contrapositive,
If an even number is not on one side of any card, then a vowel is not on the other side.
For Mary to show Jane wrong, she must find a card with an odd number on one side, and a vowel on the other side. The only card that could possibly have this property is the card with 
, which is answer choice 
Solution 2
Using the answer choices, we see that P and Q are logically equivalent (both are non vowel letters) and so are 
and 
(both are even numbers). Thus, if one of these are correct, then the other option in its pair must also be correct. The only remaining answer choice is
Solution 3
I have a pretty odd solution that I personally used when I was practicing for the AMC 8. At the time, I really didn't have any idea about how to solve this problem, but I noticed something that might give you an example on how to get the answer to a problem without even doing any math. This doesn't really have any logic or imagination behind it, so if it isn't recognized as a legitimate solution, I wouldn't argue-I simply pointed this out because I found a different way to solve this problem. 
If you really understand the problem after reading it, you'll notice that there is only one solution to this problem(Please don't think "duh". There are people out there who haven't fully mastered math yet.) Basically, the problem wants a card that is special and different from others, since that's the one that proves that Jane's theory is incorrect. Looking at the cards, notice that there are two pairs of cards that are similar to each other. The cards 
and 
are both consonants, and since there is only one solution, both of these cards can be considered wrong. The cards and are both even, and since they are similar and there can be only one solution, you can also consider these cards wrong. Therefore, 3 is the right answer.
See Also
| 1985 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
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| All AJHSME/AMC 8 Problems and Solutions | ||
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