Difference between revisions of "1985 AJHSME Problems/Problem 25"
5849206328x (talk | contribs) (New page: ==Problem== Five cards are lying on a table as shown. <cmath>\begin{matrix} & \qquad & \boxed{\tt{P}} & \qquad & \boxed{\tt{Q}} \\ \\ \boxed{\tt{3}} & \qquad & \boxed{\tt{4}} & \qquad &...) |
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==Solution== | ==Solution== | ||
| − | Logically, Jane's statement is equivalent to | + | Logically, Jane's statement is equivalent to its [[Contrapositive|contrapositive]], |
If an even number is not on one side of any card, then a vowel is not on the other side. | If an even number is not on one side of any card, then a vowel is not on the other side. | ||
Revision as of 21:15, 12 January 2009
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?
Solution
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 
