Difference between revisions of "1985 AJHSME Problems/Problem 25"
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Using the answer choices, we see that P and Q are logically equivalent (both are non vowel letters) and so are <math>4</math> and <math>6</math> (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 <math>\boxed{\text{A}}</math> | Using the answer choices, we see that P and Q are logically equivalent (both are non vowel letters) and so are <math>4</math> and <math>6</math> (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 <math>\boxed{\text{A}}</math> | ||
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| + | Solution 3 | ||
==See Also== | ==See Also== | ||
Revision as of 18:13, 27 September 2020
Contents
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
See Also
| 1985 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Problem | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
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