Difference between revisions of "1985 AJHSME Problems/Problem 25"
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\boxed{\tt{3}} & \qquad & \boxed{\tt{4}} & \qquad & \boxed{\tt{6}} \end{matrix}</cmath> | \boxed{\tt{3}} & \qquad & \boxed{\tt{4}} & \qquad & \boxed{\tt{6}} \end{matrix}</cmath> | ||
| − | 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 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) |
<math>\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ \text{P} \qquad \text{(E)}\ \text{Q}</math> | <math>\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ \text{P} \qquad \text{(E)}\ \text{Q}</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Logically, Jane's statement is equivalent to its [[Contrapositive|contrapositive]], | Logically, Jane's statement is equivalent to its [[Contrapositive|contrapositive]], | ||
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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. | ||
| − | 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 <math>3</math>, which is answer choice <math>\boxed{\ | + | 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 <math>3</math>, which is answer choice <math>\boxed{\textbf{(A)}}</math> |
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| + | ==Solution 2== | ||
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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 is correct, then the other option in its pair must also be correct. There can't be 2 answers. So, the only remaining answer choice is <math>\boxed{\textbf{(A) }3}</math> | ||
==See Also== | ==See Also== | ||
Latest revision as of 09:05, 10 June 2024
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 is correct, then the other option in its pair must also be correct. There can't be 2 answers. So, the only remaining answer choice is 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
