Difference between revisions of "1985 AJHSME Problems/Problem 25"
(→See Also) |
(→Solution) |
||
| Line 11: | Line 11: | ||
<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]], | ||
| Line 18: | Line 18: | ||
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{\text{A}}</math> | 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{\text{A}}</math> | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | 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). The only remaining answer choice is <math>\boxed{\text{A}}</math> | ||
==See Also== | ==See Also== | ||
Revision as of 01:24, 24 November 2017
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?
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). 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. 
