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Difference between revisions of "1985 AJHSME Problems/Problem 25"

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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.
  
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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>
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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>
  
 
==Solution 2==
 
==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 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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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>
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==Solution 3==
 
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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. <math>\textbf{THIS IS NOT A SHORTCUT FOR PROBLEMS YOU MIGHT FACE IN THE FUTURE.}</math> <math>\textbf{PLEASE REFER TO THE OTHER SOLUTIONS ABOVE FOR THE ACTUAL WAY TO SOLVE THIS}</math> <math>\textbf{PROBLEM IF NOT ALREADY.}</math> <math>\textbf{I REPEAT AGAIN: ONCE YOU'VE READ THIS PROBLEM, SIMPLY PUT THIS}</math> <math>\textbf{METHOD AT THE BACK OF YOUR MIND AND LEARN HOW TO ACTUALLY SOLVE THIS PROBLEMS.}</math>
 
  
 
==See Also==
 
==See Also==

Latest revision as of 09:05, 10 June 2024

Problem

Five cards are lying on a table as shown.

\[\begin{matrix} & \qquad & \boxed{\tt{P}} & \qquad & \boxed{\tt{Q}} \\ \\ \boxed{\tt{3}} & \qquad & \boxed{\tt{4}} & \qquad & \boxed{\tt{6}} \end{matrix}\]

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)

$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ \text{P} \qquad \text{(E)}\ \text{Q}$

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 $3$, which is answer choice $\boxed{\textbf{(A)}}$

Solution 2

Using the answer choices, we see that P and Q are logically equivalent (both are non-vowel letters) and so are $4$ and $6$ (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 $\boxed{\textbf{(A) }3}$

See Also

1985 AJHSME (Problems • Answer Key • Resources)
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All AJHSME/AMC 8 Problems and Solutions


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