1985 AJHSME Problems/Problem 25

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?

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

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 $\boxed{\text{A}}$

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). The only remaining answer choice is $\boxed{\text{A}}$

1985 AJHSME (Problems • Answer Key • Resources)
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1985 AJHSME Problems/Problem 25

Contents

Problem

Solution 1

Solution 2

See Also