1978 AHSME Problems/Problem 22

Revision as of 20:07, 29 December 2019 by Ryjs (talk | contribs) (solution)

There can be at most one true statement on the card, eliminating $\textbf{(A)}, \textbf{(B)},$ and $\textbf{(C)}$. If there are $0$ true on the card, statement $4$ ("On this card exactly four statements are false") will be correct, causing a contradiction. Therefore, the answer is $\textbf{(D)}\ 3$, since $3$ are false and only the third statement ("On this card exactly three statements are false") is correct.

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