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Difference between revisions of "1978 AHSME Problems/Problem 22"

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

Revision as of 21:07, 29 December 2019

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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