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

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== Solution 2==
 
== Solution 2==
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If all of them are false, that would mean that the <math>4</math>th one is false too. Therefore, <math>E</math> is not the correct answer. If exactly <math>3</math> of them are false, that would mean that only <math>1</math> statement is true. This is correct since if only <math>1</math> statement is true, the card that is true is the one that has <math>3</math> of these statements are false. Therefore, our answer is <math>\boxed {(D)}</math>.
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~Arcticturn

Revision as of 17:14, 6 November 2021

The following four statements, and only these are found on a card: [asy] pair A,B,C,D,E,F,G; A=(0,1); B=(0,5); C=(11,5); D=(11,1); E=(0,4); F=(0,3); G=(0,2); draw(A--B--C--D--cycle); label("On this card exactly one statement is false.", B, SE); label("On this card exactly two statements are false.", E, SE); label("On this card exactly three statements are false.", F, SE); label("On this card exactly four statements are false.", G, SE); [/asy]

(Assume each statement is either true or false.) Among them the number of false statements is exactly

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

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 $\boxed{\textbf{(D) } 3}$, since $3$ are false and only the third statement ("On this card exactly three statements are false") is correct.

Solution 2

If all of them are false, that would mean that the $4$th one is false too. Therefore, $E$ is not the correct answer. If exactly $3$ of them are false, that would mean that only $1$ statement is true. This is correct since if only $1$ statement is true, the card that is true is the one that has $3$ of these statements are false. Therefore, our answer is $\boxed {(D)}$.

~Arcticturn

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