# 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>\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)} | + | 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 | ||

+ | |||

+ | <math>\textbf{(A)}\ 0 \qquad | ||

+ | \textbf{(B)}\ 1 \qquad | ||

+ | \textbf{(C)}\ 2 \qquad | ||

+ | \textbf{(D)}\ 3 \qquad | ||

+ | \textbf{(E)}\ 4 </math> | ||

+ | |||

+ | == Solution == | ||

+ | |||

+ | 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>\boxed{\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 12:38, 3 February 2020

The following four statements, and only these are found on a card:

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

## Solution

There can be at most one true statement on the card, eliminating and . If there are true on the card, statement ("On this card exactly four statements are false") will be correct, causing a contradiction. Therefore, the answer is , since are false and only the third statement ("On this card exactly three statements are false") is correct.