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

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\textbf{(E)}\ 4    </math>
 
\textbf{(E)}\ 4    </math>
  
== Solution ==
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== Solutions ==
  
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.
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== Solution 1 ==
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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>\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.
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== 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. If we have <math>1</math> or <math>2</math> false statements, that would mean that there is more than <math>1</math> true statement. Therefore, our answer is <math>\boxed {(D)}</math>.
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~Arcticturn
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==See Also==
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{{AHSME box|year=1978|num-b=21|num-a=23}} 
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{{MAA Notice}}

Latest revision as of 17:17, 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

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

Solutions

Solution 1

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. If we have $1$ or $2$ false statements, that would mean that there is more than $1$ true statement. Therefore, our answer is $\boxed {(D)}$.

~Arcticturn

See Also

1978 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

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