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

(Solution 2)
(Solution 2)
Line 32: Line 32:
  
 
~Arcticturn
 
~Arcticturn
 +
 +
==See Also==
 +
{{AHSME box|year=1979|num-b=25|num-a=27}} 
 +
{{MAA Notice}}

Revision as of 17:15, 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

See Also

1979 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 25
Followed by
Problem 27
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png