Difference between revisions of "1978 AHSME Problems/Problem 22"
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\textbf{(E)}\ 4 </math> | \textbf{(E)}\ 4 </math> | ||
| − | == Solution == | + | == Solutions == |
| + | |||
| + | == Solution 1 == | ||
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. | 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. | ||
| + | |||
| + | == Solution 2== | ||
| + | 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 | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME box|year=1978|num-b=21|num-a=23}} | ||
| + | {{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]](http://latex.artofproblemsolving.com/7/a/4/7a464674d5a7868a7d9ac41c5757b1cce7500bc3.png)
(Assume each statement is either true or false.) Among them the number of false statements is exactly
Contents
Solutions
Solution 1
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.
Solution 2
If all of them are false, that would mean that the th one is false too. Therefore, 
is not the correct answer. If exactly of them are false, that would mean that only 
statement is true. This is correct since if only statement is true, the card that is true is the one that has of these statements are false. If we have or 
false statements, that would mean that there is more than true statement. Therefore, our answer is 
.
~Arcticturn
See Also
| 1978 AHSME (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
