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Difference between revisions of "1953 AHSME Problems/Problem 40"

m (→‎Solution: was completely wrong the previous one at plain sight. This also agrees with the answer key now.)
(→‎Solution: the previous solution was entirely wrong)
 
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==Solution==
 
==Solution==
  
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Note that <math>\textbf{(A)}</math> and <math>\textbf{(B)}</math> are the same, and <math>\textbf{(D)}</math> is the same as the quote in the statement "all men are honest" (that we have to found its negation). If not all men are honest, then that means that there is at least 1 dishonest (could be just one that spoils it for all the rest, or 5, for one, we definitely can't say it's all of them who are dishonest). So: <math>\boxed{\textbf{(C)} \text{some men are dishonest}}</math>.
+
Note that <math>\textbf{(A)}</math> and <math>\textbf{(B)}</math> are the same, and <math>\textbf{(D)}</math> is the same as the quote in the statement "all men are honest" (that we have to found its negation). If not all men are honest, then that means that there is at least 1 dishonest (could be just one that spoils it for all the rest, or 5, for one, we definitely can't say it's all of them who are dishonest). So: <math>\textbf{(C)}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 15:02, 1 July 2021

Problem

The negation of the statement "all men are honest," is:

$\textbf{(A)}\ \text{no men are honest} \qquad \textbf{(B)}\ \text{all men are dishonest} \\ \textbf{(C)}\ \text{some men are dishonest}\qquad \textbf{(D)}\ \text{no men are dishonest}\\ \textbf{(E)}\ \text{some men are honest}$

Solution

Note that $\textbf{(A)}$ and $\textbf{(B)}$ are the same, and $\textbf{(D)}$ is the same as the quote in the statement "all men are honest" (that we have to found its negation). If not all men are honest, then that means that there is at least 1 dishonest (could be just one that spoils it for all the rest, or 5, for one, we definitely can't say it's all of them who are dishonest). So: $\textbf{(C)}$.

See Also

1953 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 39 		Followed by
Problem 41
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions


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