Difference between revisions of "1953 AHSME Problems/Problem 40"
(Created page with "==Problem== The negation of the statement "all men are honest," is: <math>\textbf{(A)}\ \text{no men are honest} \qquad \textbf{(B)}\ \text{all men are dishonest} \ \textbf...") |
Santropedro (talk | contribs) (→Solution: the previous solution was entirely wrong) |
||
(3 intermediate revisions by 3 users not shown) | |||
Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
− | |||
− | + | 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== | |
+ | {{AHSME 50p box|year=1953|num-b=39|num-a=41}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 15:02, 1 July 2021
Problem
The negation of the statement "all men are honest," is:
Solution
Note that and are the same, and 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: .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.