Difference between revisions of "1953 AHSME Problems/Problem 40"
m |
Santropedro (talk | contribs) m (→Solution: was completely wrong the previous one at plain sight. This also agrees with the answer key now.) |
||
| 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>\boxed{\textbf{(C)} \text{some men are dishonest}}</math>. | |
| − | |||
| − | |||
| − | |||
| − | |||
==See Also== | ==See Also== | ||
Revision as of 14:59, 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. 
