Difference between revisions of "1957 AHSME Problems/Problem 12"
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| − | + | == Problem 12== | |
| + | |||
| + | Comparing the numbers <math>10^{-49}</math> and <math>2\cdot 10^{-50}</math> we may say: | ||
| + | |||
| + | <math>\textbf{(A)}\ \text{the first exceeds the second by }{8\cdot 10^{-1}}\qquad\\ | ||
| + | \textbf{(B)}\ \text{the first exceeds the second by }{2\cdot 10^{-1}}\qquad\\ | ||
| + | \textbf{(C)}\ \text{the first exceeds the second by }{8\cdot 10^{-50}}\qquad\\ | ||
| + | \textbf{(D)}\ \text{the second is five times the first}\qquad\\ | ||
| + | \textbf{(E)}\ \text{the first exceeds the second by }{5} </math> | ||
| + | |||
| + | == Solution == | ||
| + | <math>\boxed{\textbf{(C)}}</math> | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME 50p box|year=1957|num-b=11|num-a=13}} | ||
| + | {{MAA Notice}} | ||
| + | [[Category:AHSME]][[Category:AHSME Problems]] | ||
Revision as of 08:12, 25 July 2024
Problem 12
Comparing the numbers 
and 
we may say:
Solution
See Also
| 1957 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
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| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
