Difference between revisions of "1954 AHSME Problems/Problem 34"
Katzrockso (talk | contribs) (Created page with "== Problem 34== The fraction <math>\frac{1}{3}</math>: <math> \textbf{(A)}\ \text{equals 0.33333333}\qquad\textbf{(B)}\ \text{is less than 0.33333333 by }\frac{1}{3\cdot 10...") |
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<math>\frac{333333333}{10^8}-\frac{1}{3}\implies \frac{3\cdot 10^7+3\cdot 10^6+\dots+3\cdot 10^0}{10^8}-\frac{1}{3}\implies\frac{3(10^{8}-1)}{9\cdot 10^{8}}-\frac{1}{3}\implies\frac{10^8-1}{3\cdot 10^8}-\frac{1}{3}\implies\frac{10^8}{3\cdot 10^8}-\frac{1}{3\cdot 10^8}-\frac{1}{3}\implies\frac{-1}{3\cdot 10^8}</math> | <math>\frac{333333333}{10^8}-\frac{1}{3}\implies \frac{3\cdot 10^7+3\cdot 10^6+\dots+3\cdot 10^0}{10^8}-\frac{1}{3}\implies\frac{3(10^{8}-1)}{9\cdot 10^{8}}-\frac{1}{3}\implies\frac{10^8-1}{3\cdot 10^8}-\frac{1}{3}\implies\frac{10^8}{3\cdot 10^8}-\frac{1}{3\cdot 10^8}-\frac{1}{3}\implies\frac{-1}{3\cdot 10^8}</math> | ||
Because we did <math>.333333333-\frac{1}{3}</math>, it is <math>\fbox{B}</math> | Because we did <math>.333333333-\frac{1}{3}</math>, it is <math>\fbox{B}</math> | ||
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| + | ==See Also== | ||
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| + | {{AHSME 50p box|year=1954|num-b=33|num-a=35}} | ||
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| + | {{MAA Notice}} | ||
Revision as of 17:07, 22 April 2020
Problem 34
The fraction 
:
Solution
Because we did 
, it is 
See Also
| 1954 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 33 |
Followed by Problem 35 | |
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| All AHSME Problems and Solutions | ||
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