Difference between revisions of "1988 AJHSME Problems/Problem 6"
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==Solution== | ==Solution== | ||
Converting the [[decimal|decimals]] to [[fraction|fractions]] gives us <math>\frac{(.2)^3}{(.02)^2} =\frac{\left( \frac{1}{5}\right)^3}{\left(\frac{1}{50}\right)^2}=\frac{50^2}{5^3}=\frac{2500}{125}=20\Rightarrow \mathrm{(E)}</math>. | Converting the [[decimal|decimals]] to [[fraction|fractions]] gives us <math>\frac{(.2)^3}{(.02)^2} =\frac{\left( \frac{1}{5}\right)^3}{\left(\frac{1}{50}\right)^2}=\frac{50^2}{5^3}=\frac{2500}{125}=20\Rightarrow \mathrm{(E)}</math>. | ||
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| + | ==Solution 2== | ||
==See Also== | ==See Also== | ||
Revision as of 22:07, 8 July 2020
Contents
Problem
Solution
Converting the decimals to fractions gives us 
.
Solution 2
See Also
| 1988 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
