Difference between revisions of "2014 AMC 10A Problems/Problem 1"
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== Solution == | == Solution == | ||
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− | {{ | + | We have <cmath>10\cdot\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{10}\right)^{-1}</cmath> |
− | {{ | + | <cmath>\implies 10\cdot\left(\frac{5}{10}+\frac{2}{10}+\frac{1}{10}\right)^{-1}</cmath> |
− | {{ | + | <cmath>\implies 10\cdot\left(\frac{5+2+1}{10}\right)^{-1}</cmath> |
+ | <cmath>\implies 10\cdot\left(\frac{8}{10}\right)^{-1}</cmath> | ||
+ | <cmath>\implies 10\cdot\left(\frac{4}{5}\right)^{-1}</cmath> | ||
+ | <cmath>\implies 10\cdot\frac{5}{4}</cmath> | ||
+ | <cmath>\implies \frac{50}{4}</cmath> | ||
+ | <cmath>\implies \boxed{\textbf{(C)}\ \frac{25}{2}}\ \blacksquare</cmath> |
Revision as of 10:11, 9 February 2014
Solution
We have