Difference between revisions of "2014 AMC 10A Problems/Problem 1"

Revision as of 10:03, 20 May 2015

Problem

What is $10\cdot\left(\tfrac{1}{2}+\tfrac{1}{5}+\tfrac{1}{10}\right)^{-1}?$

$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ \frac{25}{2} \qquad\textbf{(D)}\ \frac{170}{3}\qquad\textbf{(E)}\ 170$

Solution

We have $10\cdot\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{10}\right)^{-1}$ Making the denominators equal gives $\implies 10\cdot\left(\frac{5}{10}+\frac{2}{10}+\frac{1}{10}\right)^{-1}$ $\implies 10\cdot\left(\frac{5+2+1}{10}\right)^{-1}$ $\implies 10\cdot\left(\frac{8}{10}\right)^{-1}$ $\implies 10\cdot\left(\frac{4}{5}\right)^{-1}$ $\implies 10\cdot\frac{5}{4}$ $\implies \frac{50}{4}$ Finally, simplifying gives $\implies \boxed{\textbf{(C)}\ \frac{25}{2}}$

2014 AMC 10A (Problems • Answer Key • Resources)
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2014 AMC 12A (Problems • Answer Key • Resources)
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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Difference between revisions of "2014 AMC 10A Problems/Problem 1"

Revision as of 10:03, 20 May 2015

Problem

Solution

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