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

Revision as of 11:11, 9 February 2014

Solution

We have $10\cdot\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{10}\right)^{-1}$ $\implies 10\cdot\left(\frac{5}{10}+\frac{2}{10}+\frac{1}{10}\right)^{-1}$ $\implies 10\cdot\left(\frac{5+2+1}{10}\right)^{-1}$ $\implies 10\cdot\left(\frac{8}{10}\right)^{-1}$ $\implies 10\cdot\left(\frac{4}{5}\right)^{-1}$ $\implies 10\cdot\frac{5}{4}$ $\implies \frac{50}{4}$ $\implies \boxed{\textbf{(C)}\ \frac{25}{2}}\ \blacksquare$

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-{{duplicate|[[2014 AMC 12A Problems|2014 AMC 12A #1]] and [[2014 AMC 10A Problems|2014 AMC 10A #1]]}}
-==Problem ==
-What is <math>10 \cdot \left(\tfrac{1}{2} + \tfrac{1}{5} + \tfrac{1}{10}\right)^{-1}?</math>
-<math> \textbf{(A)}\ 3\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ \frac{25}{2}\qquad\textbf{(D)}}\ \frac{170}{3}\qquad\textbf{(E)}\ 170</math>
 == Solution ==
-Sum the fractions over the common denominator: <math>\dfrac{1}{2}+\dfrac15+\dfrac1{10}=\dfrac{5+2+1}{10}=\dfrac45</math>
-Now the answer is just some arithmetic: <math>10\cdot\left(\dfrac45\right)^{-1}=10\cdot\dfrac{5}{4}=\boxed{\textbf{(C)}\ \dfrac{25}2}</math>
-==See Also==
-{{AMC10 box|year=2014|ab=A|before=First Problem|num-a=2}}
+We have <cmath>10\cdot\left(\frac{1}{2}+\frac{1}{5}+\frac{1}{10}\right)^{-1}</cmath>
-{{AMC12 box|year=2014|ab=A|before=First Problem|num-a=2}}
+<cmath>\implies 10\cdot\left(\frac{5}{10}+\frac{2}{10}+\frac{1}{10}\right)^{-1}</cmath>
-{{MAA Notice}}
+<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>

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

Revision as of 11:11, 9 February 2014

Solution