2014 AMC 12A Problems/Problem 1: Difference between revisions

Latest revision as of 13:17, 8 February 2014

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-==Problem ==
+#REDIRECT[[2014 AMC 10A Problems/Problem 1]]
-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*\left(\dfrac45\right)^{-1}=10*\dfrac{5}{4}=\textbf{(C)}\ \dfrac{25}2</math>
-==See Also==
-{{AMC10 box|year=2014|ab=A|before=First Problem|num-a=2}}
-{{MAA Notice}}