2015 AMC 12A Problems/Problem 1: Difference between revisions

Latest revision as of 19:38, 4 February 2015

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-==Problem==
+#REDIRECT[[2015 AMC 10A Problems/Problem 1]]
-What is the value of <math>(2^0-1+5^2-0)^{-1}\times5?</math>
-<math> \textbf{(A)}\ -125\qquad\textbf{(B)}\ -120\qquad\textbf{(C)}\ \frac{1}{5}\qquad\textbf{(D)}}\ \frac{5}{24}\qquad\textbf{(E)}\ 25</math>
-==Solution==
-Evaluating, we have <math>(2^0-1+5^2-0)^{-1}\times5</math> = <math>(1-1+25-0)^{-1}\times5</math> = <math>(25)^{-1}\times5</math> = <math>\frac{5}{25}</math> = <math>\boxed{\textbf{(C)}\frac15}</math>.
-{{AMC12 box|year=2015|ab=A|before=First Question|num-a=2}}
-{{MAA Notice}}