Difference between revisions of "2015 AMC 12A Problems/Problem 1"

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==Solution==
 
==Solution==
  
Simply evaluating gives <math>\boxed{\textbf{(C)}\frac15}</math>.
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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>.
  
  

Revision as of 14:26, 4 February 2015

Problem

What is the value of $(2^0-1+5^2-0)^{-1}\times5?$

$\textbf{(A)}\ -125\qquad\textbf{(B)}\ -120\qquad\textbf{(C)}\ \frac{1}{5}\qquad\textbf{(D)}}\ \frac{5}{24}\qquad\textbf{(E)}\ 25$ (Error compiling LaTeX. Unknown error_msg)

Solution

Evaluating, we have $(2^0-1+5^2-0)^{-1}\times5$ = $(1-1+25-0)^{-1}\times5$ = $(25)^{-1}\times5$ = $\frac{5}{25}$ = $\boxed{\textbf{(C)}\frac15}$.


2015 AMC 12A (ProblemsAnswer KeyResources)
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First Question
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Problem 2
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