Difference between revisions of "2015 AMC 10A Problems/Problem 1"

(Solution)
(Solution)
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<math>=25^{-1}\times5</math>
 
<math>=25^{-1}\times5</math>
 
<math>=\frac{1}{25}\times5</math>
 
<math>=\frac{1}{25}\times5</math>
<math>=\frac{1}{5}\implies{\textbf{(C)}\ \frac{1}{5}}</math>
+
<math>=\frac{1}{5}\implies{\boxed{\textbf{(C)}\ \frac{1}{5}}}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2015|ab=A|before=First Problem|num-a=2}}
 
{{AMC10 box|year=2015|ab=A|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:30, 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

$(2^0-1+5^2-0)^{-1}\times5$ $=(1-1+25-0)^{-1}\times5$ $=25^{-1}\times5$ $=\frac{1}{25}\times5$ $=\frac{1}{5}\implies{\boxed{\textbf{(C)}\ \frac{1}{5}}}$

See Also

2015 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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All AMC 10 Problems and Solutions

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