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

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==Solution==
 
==Solution==
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<math>(2^{0} - 1 + 5^{2} +0)^{-1} \ctimes{} 5 = (1 - 1 + 25 +0)^{-1} \ctimes{} 5 = (25)^{-1} \ctimes{} 5 = \frac{1}{25} \ctimes{} 5 = \boxed{\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}}
 
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{{MAA Notice}}

Revision as of 18:26, 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

$(2^{0} - 1 + 5^{2} +0)^{-1} \ctimes{} 5 = (1 - 1 + 25 +0)^{-1} \ctimes{} 5 = (25)^{-1} \ctimes{} 5 = \frac{1}{25} \ctimes{} 5 = \boxed{\frac{1}{5}}$ (Error compiling LaTeX. Unknown error_msg)

See Also

2015 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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