Difference between revisions of "2016 AMC 12B Problems/Problem 1"

(Solution)
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==Solution==
 
==Solution==
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We find that <math>a^{-1}</math> is the same as 2, since a number to the power of <math>-1</math> is just the reciprocal of that number. We then get the equation to be
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<math>\frac{2\times2+\frac{2}{2}}{\frac{1}{2}}</math>
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We can then simplify the equation to get <math>\boxed{\textbf{(C)}\ \frac{5}{2}}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=B|before=First Problem|num-a=2}}
 
{{AMC12 box|year=2016|ab=B|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 10:59, 21 February 2016

Problem

What is the value of $\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}$ when $a= \frac{1}{2}$?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20$

Solution

We find that $a^{-1}$ is the same as 2, since a number to the power of $-1$ is just the reciprocal of that number. We then get the equation to be

$\frac{2\times2+\frac{2}{2}}{\frac{1}{2}}$

We can then simplify the equation to get $\boxed{\textbf{(C)}\ \frac{5}{2}}$

See Also

2016 AMC 12B (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 12 Problems and Solutions

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