Difference between revisions of "2016 AMC 10B Problems/Problem 1"

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==Solution 2==
 
==Solution 2==
 
Substituting <math>\frac{1}{2}</math> for <math>a</math> in <math>\frac{\frac{1}{a}\cdot(2+\frac{1}{2})}{a}</math> gives us <math>\boxed{\textbf{(D) }10}</math>. ~peelybonehead
 
Substituting <math>\frac{1}{2}</math> for <math>a</math> in <math>\frac{\frac{1}{a}\cdot(2+\frac{1}{2})}{a}</math> gives us <math>\boxed{\textbf{(D) }10}</math>. ~peelybonehead
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==Video Solution (CREATIVE THINKING)==
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https://youtu.be/2erUXM5pD2g
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~Education, the Study of Everything
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 +
  
 
==Video Solution==
 
==Video Solution==

Revision as of 11:49, 2 July 2023

Problem

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

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

Solution

Factorizing the numerator, $\frac{\frac{1}{a}\cdot(2+\frac{1}{2})}{a}$ then becomes $\frac{\frac{5}{2}}{a^{2}}$ which is equal to $\frac{5}{2}\cdot 2^2$ which is $\boxed{\textbf{(D) }10}$.


Solution 2

Substituting $\frac{1}{2}$ for $a$ in $\frac{\frac{1}{a}\cdot(2+\frac{1}{2})}{a}$ gives us $\boxed{\textbf{(D) }10}$. ~peelybonehead

Video Solution (CREATIVE THINKING)

https://youtu.be/2erUXM5pD2g

~Education, the Study of Everything


Video Solution

https://youtu.be/1IZ3oj1iGf0

~savannahsolver

See Also

2016 AMC 10B (ProblemsAnswer KeyResources)
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Problem 2
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All AMC 10 Problems and Solutions

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