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Difference between revisions of "2016 AMC 10B Problems/Problem 1"
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==Solution==
 
==Solution==
  
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Factorizing the numerator, <math>\frac{\frac{1}{a}\cdot(2+\frac{1}{2})}{a}</math> then becomes <math>\frac{\frac{5}{2}}{a^{2}}</math> which is equal to <math>\frac{5}{2}\cdot 2^2</math> which is <math>\textbf{(D) }10</math>.
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Factorizing the numerator, <math>\frac{\frac{1}{a}\cdot(2+\frac{1}{2})}{a}</math> then becomes <math>\frac{\frac{5}{2}}{a^{2}}</math> which is equal to <math>\frac{5}{2}\cdot 2^2</math> which is <math>\boxed{\textbf{(D) }10}</math>.
 +
 
 +
==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>.
 +
 
 +
==Video Solution (CREATIVE THINKING)==
 +
https://youtu.be/2erUXM5pD2g
 +
 
 +
~Education, the Study of Everything
 +
 
 +
 
 +
 
 +
==Video Solution==
 +
https://youtu.be/1IZ3oj1iGf0
 +
 
 +
~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=B|before=-|num-a=2}}
 
{{AMC10 box|year=2016|ab=B|before=-|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:39, 6 August 2024

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 $\boxed{\textbf{(D) }10}$.

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 (Problems • Answer Key • Resources)
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Problem 2
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All AMC 10 Problems and Solutions

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