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

m (Solution)
m (Solution)
Line 10: Line 10:
 
==Solution==
 
==Solution==
  
<math>\frac{frac{1}{a}*(2+frac{1}{2})}{a}</math> then becomes <math>frac{frac{5}{2}}{a^{2}}</math> which is equal to <math>frac{5}{2}*2^{2}</math> = <math>{(D)}\ 10\qquad\textbf{(E)}\ 20</math>
+
<math>\frac{frac{1}{a}*(2+frac{1}{2})}{a}</math> then becomes <math>frac{frac{5}{2}}{a^{2}}</math> which is equal to <math>frac{5}{2}*2^{2}</math> = <math>{(D)}</math>
  
 
Solution by ngeorge
 
Solution by ngeorge

Revision as of 10:19, 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

$\frac{frac{1}{a}*(2+frac{1}{2})}{a}$ then becomes $frac{frac{5}{2}}{a^{2}}$ which is equal to $frac{5}{2}*2^{2}$ = ${(D)}$

Solution by ngeorge