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

(Video Solution)
(5 intermediate revisions by 3 users not shown)
Line 7: Line 7:
 
==Solution==
 
==Solution==
  
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>.
+
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>.
 +
 
 +
==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}}

Revision as of 17:57, 16 June 2020

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

Video Solution

https://youtu.be/1IZ3oj1iGf0

~savannahsolver

See Also

2016 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
-
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png