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

Latest revision as of 10:44, 26 July 2022

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

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

By: Dragonfly

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

2016 AMC 12B (Problems • Answer Key • Resources)
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

@@ Line 1: / Line 1: @@
-Last year we had answers by this time :(
+==Problem==
-Turns out we have to wait til sunday I think.
+What is the value of <math>\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}</math> when <math>a= \frac{1}{2}</math>?
+<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20</math>
+==Solution==
+By: Dragonfly
+We find that <math>a^{-1}</math> is the same as <math>2</math>, since a number to the power of <math>-1</math> is just the reciprocal of that number. We then get the equation to be
+<cmath>\frac{2\times2+\frac{2}{2}}{\frac{1}{2}}</cmath>
+We can then simplify the equation to get <math>\boxed{\textbf{(D)}\ 10}</math>
+==See Also==
+{{AMC12 box|year=2016|ab=B|before=First Problem|num-a=2}}
+{{MAA Notice}}