Difference between revisions of "2014 AMC 10B Problems/Problem 2"

m (Solution 2)
(Solution 1)
Line 5: Line 5:
  
 
==Solution 1==
 
==Solution 1==
We can synchronously multiply <math> {2^3} </math> to the polynomials both above and below the fraction bar. Thus,  
+
We can synchronously multiply <math> {2^3} </math> to the expresions both above and below the fraction bar. Thus,  
 
<cmath>\frac{2^3+2^3}{2^{-3}+2^{-3}}\\=\frac{2^6+2^6}{1+1}\\={2^6}. </cmath>
 
<cmath>\frac{2^3+2^3}{2^{-3}+2^{-3}}\\=\frac{2^6+2^6}{1+1}\\={2^6}. </cmath>
 
Hence, the fraction equals to <math>\boxed{{\textbf{(E) }64}}</math>.
 
Hence, the fraction equals to <math>\boxed{{\textbf{(E) }64}}</math>.

Revision as of 15:09, 18 October 2022

Problem

What is $\frac{2^3 + 2^3}{2^{-3} + 2^{-3}}$?

$\textbf {(A) } 16 \qquad \textbf {(B) } 24 \qquad \textbf {(C) } 32 \qquad \textbf {(D) } 48 \qquad \textbf {(E) } 64$

Solution 1

We can synchronously multiply ${2^3}$ to the expresions both above and below the fraction bar. Thus, \[\frac{2^3+2^3}{2^{-3}+2^{-3}}\\=\frac{2^6+2^6}{1+1}\\={2^6}.\] Hence, the fraction equals to $\boxed{{\textbf{(E) }64}}$.

Solution 2

We have \[\frac{2^3+2^3}{2^{-3}+2^{-3}} = \frac{8 + 8}{\frac{1}{8} + \frac{1}{8}} = \frac{16}{\frac{1}{4}} = 16 \cdot 4 = 64,\] so our answer is $\boxed{{(\textbf{E) }64}}$.

Video Solution

https://youtu.be/vLFULrT_7yk

~savannahsolver

See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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