Difference between revisions of "2014 AMC 10B Problems/Problem 2"
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<math> \textbf {(A) } 16 \qquad \textbf {(B) } 24 \qquad \textbf {(C) } 32 \qquad \textbf {(D) } 48 \qquad \textbf {(E) } 64 </math> | <math> \textbf {(A) } 16 \qquad \textbf {(B) } 24 \qquad \textbf {(C) } 32 \qquad \textbf {(D) } 48 \qquad \textbf {(E) } 64 </math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | We can synchronously multiply <math> {2^3} </math> to the | + | 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{{ | + | Hence, the fraction equals to <math>\boxed{{\textbf{(E) }64}}</math>. |
==Solution 2== | ==Solution 2== | ||
We have | We have | ||
| − | < | + | <cmath>\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,</cmath> so our answer is <math>\boxed{{(\textbf{E) }64}}</math>. |
| + | |||
| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/NwWf1uYFZqw | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | |||
==Video Solution== | ==Video Solution== | ||
Latest revision as of 11:05, 2 July 2023
Contents
Problem
What is 
?
Solution 1
We can synchronously multiply 
to the expresions both above and below the fraction bar. Thus,
![\[\frac{2^3+2^3}{2^{-3}+2^{-3}}\\=\frac{2^6+2^6}{1+1}\\={2^6}.\]](http://latex.artofproblemsolving.com/5/f/6/5f61712212b2338ec47df60099f80f0e4a58c157.png)
Hence, the fraction equals to 
.
Solution 2
We have
![\[\frac{2^3+2^3}{2^{-3}+2^{-3}} = \frac{8 + 8}{\frac{1}{8} + \frac{1}{8}} = \frac{16}{\frac{1}{4}} = 16 \cdot 4 = 64,\]](http://latex.artofproblemsolving.com/2/f/7/2f746975ff2b7b39e12fbe1865ec81670932b221.png)
so our answer is 
.
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
| 2014 AMC 10B (Problems • Answer Key • Resources) | ||
| 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. 
