2014 AMC 10B Problems/Problem 2

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 (CREATIVE THINKING)

https://youtu.be/NwWf1uYFZqw

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Video Solution

https://youtu.be/vLFULrT_7yk

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2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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All AMC 10 Problems and Solutions

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

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