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>\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>. | |
+ | |||
+ | ==Solution 2== | ||
+ | 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== | ||
+ | https://youtu.be/vLFULrT_7yk | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=B|num-b=1|num-a=3}} | {{AMC10 box|year=2014|ab=B|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 12:05, 2 July 2023
Contents
Problem
What is ?
Solution 1
We can synchronously multiply to the expresions both above and below the fraction bar. Thus,
Hence, the fraction equals to
.
Solution 2
We have
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.