Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2014 AMC 10B Problems/Problem 2"
(→‎Video Solution)
 
(10 intermediate revisions by 4 users not shown)
Line 4: Line 4:
 
<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 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{{64 (\textbf{E})}}</math>.
+
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==
 
==Video Solution==
Line 17: Line 28:
 
{{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}}
−
We can factor the numerator and the denominator. The numerator becomes 2^3(1+1) and the denominator becomes 2^-3(1+1). The (1+1)'s cancel so we are left with 2^3 over 2^-3. This leaves us with 2^6 which equals to E) 64
 

Latest revision as of 11:05, 2 July 2023

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

~Education, the Study of Everything


Video Solution

https://youtu.be/vLFULrT_7yk

~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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_2&oldid=195093"