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Difference between revisions of "2016 AMC 10B Problems/Problem 2"

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<math>\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4</math>
 
<math>\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4</math>
  
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==Solution 1==
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<math>\frac{2^3(2^2)^2}{(2^2)^32^2}=\frac{2^7}{2^8}=\frac12</math> which is <math>\textbf{(B)}</math>.
  
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==Solution 2==
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We can replace <math>2</math> and <math>4</math> with <math>a</math> and <math>b</math> respectively. Then substituting with <math>n</math> and <math>m</math> we can get <math>\dfrac{a^3b^2}{b^3a^2}=\dfrac{a}{b}</math> and substitute to get <math>\dfrac{2}{4}=\boxed{\dfrac{1}{2}}</math> which is <math>\boxed{\textbf{(B)}}</math>
  
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==Solution==
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==Video Solution (CREATIVE THINKING)==
−
<math>\textbf{(B)}\ \frac{1}{2}</math>
+
https://youtu.be/boCvD0Hb6h0
 +
 
 +
~Education, the Study of Everything
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 +
 
 +
 
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==Video Solution==
 +
https://youtu.be/W7IwD6sZWco
 +
 
 +
~savannahsolver
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==See Also==
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{{AMC10 box|year=2016|ab=B|num-b=1|num-a=3}}
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{{MAA Notice}}

Latest revision as of 11:48, 2 July 2023

Problem

If $n\heartsuit m=n^3m^2$, what is $\frac{2\heartsuit 4}{4\heartsuit 2}$?

$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4$

Solution 1

$\frac{2^3(2^2)^2}{(2^2)^32^2}=\frac{2^7}{2^8}=\frac12$ which is $\textbf{(B)}$.

Solution 2

We can replace $2$ and $4$ with $a$ and $b$ respectively. Then substituting with $n$ and $m$ we can get $\dfrac{a^3b^2}{b^3a^2}=\dfrac{a}{b}$ and substitute to get $\dfrac{2}{4}=\boxed{\dfrac{1}{2}}$ which is $\boxed{\textbf{(B)}}$

Video Solution (CREATIVE THINKING)

https://youtu.be/boCvD0Hb6h0

~Education, the Study of Everything


Video Solution

https://youtu.be/W7IwD6sZWco

~savannahsolver

See Also

2016 AMC 10B (Problems • Answer Key • Resources)
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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

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