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

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==Problem==
 
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If <math>n\heartsuit m=n^3m^2</math>, what is <math>\frac{2\heartsuit 4}{4\heartsuit 2}</math>?
 
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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>
 
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==Solution 1==
 
==Solution 1==
 
<math>\frac{2^3(2^2)^2}{(2^2)^32^2}=\frac{2^7}{2^8}=\frac12</math> which is <math>\textbf{(B)}</math>.
 
<math>\frac{2^3(2^2)^2}{(2^2)^32^2}=\frac{2^7}{2^8}=\frac12</math> which is <math>\textbf{(B)}</math>.

Revision as of 20:49, 23 January 2020

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 $\boxed{\textbf{(B)}}$

See Also

2016 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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All AMC 10 Problems and Solutions

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