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
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

Difference between revisions of "2016 AMC 10B Problems/Problem 2"
(→‎See Also)
 
(One intermediate revision by one other user not shown)
Line 10: Line 10:
 
==Solution 2==
 
==Solution 2==
 
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>
 
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>
 +
 +
==Video Solution (CREATIVE THINKING)==
 +
https://youtu.be/boCvD0Hb6h0
 +
 +
~Education, the Study of Everything
 +
 +
  
 
==Video Solution==
 
==Video Solution==
Line 19: Line 26:
 
{{AMC10 box|year=2016|ab=B|num-b=1|num-a=3}}
 
{{AMC10 box|year=2016|ab=B|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
[[Category: Introductory Algebra Problems]]

Latest revision as of 16:55, 18 October 2025

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)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_2&oldid=263453"