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> | ||
| − | ==Solution== | + | ==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>. | ||
| + | |||
| + | ==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>\textbf{(B)}</math> | ||
==See Also== | ==See Also== | ||
{{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}} | ||
Revision as of 14:17, 28 December 2017
Contents
Problem
If 
, what is 
?
Solution 1
which is 
.
Solution 2
We can replace 
and 
with 
and 
respectively. Then substituting with 
and 
we can get 
and substitute to get 
which is
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
