Difference between revisions of "2021 AMC 12B Problems/Problem 3"
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<math>\textbf{(A) }\frac34 \qquad \textbf{(B) }\frac78 \qquad \textbf{(C) }\frac{14}{15} \qquad \textbf{(D) }\frac{37}{38} \qquad \textbf{(E) }\frac{52}{53}</math> | <math>\textbf{(A) }\frac34 \qquad \textbf{(B) }\frac78 \qquad \textbf{(C) }\frac{14}{15} \qquad \textbf{(D) }\frac{37}{38} \qquad \textbf{(E) }\frac{52}{53}</math> | ||
| − | ==Solution | + | ==Solution== |
Subtracting <math>2</math> from both sides and taking reciprocals gives <math>1+\frac{1}{2+\frac{2}{3+x}}=\frac{53}{38}</math>. Subtracting <math>1</math> from both sides and taking reciprocals again gives <math>2+\frac{2}{3+x}=\frac{38}{15}</math>. Subtracting <math>2</math> from both sides and taking reciprocals for the final time gives <math>\frac{x+3}{2}=\frac{15}{8}</math> or <math>x=\frac{3}{4} \implies \boxed{\text{A}}</math>. | Subtracting <math>2</math> from both sides and taking reciprocals gives <math>1+\frac{1}{2+\frac{2}{3+x}}=\frac{53}{38}</math>. Subtracting <math>1</math> from both sides and taking reciprocals again gives <math>2+\frac{2}{3+x}=\frac{38}{15}</math>. Subtracting <math>2</math> from both sides and taking reciprocals for the final time gives <math>\frac{x+3}{2}=\frac{15}{8}</math> or <math>x=\frac{3}{4} \implies \boxed{\text{A}}</math>. | ||
Revision as of 11:41, 16 February 2021
Contents
Problem
Suppose![\[2+\frac{1}{1+\frac{1}{2+\frac{2}{3+x}}}=\frac{144}{53}.\]](http://latex.artofproblemsolving.com/9/f/2/9f2bfdf62a315bd20a6a692df961e22f22f7611d.png)
What is the value of 
Solution
Subtracting 
from both sides and taking reciprocals gives 
. Subtracting 
from both sides and taking reciprocals again gives 
. Subtracting from both sides and taking reciprocals for the final time gives 
or 
.
~ OlympusHero
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=qpvS2PVkI8A&t=127s
Video Solution by OmegaLearn (Algebraic Manipulations)
~ pi_is_3.14
Video Solution by Hawk Math
https://www.youtube.com/watch?v=VzwxbsuSQ80
See Also
| 2021 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 2 |
Followed by Problem 4 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
