2021 AMC 12B Problems/Problem 3: Difference between revisions

Latest revision as of 22:50, 18 July 2023

Problem

Suppose $2+\frac{1}{1+\frac{1}{2+\frac{2}{3+x}}}=\frac{144}{53}.$ What is the value of $x?$

$\textbf{(A) }\frac34 \qquad \textbf{(B) }\frac78 \qquad \textbf{(C) }\frac{14}{15} \qquad \textbf{(D) }\frac{37}{38} \qquad \textbf{(E) }\frac{52}{53}$

Solution

Subtracting $2$ from both sides and taking reciprocals gives $1+\frac{1}{2+\frac{2}{3+x}}=\frac{53}{38}$ . Subtracting $1$ from both sides and taking reciprocals again gives $2+\frac{2}{3+x}=\frac{38}{15}$ . Subtracting $2$ from both sides and taking reciprocals for the final time gives $\frac{x+3}{2}=\frac{15}{8}$ or $x=\frac{3}{4} \implies \boxed{\text{A}}$ .

~ OlympusHero

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=qpvS2PVkI8A&t=127s

Video Solution by OmegaLearn (Algebraic Manipulations)

https://youtu.be/WskJI8_7Gk0

~ pi_is_3.14

Video Solution by Hawk Math

https://www.youtube.com/watch?v=VzwxbsuSQ80

Video Solution by TheBeautyofMath

https://youtu.be/EMzdnr1nZcE?t=384

~IceMatrix

Video Solution (Just 1 min!)

https://youtu.be/UhoVUqZjfhg

~Education, the Study of Everything

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

@@ Line 24: / Line 24: @@
 ~IceMatrix
+==Video Solution (Just 1 min!)==
+https://youtu.be/UhoVUqZjfhg
+~Education, the Study of Everything
 ==See Also==
 {{AMC12 box|year=2021|ab=B|num-b=2|num-a=4}}
 {{MAA Notice}}

Page

Toolbox

Search