Difference between revisions of "2020 AMC 10A Problems/Problem 1"

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==Solution 2==
 
==Solution 2==
Multiplying <math>12</math> on both sides gets us <math>12x-9=1</math>, therefore <math>\boxed{x=\textbf{(E)}\frac{5}{6}}</math>. ~CoolJupiter
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Multiplying <math>12</math> on both sides gets us <math>12x-9=1 \Rightarrow 12x=10</math>, therefore <math>\boxed{x=\textbf{(E)}~\frac{5}{6}}</math>.
  
==Video Solution==
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==Video Solution 1==
https://youtu.be/WUcbVNy2uv0
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Education, The Study of Everything
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https://youtu.be/4lsmGWDYusk
  
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==Video Solution 2==
 
~IceMatrix
 
~IceMatrix
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https://youtu.be/WUcbVNy2uv0
  
 
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==Video Solution 3==
 
https://www.youtube.com/watch?v=7-3sl1pSojc
 
https://www.youtube.com/watch?v=7-3sl1pSojc
  
 
~bobthefam
 
~bobthefam
  
 
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==Video Solution 4==
 
https://youtu.be/OKoBg15l8ro
 
https://youtu.be/OKoBg15l8ro
  
~savannahsolver
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~savannahsolve
  
 
== See Also ==
 
== See Also ==

Latest revision as of 16:04, 31 March 2021

Problem

What value of $x$ satisfies \[x- \frac{3}{4} = \frac{5}{12} - \frac{1}{3}?\]

$\textbf{(A)}\ -\frac{2}{3}\qquad\textbf{(B)}\ \frac{7}{36}\qquad\textbf{(C)}\ \frac{7}{12}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{5}{6}$

Solution

Adding $\frac{3}{4}$ to both sides, $x= \frac{5}{12} - \frac{1}{3} + \frac{3}{4} = \frac{5}{12} - \frac{4}{12} + \frac{9}{12}=\boxed{\textbf{(E) }\frac{5}{6}}$.

Solution 2

Multiplying $12$ on both sides gets us $12x-9=1 \Rightarrow 12x=10$, therefore $\boxed{x=\textbf{(E)}~\frac{5}{6}}$.

Video Solution 1

Education, The Study of Everything

https://youtu.be/4lsmGWDYusk

Video Solution 2

~IceMatrix https://youtu.be/WUcbVNy2uv0

Video Solution 3

https://www.youtube.com/watch?v=7-3sl1pSojc

~bobthefam

Video Solution 4

https://youtu.be/OKoBg15l8ro

~savannahsolve

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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

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