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

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

Latest revision as of 01:03, 11 November 2024

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 1

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 $x=\boxed{\textbf{(E)}~\frac{5}{6}}$.

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 (Problems • Answer Key • Resources)
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First Problem 		Followed by
Problem 2
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All AMC 10 Problems and Solutions

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