Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Video Solution)
(→‎Solution 2)
 
(21 intermediate revisions by 10 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
What value of <math>x</math> satisfies <cmath>x- \frac{3}{4} = \frac{5}{12} - \frac{1}{3}?</cmath>
+
What value of <math>x</math> satisfies
 +
<cmath>x- \frac{3}{4} = \frac{5}{12} - \frac{1}{3}?</cmath>
 +
<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>
+
== Solution 1 ==
  
−
== Solution ==  
+
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>.
  
−
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>.
+
==Video Solution 1==
 +
Education, The Study of Everything
  
−
==Video Solution==
+
https://youtu.be/4lsmGWDYusk
−
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
 +
 +
~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)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_1&oldid=233204"