Difference between revisions of "2020 AMC 10A Problems/Problem 1"
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==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> | ||
− | + | == 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>. | 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== | ==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>. | + | 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 1== | ==Video Solution 1== | ||
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https://youtu.be/OKoBg15l8ro | https://youtu.be/OKoBg15l8ro | ||
− | ~ | + | ~savannahsolve |
== See Also == | == See Also == |
Latest revision as of 12:05, 29 December 2022
Contents
Problem
What value of satisfies
Solution 1
Adding to both sides, .
Solution 2
Multiplying on both sides gets us , therefore .
Video Solution 1
Education, The Study of Everything
Video Solution 2
~IceMatrix https://youtu.be/WUcbVNy2uv0
Video Solution 3
https://www.youtube.com/watch?v=7-3sl1pSojc
~bobthefam
Video Solution 4
~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.