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

Revision as of 07:56, 1 September 2020

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 $12x-9=1$ , therefore $\boxed{x=\textbf{(E)}~\frac{5}{6}}$ . ~CoolJupiter

Video Solution

https://youtu.be/WUcbVNy2uv0

~IceMatrix

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

~bobthefam

https://youtu.be/OKoBg15l8ro

~savannahsolver

2020 AMC 10A (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
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All AMC 10 Problems and Solutions

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

@@ Line 9: / Line 9: @@
 ==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
+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
 ==Video Solution==

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