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

## 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

~IceMatrix

~bobthefam

~savannahsolver

## See Also

 2020 AMC 10A (Problems • Answer Key • Resources) Preceded byFirst Problem Followed byProblem 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.

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