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

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

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.

Page

Toolbox

Search