2020 AMC 10A Problems/Problem 1

Problem 1

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{\text{(E) }\frac{5}{5}}$.

2020 AMC 10A (Problems • Answer Key • Resources)
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2020 AMC 10A Problems/Problem 1

Problem 1

Solution

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