Difference between revisions of "2020 AMC 10A Problems/Problem 1"
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| − | + | ==Problem 1== | |
| + | What value of <math>x</math> satisfies <cmath>x- \frac{3}{4} = \frac{5}{12} - \frac{1}{3}?</cmath> | ||
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| + | <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> | ||
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| + | == Solution == | ||
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| + | 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}=</math>\boxed{\text{(E) }\frac{5}{5}}$. | ||
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| + | == See Also == | ||
| + | |||
| + | {{AMC10 box|year=2020|ab=A|before=First Problem|num-a=2}} | ||
| + | {{MAA Notice}} | ||
Revision as of 20:53, 31 January 2020
Problem 1
What value of 
satisfies ![\[x- \frac{3}{4} = \frac{5}{12} - \frac{1}{3}?\]](http://latex.artofproblemsolving.com/5/b/f/5bfe02ab9c82120a6496887cf9a11c3438b9be67.png)
Solution
Adding 
to both sides, 
\boxed{\text{(E) }\frac{5}{5}}$.
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. 
