Difference between revisions of "2020 AMC 10A Problems/Problem 1"
(Created page with "Will come out when 2020 amc 10 comes out.") |
Advancedjus (talk | contribs) |
||
Line 1: | Line 1: | ||
− | + | ==Problem 1== | |
+ | What value of <math>x</math> satisfies <cmath>x- \frac{3}{4} = \frac{5}{12} - \frac{1}{3}?</cmath> | ||
+ | |||
+ | <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> | ||
+ | |||
+ | == Solution == | ||
+ | |||
+ | 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}}$. | ||
+ | |||
+ | == See Also == | ||
+ | |||
+ | {{AMC10 box|year=2020|ab=A|before=First Problem|num-a=2}} | ||
+ | {{MAA Notice}} |
Revision as of 21:53, 31 January 2020
Problem 1
What value of satisfies
Solution
Adding to both sides, \boxed{\text{(E) }\frac{5}{5}}$.
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.