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

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Will come out when 2020 amc 10 comes out.
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==Problem 1==
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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 ==
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{{AMC10 box|year=2020|ab=A|before=First Problem|num-a=2}}
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{{MAA Notice}}

Revision as of 20:53, 31 January 2020

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}}$.

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
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

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