Difference between revisions of "2010 AMC 12A Problems/Problem 1"

Revision as of 16:18, 16 February 2015

What is $\left(20-\left(2010-201\right)\right)+\left(2010-\left(201-20\right)\right)$ ?

$\textbf{(A)}\ -4020 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 401 \qquad \textbf{(E)}\ 4020$

$20-2010+201+2010-201+20=20+20=\boxed{\textbf{(C)}\,40}$ .

2010 AMC 12A (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 12 Problems and Solutions

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 == Solution ==
-<math>20-2010+201+2010-201+20=20+20=40; \boxed{\textbf{(C)}}</math>.
+<math>20-2010+201+2010-201+20=20+20=\boxed{\textbf{(C)}\,40}</math>.
 == See also ==