2010 AMC 12A Problems/Problem 1

Problem

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

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

Solution

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

See Also

2010 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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All AMC 12 Problems and Solutions

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