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

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== Solution ==
 
== Solution ==
 
<math>20-2010+201+2010-201+20=20+20=\boxed{\textbf{(C)}\,40}</math>.
 
<math>20-2010+201+2010-201+20=20+20=\boxed{\textbf{(C)}\,40}</math>.
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== Video Solution ==
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https://youtu.be/KBlf0TKdI4I?si=XFer3NW2lpMxgjW_
  
 
== See Also ==
 
== See Also ==

Revision as of 09:50, 18 September 2024

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

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

Video Solution

https://youtu.be/KBlf0TKdI4I?si=XFer3NW2lpMxgjW_

See Also

2010 AMC 12A (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 12 Problems and Solutions

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