2020 AMC 10B Problems/Problem 1

Problem

What is the value of $1-(-2)-3-(-4)-5-(-6)?$

$\textbf{(A)}\ -20 \qquad\textbf{(B)}\ -3 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 21$

Solution 1

We know that when we subtract negative numbers, $a-(-b)=a+b$ .

The equation becomes $1+2-3+4-5+6 = \boxed{\textbf{(D)}\ 5}$ ~quacker88

Solution 2

We know that when we subtract negative numbers, a-(-b), this will equal a+b, because a negative of a negative is a positive. Because we know this, we can rewrite the expression as 1+2-3+4-5+6. Now, we can simplify this expression. You will get 5. So, you will choose D. - BrightPorcupine

Solution 3

Like Solution 1, we know that when we subtract $a-(-b)$ , that will equal $a+b$ as the opposite/negative of a negative is a positive. Thus, $1-(-2)-3-(-4)-5-(-6)=1+2-3+4-5+6$ . We can group together a few terms to make our computation a bit simpler. $1+(2-3)+4+(-5+6)= 1+(-1)+4+1=\boxed{\textbf{(D) }5}$ .

~BakedPotato66

Video Solution

Check It Out! :) Education, the Study of Everything (wow!) https://www.youtube.com/watch?v=NpDVTLSi-Ik

https://youtu.be/Gkm5rU5MlOU

~IceMatrix

https://youtu.be/-wciFhP5h3I

~savannahsolver

https://www.youtube.com/watch?v=GNPAgQ8fSP0&t=1s

~AlexExplains

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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