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

(See Also =)
(Video Solution)
(5 intermediate revisions by 4 users not shown)
Line 11: Line 11:
 
The equation becomes <math>1+2-3+4-5+6 = \boxed{\textbf{(D)}\ 5}</math> ~quacker88
 
The equation becomes <math>1+2-3+4-5+6 = \boxed{\textbf{(D)}\ 5}</math> ~quacker88
  
== Solution ==
 
Solution
 
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/Gkm5rU5MlOU
 
https://youtu.be/Gkm5rU5MlOU
Line 18: Line 16:
 
~IceMatrix
 
~IceMatrix
  
= See Also =
+
 
 +
https://youtu.be/-wciFhP5h3I
 +
 
 +
~savannahsolver
 +
 
 +
==See Also==
  
 
{{AMC10 box|year=2020|ab=B|before=First Problem|num-a=2}}
 
{{AMC10 box|year=2020|ab=B|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:52, 16 June 2020

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

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

Video Solution

https://youtu.be/Gkm5rU5MlOU

~IceMatrix


https://youtu.be/-wciFhP5h3I

~savannahsolver

See Also

2020 AMC 10B (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

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