2020 AMC 10B Problems/Problem 1: Difference between revisions
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<math>\textbf{(A)}\ -20 \qquad\textbf{(B)}\ -3 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 21</math> | <math>\textbf{(A)}\ -20 \qquad\textbf{(B)}\ -3 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 21</math> | ||
==Solution== | ==Solution 1== | ||
We know that when we subtract negative numbers, <math>a-(-b)=a+b</math>. | We know that when we subtract negative numbers, <math>a-(-b)=a+b</math>. | ||
The equation becomes <math>1+2-3+4-5+6 = \boxed{\textbf{(D)}\ 5}</math> | The equation becomes <math>1+2-3+4-5+6 = \boxed{\textbf{(D)}\ 5}</math>. | ||
==Video Solution== | ~quacker88 | ||
==Solution 2== | |||
Like Solution 1, we know that when we subtract <math>a-(-b)</math>, that will equal <math>a+b</math> as the opposite/negative of a negative is a positive. Thus, <math>1-(-2)-3-(-4)-5-(-6)=1+2-3+4-5+6</math>. We can group together a few terms to make our computation a bit simpler. <math>1+(2-3)+4+(-5+6)= 1+(-1)+4+1=\boxed{\textbf{(D)}\ 5}</math>. | |||
~BakedPotato66 | |||
== Solution 3 == | |||
Notice <math>1-(-2)-3-(-4)-5-(-6)</math> has three groups: <math>1-(-2)</math>, <math>-3-(-4)</math>, <math>-5-(-6)</math> | |||
The first group, <math>1-(-2)</math>, can be expressed as <math>a-[-(a+1)]</math>. | |||
Simplify, | |||
<math>a-[-(a+1)]</math> is <math>2a+1</math> | |||
The second and third groups can be expressed as <math>-c-[-(c+1)]</math>. | |||
Simplify, | |||
<math>-c-[-(c+1)]</math> is <math>1</math> | |||
Thus, the sum of the three groups is <math>2 \cdot 1 + 1 + 1 + 1 = 5</math> or <math>\boxed{\textbf{(D)}\ 5}</math> | |||
~ lovelearning999 | |||
==Video Solutions== | |||
===Video Solution by Education, the study of everything=== | |||
https://www.youtube.com/watch?v=NpDVTLSi-Ik | |||
~Education, the Study of Everything | |||
===Video Solution by TheBeautyofMath=== | |||
https://youtu.be/Gkm5rU5MlOU | https://youtu.be/Gkm5rU5MlOU | ||
~IceMatrix | ~IceMatrix | ||
===Video Solution by WhyMath=== | |||
https://youtu.be/-wciFhP5h3I | |||
~savannahsolver | |||
===Video Solution by Alex Explains=== | |||
https://www.youtube.com/watch?v=GNPAgQ8fSP0&t | |||
~AlexExplains | |||
==See Also== | ==See Also== | ||
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{{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}} | ||
[[Category: Introductory Algebra Problems]] | |||
Latest revision as of 16:16, 16 October 2025
Problem
What is the value of
Solution 1
We know that when we subtract negative numbers,
.
The equation becomes
.
~quacker88
Solution 2
Like Solution 1, we know that when we subtract
, that will equal
as the opposite/negative of a negative is a positive. Thus,
. We can group together a few terms to make our computation a bit simpler.
.
~BakedPotato66
Solution 3
Notice
has three groups:
,
,
The first group,
, can be expressed as
.
Simplify,
is
The second and third groups can be expressed as
.
Simplify,
is
Thus, the sum of the three groups is
or
~ lovelearning999
Video Solutions
Video Solution by Education, the study of everything
https://www.youtube.com/watch?v=NpDVTLSi-Ik
~Education, the Study of Everything
Video Solution by TheBeautyofMath
~IceMatrix
Video Solution by WhyMath
~savannahsolver
Video Solution by Alex Explains
https://www.youtube.com/watch?v=GNPAgQ8fSP0&t
~AlexExplains
See Also
| 2020 AMC 10B (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 10 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America.