Difference between revisions of "2010 AMC 10B Problems/Problem 1"

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<math>
 
<math>
\mathrm{(A)}\ -20,000
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\textbf{(A)}\ -20,000
 
\qquad
 
\qquad
\mathrm{(B)}\ -10,000
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\textbf{(B)}\ -10,000
 
\qquad
 
\qquad
\mathrm{(C)}\ -297
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\textbf{(C)}\ -297
 
\qquad
 
\qquad
\mathrm{(D)}\ -6
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\textbf{(D)}\ -6
 
\qquad
 
\qquad
\mathrm{(E)}\ 0
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\textbf{(E)}\ 0
 
</math>
 
</math>
 
== Solution ==
 
== Solution ==
We first expand the first term, simplify, and then compute to get an answer of <math>\boxed{\mathrm {(C)} -297}</math>.
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<math>100(100-3)-(100\cdot{100}-3)=10000-300-10000+3=-300+3=\boxed{\textbf{(C)}\ -297}</math>.
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==Video Solution==
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https://youtu.be/uAc9VHtRRPg
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 +
~IceMatrix
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2010|ab=B|before=First Problem|num-a=2}}
 
{{AMC10 box|year=2010|ab=B|before=First Problem|num-a=2}}
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{{MAA Notice}}

Revision as of 03:00, 26 September 2020

Problem

What is $100(100-3)-(100\cdot100-3)$?

$\textbf{(A)}\ -20,000 \qquad \textbf{(B)}\ -10,000 \qquad \textbf{(C)}\ -297 \qquad \textbf{(D)}\ -6 \qquad \textbf{(E)}\ 0$

Solution

$100(100-3)-(100\cdot{100}-3)=10000-300-10000+3=-300+3=\boxed{\textbf{(C)}\ -297}$.

Video Solution

https://youtu.be/uAc9VHtRRPg

~IceMatrix

See Also

2010 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

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