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

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<math>
 
<math>
\mathrm{(A)}\ -20,000
+
\textbf{(A)}\ -20,000
 
\qquad
 
\qquad
\mathrm{(B)}\ -10,000
+
\textbf{(B)}\ -10,000
 
\qquad
 
\qquad
\mathrm{(C)}\ -297
+
\textbf{(C)}\ -297
 
\qquad
 
\qquad
\mathrm{(D)}\ -6
+
\textbf{(D)}\ -6
 
\qquad
 
\qquad
\mathrm{(E)}\ 0
+
\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>.
+
We first expand the first term, simplify, and then compute to get an answer of <math>\boxed{\textbf{(C)}\ -297}</math>.
  
 
==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}}

Revision as of 14:01, 26 November 2011

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

We first expand the first term, simplify, and then compute to get an answer of $\boxed{\textbf{(C)}\ -297}$.

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