Difference between revisions of "2013 AMC 10B Problems/Problem 1"

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This expression is equivalent to <math>\frac{12}{9} - \frac{9}{12} = \frac{16}{12} - \frac{9}{12} = \boxed{\textbf{(C) }\frac{7}{12}}</math>
 
This expression is equivalent to <math>\frac{12}{9} - \frac{9}{12} = \frac{16}{12} - \frac{9}{12} = \boxed{\textbf{(C) }\frac{7}{12}}</math>
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For the exact same problem but with answer choice D unsimplified, see http://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_1
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==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2013|ab=B|before=First Problem|num-a=2}}
 
{{AMC10 box|year=2013|ab=B|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:04, 13 February 2017

Problem

What is $\frac{2+4+6}{1+3+5} - \frac{1+3+5}{2+4+6}$?

$\textbf{(A)}\ -1 \qquad\textbf{(B)}\ \frac{5}{36}  \qquad\textbf{(C)}\ \frac{7}{12} \qquad\textbf{(D)}\ \frac{49}{20} \qquad\textbf{(E)}\ \frac{43}{3}$

Solution

This expression is equivalent to $\frac{12}{9} - \frac{9}{12} = \frac{16}{12} - \frac{9}{12} = \boxed{\textbf{(C) }\frac{7}{12}}$

For the exact same problem but with answer choice D unsimplified, see http://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_1

See Also

2013 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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All AMC 10 Problems and Solutions

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