Difference between revisions of "2011 AMC 10B Problems/Problem 1"

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== Solution ==
 
== Solution ==
  
First, simplify the fractions.
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<math>\dfrac{2+4+6}{1+3+5} - \dfrac{1+3+5}{2+4+6} = \dfrac{12}{9} - \dfrac{9}{12} = \dfrac{4}{3} - \dfrac{3}{4} = \boxed{\dfrac{7}{12}\; \textbf{(C)}}</math>
  
<math>\dfrac{2+4+6}{1+3+5} - \dfrac{1+3+5}{2+4+6} = \dfrac{12}{9} - \dfrac{9}{12}</math>
 
  
<math>\dfrac{12}{9} - \dfrac{9}{12} = \dfrac{48}{36} - \dfrac{27}{36} = \dfrac{21}{36} = \boxed{\dfrac{7}{12} \textbf{(C)}}</math>
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Note: This exact problem was reused in 2013 AMC 10B:
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https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_1
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==Video Solution==
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https://youtu.be/bkRNTz2IJE8
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~savannahsolver
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2011|ab=B|before=First Problem|num-a=2}}
 
{{AMC10 box|year=2011|ab=B|before=First Problem|num-a=2}}
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{{MAA Notice}}

Latest revision as of 14:35, 8 November 2020

Problem

What is \[\dfrac{2+4+6}{1+3+5} - \dfrac{1+3+5}{2+4+6} ?\]

$\textbf{(A)}\ -1\qquad\textbf{(B)}\ \frac{5}{36}\qquad\textbf{(C)}\ \frac{7}{12}\qquad\textbf{(D)}\ \frac{147}{60}\qquad\textbf{(E)}\ \frac{43}{3}$

Solution

$\dfrac{2+4+6}{1+3+5} - \dfrac{1+3+5}{2+4+6} = \dfrac{12}{9} - \dfrac{9}{12} = \dfrac{4}{3} - \dfrac{3}{4} = \boxed{\dfrac{7}{12}\; \textbf{(C)}}$


Note: This exact problem was reused in 2013 AMC 10B:

https://artofproblemsolving.com/wiki/index.php/2013_AMC_10B_Problems/Problem_1

Video Solution

https://youtu.be/bkRNTz2IJE8

~savannahsolver

See Also

2011 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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