Art of Problem Solving

2011 AMC 10B Problems/Problem 1: Difference between revisions

Created page with 'First, simplify the fractions. <math>\dfrac{2+4+6}{1+3+5} - \dfrag{1+3+5}{2+4+6} = \dfrac{12}{9} - \dfrac{9}{12}</math> <math>\dfrac{12}{9} - \dfrac{9}{12} = \dfrac{48}{36} - \…'
 
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First, simplify the fractions.
First, simplify the fractions.


<math>\dfrac{2+4+6}{1+3+5} - \dfrag{1+3+5}{2+4+6} = \dfrac{12}{9} - \dfrac{9}{12}</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} = \dfrac{7}{12}</math>
<math>\dfrac{12}{9} - \dfrac{9}{12} = \dfrac{48}{36} - \dfrac{27}{36} = \dfrac{21}{36} = \dfrac{7}{12}</math>


<math>(C) \dfrac{7}{12}</math>
<math>(C) \dfrac{7}{12}</math>

Revision as of 00:00, 27 February 2011

First, simplify the fractions.

$\dfrac{2+4+6}{1+3+5} - \dfrac{1+3+5}{2+4+6} = \dfrac{12}{9} - \dfrac{9}{12}$

$\dfrac{12}{9} - \dfrac{9}{12} = \dfrac{48}{36} - \dfrac{27}{36} = \dfrac{21}{36} = \dfrac{7}{12}$

$(C) \dfrac{7}{12}$