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

Revision as of 17:53, 25 May 2011

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

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

First, simplify the fractions.

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

$\dfrac{12}{9} - \dfrac{9}{12} = \dfrac{48}{36} - \dfrac{27}{36} = \dfrac{21}{36} = \boxed{\dfrac{7}{12} \textbf{(C)}}$

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+== Problem 1 ==
+What is <cmath>\dfrac{2+4+6}{1+3+5} - \dfrac{1+3+5}{2+4+6} ?</cmath>
+<math> \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} </math>
+== Solution ==
 First, simplify the fractions.
 <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} = \boxed{\dfrac{7}{12} \textbf{(C)}}</math>
-<math>(C) \dfrac{7}{12}</math>