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

Revision as of 07:39, 1 April 2015

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

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

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

2011 AMC 10B (Problems • Answer Key • Resources)
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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 == 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} = \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{4}{3} - \dfrac{3}{4} = \boxed{\dfrac{7}{12} \textbf{(C)}}</math>
-{thequantumguy}
 == See Also ==