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

Latest revision as of 13:35, 8 November 2020

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

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

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

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

~savannahsolver

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

@@ Line 1: / Line 1: @@
-First, simplify the fractions.
+== Problem==
-<math>\dfrac{2+4+6}{1+3+5} - \dfrag{1+3+5}{2+4+6} = \dfrac{12}{9} - \dfrac{9}{12}</math>
+What is <cmath>\dfrac{2+4+6}{1+3+5} - \dfrac{1+3+5}{2+4+6} ?</cmath>
-<math>\dfrac{12}{9} - \dfrac{9}{12} = \dfrac{48}{36} - \dfrac{27}{36} = \dfrac{21}{36} = \dfrac{7}{12}</math>
+<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>
-<math>(C) \dfrac{7}{12}</math>
+== Solution ==
+<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>
+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 ==
+{{AMC10 box|year=2011|ab=B|before=First Problem|num-a=2}}
+{{MAA Notice}}