Difference between revisions of "2011 AMC 12B Problems/Problem 1"

m (Solution)
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== Solution ==
 
== Solution ==
Add up the numbers in each fraction to get <math>\frac{12}{9} - \frac{9}{12}</math>, which equals <math>\frac{4}{3} - \frac{3}{4}</math>. Doing the subtraction yields
+
Add up the numbers in each fraction to get <math>\frac{12}{9} - \frac{9}{12}</math>, which equals <math>\frac{4}{3} - \frac{3}{4}</math>. Doing the subtraction yields <math>\boxed{\frac{7}{12}\  \textbf{(C)}}</math>
  
<math>
 
\boxed{\frac{7}{12}\  \textbf{(C)}}
 
</math>
 
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|before=First Problem|num-a=2|ab=B}}
 
{{AMC12 box|year=2011|before=First Problem|num-a=2|ab=B}}

Revision as of 02:56, 9 March 2012

Problem

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

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


Solution

Add up the numbers in each fraction to get $\frac{12}{9} - \frac{9}{12}$, which equals $\frac{4}{3} - \frac{3}{4}$. Doing the subtraction yields $\boxed{\frac{7}{12}\   \textbf{(C)}}$

See also

2011 AMC 12B (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 12 Problems and Solutions