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

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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 <math>\boxed{\frac{7}{12}\  \textbf{(C)}}</math>
 
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>
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= Solution 2 ==
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Notice that the numerators and denominators of each expression are 3-term arithmetic series. The sum of an arithmetic series is the middle term multiplied with the number of terms. Since each of the arithmetic series have the same number of terms, we can replace the fractions with the middle terms, which gives us <math>\frac{4}{3} - \frac{3}{4} = \frac{7}{12}</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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:13, 3 January 2019

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 $\boxed{\frac{7}{12}\   \textbf{(C)}}$


Solution 2 =

Notice that the numerators and denominators of each expression are 3-term arithmetic series. The sum of an arithmetic series is the middle term multiplied with the number of terms. Since each of the arithmetic series have the same number of terms, we can replace the fractions with the middle terms, which gives us $\frac{4}{3} - \frac{3}{4} = \frac{7}{12}$

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

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