Difference between revisions of "2011 AMC 12B Problems/Problem 1"
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What is <math>\frac{2+4+6}{1+3+5} - \frac{1+3+5}{2+4+6}?</math> | What is <math>\frac{2+4+6}{1+3+5} - \frac{1+3+5}{2+4+6}?</math> | ||
| − | <math> | + | <math>\textbf{(A)}\ -1 \qquad |
| − | \textbf{(A)}\ -1 \qquad | ||
\textbf{(B)}\ \frac{5}{36} \qquad | \textbf{(B)}\ \frac{5}{36} \qquad | ||
\textbf{(C)}\ \frac{7}{12} \qquad | \textbf{(C)}\ \frac{7}{12} \qquad | ||
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\textbf{(E)}\ \frac{43}{3} </math> | \textbf{(E)}\ \frac{43}{3} </math> | ||
| + | == Solutions == | ||
| + | === Solution 1 === | ||
| + | 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> | ||
| − | == Solution == | + | === 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 <math>\frac{4}{3} - \frac{3}{4} = \frac{7}{12}</math> | |
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== 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}} | ||
Latest revision as of 13:12, 19 January 2021
Problem
What is 
Solutions
Solution 1
Add up the numbers in each fraction to get 
, which equals 
. Doing the subtraction yields 
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 
See also
| 2011 AMC 12B (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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
