Difference between revisions of "2012 AMC 8 Problems/Problem 20"
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<math> \textbf{(D)}\hspace{.05in}\frac{5}{19}<\frac{9}{23}<\frac{7}{21}\quad\textbf{(E)}\hspace{.05in}\frac{7}{21}<\frac{5}{19}<\frac{9}{23} </math> | <math> \textbf{(D)}\hspace{.05in}\frac{5}{19}<\frac{9}{23}<\frac{7}{21}\quad\textbf{(E)}\hspace{.05in}\frac{7}{21}<\frac{5}{19}<\frac{9}{23} </math> | ||
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| + | ==Solution== | ||
| + | The value of <math> \frac{7}{21} </math> is <math> \frac{1}{3} </math>. Now we give all the fractions a common denominator. | ||
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| + | <math> \frac{5}{19} \implies \frac{345}{1311} </math> | ||
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| + | <math> \frac{1}{3} \implies \frac{437}{1311} </math> | ||
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| + | <math> \frac{9}{23} \implies \frac{513}{1311} </math> | ||
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| + | Ordering the fractions from least to greatest, we find that they are in the order listed. Therefore, our final answer is <math> \boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} </math>. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2012|num-b=19|num-a=21}} | {{AMC8 box|year=2012|num-b=19|num-a=21}} | ||
Revision as of 11:54, 24 November 2012
What is the correct ordering of the three numbers 
, 
, and 
, in increasing order?
Solution
The value of is 
. Now we give all the fractions a common denominator.
Ordering the fractions from least to greatest, we find that they are in the order listed. Therefore, our final answer is 
.
See Also
| 2012 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
