2012 AMC 8 Problems/Problem 20: Difference between revisions
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<math>1-\dfrac{9}{23}=\dfrac{14}{23}</math> | <math>1-\dfrac{9}{23}=\dfrac{14}{23}</math> | ||
All three fractions have common | All three fractions have common numerator <math>14</math>. Now it is obvious the order of the fractions. <math>\dfrac{14}{19}>\dfrac{14}{21}>\dfrac{14}{23}\implies\dfrac{5}{19}<\dfrac{7}{21}<\dfrac{9}{23}</math>. Therefore, our 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}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 22:36, 20 November 2013
Problem
What is the correct ordering of the three numbers 
, 
, and 
, in increasing order?
Solution 1
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 
.
Solution 2
Instead of finding the LCD, we can subtract each fraction from 
to get a common numerator. Thus,
All three fractions have common numerator 
. Now it is obvious the order of the fractions. 
. Therefore, our 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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
