2012 AMC 8 Problems/Problem 20: Difference between revisions
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Therefore, our answer is <math> \boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} </math>. | Therefore, our answer is <math> \boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} </math>. | ||
==Solution 3== | ==Solution 3 (quick and easy)== | ||
We know that <math>\frac{5}{19}</math> is <math>\frac{14}{19}</math> away from 0, <math>\frac{7}{21}</math> is <math>\frac{14}{21}</math> away from 0, and <math>\frac{9}{23}</math> is <math>\frac{14}{23}</math> away from 0. Since <math>\frac{14}{19}</math> is the largest, we know that it is the farthest away from 0, and <math>\frac{14}{23}</math> is the smallest, so it is the closest to 0. Therefore, our answer is <math> \boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} </math>. | We know that <math>\frac{5}{19}</math> is <math>\frac{14}{19}</math> away from 0, <math>\frac{7}{21}</math> is <math>\frac{14}{21}</math> away from 0, and <math>\frac{9}{23}</math> is <math>\frac{14}{23}</math> away from 0. Since <math>\frac{14}{19}</math> is the largest, we know that it is the farthest away from 0, and <math>\frac{14}{23}</math> is the smallest, so it is the closest to 0. Therefore, our answer is <math> \boxed{\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} </math>. | ||
Revision as of 11:22, 16 November 2024
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
Change 
into 
;
![\[\frac{1}{3}\cdot\frac{5}{5}=\frac{5}{15}\]](http://latex.artofproblemsolving.com/5/1/f/51ff65c8d23f72a31b94d65d657981c33a23753e.png)
![\[\frac{5}{15}>\frac{5}{19}\]](http://latex.artofproblemsolving.com/a/e/d/aed69704eee407e6b45f914272f14f9fbda8dcd7.png)
![\[\frac{7}{21}>\frac{5}{19}\]](http://latex.artofproblemsolving.com/9/9/1/991bcb813a14be818141a0aeed6dff7b78e7fe87.png)
And
![\[\frac{1}{3}\cdot\frac{9}{9}=\frac{9}{27}\]](http://latex.artofproblemsolving.com/3/7/c/37cfd4a051dc8159cd0ef6c5511ecdd185d8ff1e.png)
![\[\frac{9}{27}<\frac{9}{23}\]](http://latex.artofproblemsolving.com/2/8/2/282cce073d0e86fb997b827ff33b7e25e26a8721.png)
![\[\frac{7}{21}<\frac{9}{23}\]](http://latex.artofproblemsolving.com/4/1/f/41f1c28eaf5a9a565b0250312fa1f546af104412.png)
Therefore, our answer is .
Solution 3 (quick and easy)
We know that is 
away from 0, is 
away from 0, and is 
away from 0. Since is the largest, we know that it is the farthest away from 0, and is the smallest, so it is the closest to 0. Therefore, our answer is .
~monkey_land
Video Solution
https://youtu.be/pU1zjw--K8M ~savannahsolver
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, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
