2012 AMC 8 Problems/Problem 20: Difference between revisions
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<cmath>\frac{7}{21}<\frac{9}{23}</cmath> | <cmath>\frac{7}{21}<\frac{9}{23}</cmath> | ||
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 4== | |||
When <math>\frac{x}{y}<1</math> and <math>z>0</math>, <math>\frac{x+z}{y+z}>\frac{x}{y}</math>. Hence, the answer is {\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} $. | |||
~ ryjs | |||
This is also similar to Problem 3 on the AMC 8 2019, but with the rule switched. | |||
==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 02:18, 24 December 2019
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 .
Solution 3
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 4
When 
and 
, 
. Hence, the answer is {\textbf{(B)}\ \frac{5}{19}<\frac{7}{21}<\frac{9}{23}} $.
~ ryjs
This is also similar to Problem 3 on the AMC 8 2019, but with the rule switched.
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
