Difference between revisions of "2012 AMC 8 Problems/Problem 20"
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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 03:18, 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 ; And 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.