Difference between revisions of "2012 AMC 8 Problems/Problem 20"
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==Solution 2== | ==Solution 2== | ||
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Change <math>7/21</math> into <math>1/3</math>; | Change <math>7/21</math> into <math>1/3</math>; | ||
<cmath>\frac{1}{3}\cdot\frac{5}{5}=\frac{5}{15}</cmath> | <cmath>\frac{1}{3}\cdot\frac{5}{5}=\frac{5}{15}</cmath> | ||
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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 | + | ==Video Solution== |
− | + | https://youtu.be/pU1zjw--K8M ~savannahsolver | |
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==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}} |
Latest revision as of 09:12, 16 July 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
;
And
Therefore, our answer is
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.