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>. | ||
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==Solution 4== | ==Solution 4== |
Revision as of 09:11, 16 July 2024
Contents
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
.
Solution 4
By dividing, we see that 5/19 ≈ 0.26, 7/21 ≈ 0.33, and 9/23 ≈ 0.39. When we put this in order, <
<
. So our answer is
~ math_genius_11
Video Solution
https://youtu.be/pU1zjw--K8M ~savannahsolver
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=TpsuRedYOiM&t=250s
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.