Difference between revisions of "2019 AMC 8 Problems/Problem 3"
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<math>\textbf{(A) }\frac{15}{11}< \frac{17}{13}< \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}< \frac{19}{15}<\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}<\frac{19}{15}<\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}<\frac{15}{11}<\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}<\frac{17}{13}<\frac{15}{11}</math> | <math>\textbf{(A) }\frac{15}{11}< \frac{17}{13}< \frac{19}{15} \qquad\textbf{(B) }\frac{15}{11}< \frac{19}{15}<\frac{17}{13} \qquad\textbf{(C) }\frac{17}{13}<\frac{19}{15}<\frac{15}{11} \qquad\textbf{(D) } \frac{19}{15}<\frac{15}{11}<\frac{17}{13} \qquad\textbf{(E) } \frac{19}{15}<\frac{17}{13}<\frac{15}{11}</math> | ||
==Solution 1== | ==Solution 1== | ||
− | + | Each one is in the form <math>\frac{x+4}{x}</math> so we are really comparing <math>\frac{4}{11}, \frac{4}{15},</math> and <math>\frac{4}{13}</math> where you can see <math>\frac{4}{11}>\frac{4}{13}>\frac{4}{15}</math> so the answer is <math>\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>. | |
==Solution 2== | ==Solution 2== |
Revision as of 18:03, 15 July 2020
Contents
Problem 3
Which of the following is the correct order of the fractions and from least to greatest?
Solution 1
Each one is in the form so we are really comparing and where you can see so the answer is .
Solution 2
We take a common denominator:
Since it follows that the answer is .
-xMidnightFirex
~ dolphin7 - I took your idea and made it an explanation.
Solution 3
When and , . Hence, the answer is . ~ ryjs
This is also similar to Problem 20 on the AMC 2012.
Solution 4(probably won't use this solution)
We use our insane mental calculator to find out that , , and . Thus, our answer is .
~~ by an insane math guy
See Also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.
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