Difference between revisions of "2019 AMC 8 Problems/Problem 3"
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+ | ==Problem 3== | ||
+ | Which of the following is the correct order of the fractions <math>\frac{15}{11},\frac{19}{15},</math> and <math>\frac{17}{13},</math> from least to greatest? | ||
+ | <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== | ||
+ | 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== | ||
+ | We take a common denominator: | ||
+ | <cmath>\frac{15}{11},\frac{19}{15}, \frac{17}{13} = \frac{15\cdot 15 \cdot 13}{11\cdot 15 \cdot 13},\frac{19 \cdot 11 \cdot 13}{15\cdot 11 \cdot 13}, \frac{17 \cdot 11 \cdot 15}{13\cdot 11 \cdot 15} = \frac{2925}{2145},\frac{2717}{2145},\frac{2805}{2145}.</cmath> | ||
+ | |||
+ | Since <math>2717<2805<2925</math> it follows that the answer is <math>\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>. | ||
+ | |||
+ | -xMidnightFirex | ||
+ | |||
+ | ~ dolphin7 - I took your idea and made it an explanation. | ||
+ | |||
+ | ==Solution 3== | ||
+ | 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 <math>\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>. | ||
+ | ~ 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 <math>\frac{15}{11} \approx 1.36</math>, <math>\frac{19}{15} \approx 1.27</math>, and <math>\frac{17}{13} \approx 1.31</math>. Thus, our answer is <math>\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>. | ||
+ | |||
+ | ~~ by an insane math guy | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC8 box|year=2019|num-b=2|num-a=4}} | ||
+ | |||
+ | {{MAA Notice}} The butterfly method is a method when you multiply the denominator of the second fraction and multiply it by the numerator from the first fraction. |
Revision as of 19:53, 15 October 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.
The butterfly method is a method when you multiply the denominator of the second fraction and multiply it by the numerator from the first fraction.