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> | ||
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− | ==Solution | + | ==Solution 1 (Bashing)== |
We take a common denominator: | 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> | <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> | ||
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~ dolphin7 - I took your idea and made it an explanation. | ~ dolphin7 - I took your idea and made it an explanation. | ||
− | ==Solution | + | - Clearness by doulai1 |
+ | |||
+ | ==Solution 2== | ||
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>. | 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 | ~ ryjs | ||
− | This is also similar to Problem 20 on the AMC | + | This is also similar to Problem 20 on the 2012 AMC 8. |
− | ==Solution | + | ==Solution 3 (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>. | 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 | + | ~~ by an insane math guy. |
+ | |||
+ | ==Solution 4== | ||
+ | Suppose each fraction is expressed with denominator <math>2145</math>: <math>\frac{2925}{2145}, \frac{2717}{2145}, \frac{2805}{2145}</math>. Clearly <math>2717<2805<2925</math> so the answer is <math>\boxed{\textbf{(E)}}</math>. | ||
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+ | ==Solution 5 -SweetMango77== | ||
+ | We notice that each of these fraction's numerator <math>-</math> denominator <math>=4</math>. If we take each of the fractions, and subtract <math>1</math> from each, we get <math>\frac{4}{11}</math>, <math>\frac{4}{15}</math>, and <math>\frac{4}{19}</math>. These are easy to order because the numerators are the same, we get <math>\frac{4}{15}<\frac{4}{13}<\frac{4}{11}</math>. Because it is a subtraction by a constant, in order to order them, we keep the inequality signs to get <math>\boxed{\textbf{(E)}\;\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>. | ||
+ | |||
+ | == Video Solution == | ||
+ | |||
+ | Solution detailing how to solve the problem: https://www.youtube.com/watch?v=q27qEcr7TbQ&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=4 | ||
− | ==See | + | ==See also== |
{{AMC8 box|year=2019|num-b=2|num-a=4}} | {{AMC8 box|year=2019|num-b=2|num-a=4}} | ||
− | {{MAA Notice}} | + | {{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. |
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Latest revision as of 21:02, 18 September 2021
Contents
Problem 3
Which of the following is the correct order of the fractions and from least to greatest?
Solution 1 (Bashing)
We take a common denominator:
Since it follows that the answer is .
-xMidnightFirex
~ dolphin7 - I took your idea and made it an explanation.
- Clearness by doulai1
Solution 2
When and , . Hence, the answer is . ~ ryjs
This is also similar to Problem 20 on the 2012 AMC 8.
Solution 3 (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.
Solution 4
Suppose each fraction is expressed with denominator : . Clearly so the answer is .
Solution 5 -SweetMango77
We notice that each of these fraction's numerator denominator . If we take each of the fractions, and subtract from each, we get , , and . These are easy to order because the numerators are the same, we get . Because it is a subtraction by a constant, in order to order them, we keep the inequality signs to get .
Video Solution
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=q27qEcr7TbQ&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=4
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.