Difference between revisions of "2019 AMC 8 Problems/Problem 3"
(→Solution 6(similar to Solution 1)) |
(→Solution 1 (Bashing)) |
||
(34 intermediate revisions by 22 users not shown) | |||
Line 3: | Line 3: | ||
<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 | + | ==Solution 1 (Bashing/Butterfly Method)== |
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> | ||
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>. | 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>. | ||
+ | |||
+ | Another approach to this problem is using the properties of one fraction being greater than another, also known as the <b>butterfly method</b>. That is, if | ||
+ | <math>\frac{a}{b}>\frac{c}{d}</math>, then it must be true that <math>a * d</math> is greater than <math>b * c</math>. Using this approach, we can check for <b>at least</b> two distinct pairs of fractions and find out the greater one of those two, logically giving us the expected answer of <math>\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>. | ||
-xMidnightFirex | -xMidnightFirex | ||
Line 16: | Line 17: | ||
~ dolphin7 - I took your idea and made it an explanation. | ~ dolphin7 - I took your idea and made it an explanation. | ||
− | ==Solution | + | - Clearness by doulai1 |
+ | |||
+ | - Alternate Solution by Nivaar | ||
+ | |||
+ | ==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. |
+ | ~~ random text that is here to distract you. | ||
− | ==Solution | + | ==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>. | 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>. | ||
− | ==See | + | *Note: Duplicate of Solution 1 |
+ | |||
+ | ==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}{13}</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>. | ||
+ | |||
+ | == Solution 6 == | ||
+ | Adding on to Solution 5, we can turn each of the fractions <math>\frac{15}{11}</math>, <math>\frac{17}{13}</math>, and <math>\frac{19}{15}</math> into <math>1</math><math>\frac{4}{11}</math>, <math>1</math><math>\frac{4}{13}</math>, and <math>1</math><math>\frac{4}{15}</math>, respectively. We now subtract <math>1</math> from each to get <math>\frac{4}{11}</math>, <math>\frac{4}{15}</math>, and <math>\frac{4}{13}</math>. Since their numerators are all 4, this is easy because we know that <math>\frac{1}{15}<\frac{1}{13}<\frac{1}{11}</math> and therefore <math>\frac{4}{15}<\frac{4}{13}<\frac{4}{11}</math>. Reverting them back to their original fractions, we can now see that the answer is <math>\boxed{\textbf{(E)}\;\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}</math>. | ||
+ | |||
+ | ~by ChipmunkT | ||
+ | |||
+ | |||
+ | == Video Solution == | ||
+ | |||
+ | ==Video Solution by Math-X (First fully understand the problem!!!)== | ||
+ | https://youtu.be/IgpayYB48C4?si=-TpVe8QyZbbc6yKr&t=266 | ||
+ | |||
+ | ~Math-X | ||
+ | |||
+ | The Learning Royal: https://youtu.be/IiFFDDITE6Q | ||
+ | |||
+ | == Video Solution 2 == | ||
+ | |||
+ | Solution detailing how to solve the problem: https://www.youtube.com/watch?v=q27qEcr7TbQ&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=4 | ||
+ | |||
+ | ==Video Solution 3== | ||
+ | https://youtu.be/xNrhMAbaEcI | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution == | ||
+ | https://youtu.be/zI2f4GpQPIo | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ====Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)== | ||
+ | https://youtu.be/Xm4ZGND9WoY | ||
+ | |||
+ | ~Hayabusa1 | ||
+ | |||
+ | ==See also== | ||
{{AMC8 box|year=2019|num-b=2|num-a=4}} | {{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. | {{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. |
Latest revision as of 10:28, 23 January 2024
Contents
- 1 Problem 3
- 2 Solution 1 (Bashing/Butterfly Method)
- 3 Solution 2
- 4 Solution 3 (probably won't use this solution)
- 5 Solution 4
- 6 Solution 5 -SweetMango77
- 7 Solution 6
- 8 Video Solution
- 9 Video Solution by Math-X (First fully understand the problem!!!)
- 10 Video Solution 2
- 11 Video Solution 3
- 12 Video Solution
- 13 ==Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
- 14 See also
Problem 3
Which of the following is the correct order of the fractions and from least to greatest?
Solution 1 (Bashing/Butterfly Method)
We take a common denominator:
Since it follows that the answer is .
Another approach to this problem is using the properties of one fraction being greater than another, also known as the butterfly method. That is, if , then it must be true that is greater than . Using this approach, we can check for at least two distinct pairs of fractions and find out the greater one of those two, logically giving us the expected answer of .
-xMidnightFirex
~ dolphin7 - I took your idea and made it an explanation.
- Clearness by doulai1
- Alternate Solution by Nivaar
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. ~~ random text that is here to distract you.
Solution 4
Suppose each fraction is expressed with denominator : . Clearly so the answer is .
- Note: Duplicate of Solution 1
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 .
Solution 6
Adding on to Solution 5, we can turn each of the fractions , , and into , , and , respectively. We now subtract from each to get , , and . Since their numerators are all 4, this is easy because we know that and therefore . Reverting them back to their original fractions, we can now see that the answer is .
~by ChipmunkT
Video Solution
Video Solution by Math-X (First fully understand the problem!!!)
https://youtu.be/IgpayYB48C4?si=-TpVe8QyZbbc6yKr&t=266
~Math-X
The Learning Royal: https://youtu.be/IiFFDDITE6Q
Video Solution 2
Solution detailing how to solve the problem: https://www.youtube.com/watch?v=q27qEcr7TbQ&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=4
Video Solution 3
~savannahsolver
Video Solution
~Education, the Study of Everything
==Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
~Hayabusa1
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.