2019 AMC 8 Problems/Problem 3
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.