2024 AMC 8 Problems/Problem 21
Contents
- 1 Problem
- 2 Solution 2
- 3 Video Solution
- 4 Video Solution by Power Solve (crystal clear)
- 5 Video Solution 1 by Math-X (First fully understand the problem!!!)
- 6 Video Solution 2 by OmegaLearn.org
- 7 Video Solution 3 by SpreadTheMathLove
- 8 Video Solution by NiuniuMaths (Easy to understand!)
- 9 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 10 Video Solution by Interstigation
- 11 See Also
Problem
A group of frogs (called an army) is living in a tree. A frog turns green when in the shade and turns yellow when in the sun. Initially, the ratio of green to yellow frogs was . Then green frogs moved to the sunny side and yellow frogs moved to the shady side. Now the ratio is . What is the difference between the number of green frogs and the number of yellow frogs now?
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Solution 2
Since the original ratio is and the new ratio is , the number of frogs must be a multiple of , the only solutions left are and .
Let's start with frogs:
We must have frogs in the shade and frogs in the sun. After the change, there would be frogs in the shade and frog in the sun, which is not a ratio.
Therefore the answer is: .
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Video Solution
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Video Solution by Power Solve (crystal clear)
https://www.youtube.com/watch?v=HodW9H55ZsE
Video Solution 1 by Math-X (First fully understand the problem!!!)
https://youtu.be/BaE00H2SHQM?si=yTyYiS2S75HoSfmI&t=6313
~Math-X
Video Solution 2 by OmegaLearn.org
Video Solution 3 by SpreadTheMathLove
https://www.youtube.com/watch?v=3ItvjukLqK0
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=looAMewBACY
~NiuniuMaths
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=3SUTUr1My7c&t=1s
Video Solution by Interstigation
https://youtu.be/ktzijuZtDas&t=2562
See Also
2024 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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.