Difference between revisions of "2024 AMC 8 Problems/Problem 21"
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Line 20: | Line 20: | ||
<math>x=4</math> | <math>x=4</math> | ||
− | If we reverse the move, 16:4 becomes 14:6, which isn't a 3:1 ratio. | + | If we reverse the move, <math>16:4</math> becomes <math>14:6</math>, which isn't a <math>3:1</math> ratio. |
Therefore the answer is: <math>\boxed{E}</math> | Therefore the answer is: <math>\boxed{E}</math> | ||
− | Sidenote: <math>x+24=4x => x=8</math> | + | Sidenote: <math>x+24=4x => x=8</math> This goes from <math>32:8</math> to <math>30:10</math>, which does fit what the question is asking. |
==Video Solution by Power Solve (crystal clear)== | ==Video Solution by Power Solve (crystal clear)== |
Revision as of 21:23, 26 January 2024
Contents
[hide]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?
Solution 1
Let the initial number of green frogs be and the initial number of yellow frogs be . Since the ratio of the number of green frogs to yellow frogs is initially , . Now, green frogs move to the sunny side and yellow frogs move to the shade side, thus the new number of green frogs is and the new number of yellow frogs is . We are given that , so , since , we have , so and . Thus the answer is .
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:
If we reverse the move, becomes , which isn't a ratio.
Therefore the answer is:
Sidenote: This goes from to , which does fit what the question is asking.
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://www.youtube.com/watch?v=zBe5vrQbn2A
~Math-X
Video Solution 2 by OmegaLearn.org
Video Solution 3 by SpreadTheMathLove
https://www.youtube.com/watch?v=3ItvjukLqK0
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=3SUTUr1My7c&t=1s
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.