2024 AMC 8 Problems/Problem 21

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 $3 : 1$ . Then $3$ green frogs moved to the sunny side and $5$ yellow frogs moved to the shady side. Now the ratio is $4 : 1$ . What is the difference between the number of green frogs and the number of yellow frogs now?

$\textbf{(A) } 10\qquad\textbf{(B) } 12\qquad\textbf{(C) } 16\qquad\textbf{(D) } 20\qquad\textbf{(E) } 24$

Solution 1

Let the initial number of green frogs be $g$ and the initial number of yellow frogs be $y$ . Since the ratio of the number of green frogs to yellow frogs is initially $3 : 1$ , $g = 3y$ . Now, $3$ green frogs move to the sunny side and $5$ yellow frogs move to the shade side, thus the new number of green frogs is $g + 2$ and the new number of yellow frogs is $y - 2$ . We are given that $\frac{g + 2}{y - 2} = \frac{4}{1}$ , so $g + 2 = 4y - 8$ , since $g = 3y$ , we have $3y + 2 = 4y - 8$ , so $y = 10$ and $g = 30$ . Thus the answer is $(g + 2) - (y - 2) = 32 - 8 = \textbf{(E) } 24$ .

Solution 2

Since the original ratio is $3:1$ and the new ratio is $4:1$ , the number of frogs must be a multiple of $12$ , the only solutions left are $(B)$ and $(E)$ .

Let's start with $12$ frogs:

$x+12=4x$

$x=4$

If we reverse the move, 16:4 becomes 14:6, which isn't a 3:1 ratio.

Therefore the answer is: $\boxed{E}$

Sidenote: $x+24=4x => x=8$ this goes from 32:8 to 30:10, 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

https://youtu.be/Ah1WTdk8nuA

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

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

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