2024 AMC 8 Problems/Problem 21

1 Problem
2 Solution 1
3 Solution 2
4 Solution 3 (Simple and easy to make sense of)
5 Video Solution 1 by Math-X (First fully understand the problem!!!)
6 Video Solution (A Clever Explanation You’ll Get Instantly)
7 Video Solution by Power Solve (crystal clear)
8 Video Solution 2 by OmegaLearn.org
9 Video Solution 3 by SpreadTheMathLove
10 Video Solution by NiuniuMaths (Easy to understand!)
11 Video Solution by CosineMethod [🔥Fast and Easy🔥]
12 Video Solution by Interstigation
13 Video Solution by Dr. David
14 Video Solution by WhyMath
15 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 $3 : 1$ . 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 $(g + 2) - (y - 2) = 32 - 8 = 24$

-anonchalantdreadhead

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:

We must have $9$ frogs in the shade and $3$ frogs in the sun. After the change, there would be $11$ frogs in the shade and $1$ frog in the sun, which is not a $4:1$ ratio.

Therefore the answer is: $\boxed{(E) \hspace{1 mm} 24}$ .

-ILoveMath31415926535

Solution 3 (Simple and easy to make sense of)

The ratio of $g$ (green) to $y$ (yellow) frogs is $3:1$ . When $3$ green frogs move to the sunny side and $5$ yellow frogs move to the shady side, the ratio becomes $g + 2:y - 2$ which is $4:1$ .

So earlier, $\frac{3}{4}$ , or $\frac{15}{20}$ , of the total frogs were green. Now, $\frac{4}{5}$ , or $\frac{16}{20}$ , of the total frogs are green. When the $2$ frogs transferred from the yellow side to the green, the green side gained $\frac{1}{20}$ of the total amount of frogs. So, $2$ = $\frac{1}{20}$ $a$ , where $a$ is the total number of frogs. Solving for $a$ we get $a$ = $40$ .

If the ratio $4: 1$ has a total of $40$ , then we can multiply each of them by $\frac{40}{(4 + 1)}$ , or $8$ , and find that there are $32$ green frogs and $8$ yellow frogs. Therefore, the difference between the green and yellow frogs is $\boxed{\textbf{(E) }24}$

~mihikamishra

Video Solution 1 by Math-X (First fully understand the problem!!!)

https://youtu.be/BaE00H2SHQM?si=yTyYiS2S75HoSfmI&t=6313

~Math-X

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/5ZIFnqymdDQ?si=l3dF11eyLXoyI9Ow&t=3107 ~hsnacademy

https://youtu.be/H7d8c_YnvqE

Video Solution by Power Solve (crystal clear)

https://www.youtube.com/watch?v=HodW9H55ZsE

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 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

Video Solution by Dr. David

https://youtu.be/d6Xtre2bwro

Video Solution by WhyMath

https://youtu.be/GtyynnIF1ZM

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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