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Difference between revisions of "2024 AMC 8 Problems/Problem 21"

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

Revision as of 00:08, 7 June 2024

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

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

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

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

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