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

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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 <math>(B)</math> and <math>(E)</math>.   
 
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 <math>(B)</math> and <math>(E)</math>.   
  
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Let's start with <math>12</math>. If the starting difference is <math>3x:x</math>, then the difference after the move is <math>4x+2:x-2</math>.
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Let's start with <math>12</math>. If the starting difference is <math>3x:x</math>.
  
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<math>4x+2-(x-2)=3x+4=12</math>
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Using the starting ratio, there are 9 green frogs and three yellow. Afterwards there are 11 green frogs and 1 yellow, this doesn't work
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If you solve, <math>x</math> isn't an integer.
 
  
 
Therefore the answer must be <math>\boxed{E}</math>.
 
Therefore the answer must be <math>\boxed{E}</math>.

Revision as of 17:41, 26 January 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 $(g + 2) - (y - 2) = 32 - 8 = \textbf{(E) } 24$.

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 $12$. If the starting difference is $3x:x$.

Using the starting ratio, there are 9 green frogs and three yellow. Afterwards there are 11 green frogs and 1 yellow, this doesn't work

Therefore the answer must be $\boxed{E}$.

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

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