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Difference between revisions of "2024 AMC 8 Problems/Problem 21"
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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 = \textbf{(E) } 24</math>.
 
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 = \textbf{(E) } 24</math>.
  
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==Solution 2==
  
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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. Therefore, we can try all multiples of 12 and we get <math>(E) \boxed{24}</math>.
  
 
==Video Solution by Math-X (First fully understand the problem!!!)==
 
==Video Solution by Math-X (First fully understand the problem!!!)==

Revision as of 16:16, 25 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. Therefore, we can try all multiples of 12 and we get $(E) \boxed{24}$.

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

https://www.youtube.com/watch?v=zBe5vrQbn2A

~Math-X

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