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. 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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2024 AMC 8 Problems/Problem 21

Contents

Problem

Solution 1

Solution 2

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