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

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==Solution 2==
 
==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, the only solutions left are <math>(B)</math> and <math>(E)</math>.   
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Since the original ratio is <math>3:1</math> and the new ratio is <math>4:1</math>, the number of frogs must be a multiple of <math>12</math>, 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> frogs:  
  
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<math>4x+2-(x-2)=3x+4=12</math>
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We must have <math>9</math> frogs in the shade and <math>3</math> frogs in the sun. After the change, there would be <math>11</math> frogs in the shade and <math>1</math> frog in the sun, which is not a <math>4:1</math> ratio.
  
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If you solve, x isn't an integer.
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Therefore the answer is: <math>\boxed{(E) \hspace{1 mm} 24}</math>.
  
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Therefore the answer must be <math>\boxed{E}</math>.
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-ILoveMath31415926535
 +
== Video Solution ==
 +
https://youtu.be/H7d8c_YnvqE
  
−
==Video Solution by Math-X (First fully understand the problem!!!)==
+
Please like and subscribe!
−
https://www.youtube.com/watch?v=zBe5vrQbn2A
+
 
 +
==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
 
~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==
 +
{{AMC8 box|year=2024|num-b=20|num-a=22}}
 +
{{MAA Notice}}

Latest revision as of 13:29, 21 April 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 $12$ 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

https://youtu.be/H7d8c_YnvqE

Please like and subscribe!

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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