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

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<math>\textbf{(A) } 10\qquad\textbf{(B) } 12\qquad\textbf{(C) } 16\qquad\textbf{(D) } 20\qquad\textbf{(E) } 24</math>
 
<math>\textbf{(A) } 10\qquad\textbf{(B) } 12\qquad\textbf{(C) } 16\qquad\textbf{(D) } 20\qquad\textbf{(E) } 24</math>
 +
  
 
==Solution 1==
 
==Solution 1==
−
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 = 24</math>
 +
 
 +
-anonchalantdreadhead
  
 
==Solution 2==
 
==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 <math>(B)</math> and <math>(E)</math>.   
+
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>.   
  
−
Let's start with <math>12</math>. If the starting difference is <math>3x:x</math>.
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Let's start with <math>12</math> frogs:  
  
−
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
+
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.
  
−
Therefore the answer must be <math>\boxed{E}</math>.
+
Therefore the answer is: <math>\boxed{(E) \hspace{1 mm} 24}</math>.
 +
 
 +
-ILoveMath31415926535
 +
 
 +
==Solution 3 (Simple and easy to make sense of)==
 +
 
 +
The ratio of <math>g</math> (green) to <math>y</math> (yellow) frogs is <math>3:1</math>. When <math>3</math> green frogs move to the sunny side and <math>5</math> yellow frogs move to the shady side, the ratio becomes <math>g + 2:y - 2</math> which is <math>4:1</math>.
 +
 
 +
So earlier, <math>\frac{3}{4}</math>, or <math>\frac{15}{20}</math>, of the total frogs were green. Now, <math>\frac{4}{5}</math>, or <math>\frac{16}{20}</math>, of the total frogs are green. When the <math>2</math> frogs transferred from the yellow side to the green, the green side gained <math>\frac{1}{20}</math> of the total amount of frogs. So, <math>2</math> = <math>\frac{1}{20}</math><math>a</math>, where <math>a</math> is the total number of frogs. Solving for <math>a</math> we get <math>a</math> = <math>40</math>.
 +
 
 +
If the ratio <math>4: 1</math> has a total of <math>40</math>, then we can multiply each of them by <math>\frac{40}{(4 + 1)}</math>, or <math>8</math>, and find that there are <math>32</math> green frogs and <math>8</math> yellow frogs. Therefore, the difference between the green and yellow frogs is <math>\boxed{\textbf{(E) }24}</math>
 +
 
 +
~mihikamishra
  
 
==Video Solution 1 by Math-X (First fully understand the problem!!!)==
 
==Video Solution 1 by Math-X (First fully understand the problem!!!)==
−
https://www.youtube.com/watch?v=zBe5vrQbn2A
+
https://youtu.be/BaE00H2SHQM?si=yTyYiS2S75HoSfmI&t=6313
  
 
~Math-X
 
~Math-X
 +
 +
==Video Solution (A Clever Explanation You’ll Get Instantly)==
 +
https://youtu.be/5ZIFnqymdDQ?si=l3dF11eyLXoyI9Ow&t=3107
 +
~hsnacademy
 +
 +
https://youtu.be/H7d8c_YnvqE
 +
 +
==Video Solution by Power Solve (crystal clear)==
 +
https://www.youtube.com/watch?v=HodW9H55ZsE
  
 
==Video Solution 2 by OmegaLearn.org==
 
==Video Solution 2 by OmegaLearn.org==
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==Video Solution 3 by SpreadTheMathLove==
 
==Video Solution 3 by SpreadTheMathLove==
 
https://www.youtube.com/watch?v=3ItvjukLqK0
 
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🔥]==
 
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
  
 
https://www.youtube.com/watch?v=3SUTUr1My7c&t=1s
 
https://www.youtube.com/watch?v=3SUTUr1My7c&t=1s
 +
 +
==Video Solution by Interstigation==
 +
https://youtu.be/ktzijuZtDas&t=2562
 +
 +
==Video Solution by Dr. David==
 +
https://youtu.be/d6Xtre2bwro
 +
 +
==Video Solution by WhyMath==
 +
https://youtu.be/GtyynnIF1ZM
 +
 +
==See Also==
 +
{{AMC8 box|year=2024|num-b=20|num-a=22}}
 +
{{MAA Notice}}

Latest revision as of 06:52, 15 November 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

Solution 3 (Simple and easy to make sense of)

The ratio of $g$ (green) to $y$ (yellow) frogs is $3:1$. When $3$ green frogs move to the sunny side and $5$ yellow frogs move to the shady side, the ratio becomes $g + 2:y - 2$ which is $4:1$.

So earlier, $\frac{3}{4}$, or $\frac{15}{20}$, of the total frogs were green. Now, $\frac{4}{5}$, or $\frac{16}{20}$, of the total frogs are green. When the $2$ frogs transferred from the yellow side to the green, the green side gained $\frac{1}{20}$ of the total amount of frogs. So, $2$ = $\frac{1}{20}$$a$, where $a$ is the total number of frogs. Solving for $a$ we get $a$ = $40$.

If the ratio $4: 1$ has a total of $40$, then we can multiply each of them by $\frac{40}{(4 + 1)}$, or $8$, and find that there are $32$ green frogs and $8$ yellow frogs. Therefore, the difference between the green and yellow frogs is $\boxed{\textbf{(E) }24}$

~mihikamishra

Video Solution 1 by Math-X (First fully understand the problem!!!)

https://youtu.be/BaE00H2SHQM?si=yTyYiS2S75HoSfmI&t=6313

~Math-X

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/5ZIFnqymdDQ?si=l3dF11eyLXoyI9Ow&t=3107 ~hsnacademy

https://youtu.be/H7d8c_YnvqE

Video Solution by Power Solve (crystal clear)

https://www.youtube.com/watch?v=HodW9H55ZsE

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

Video Solution by Dr. David

https://youtu.be/d6Xtre2bwro

Video Solution by WhyMath

https://youtu.be/GtyynnIF1ZM

See Also

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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