Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
m (→‎Solution 3)
Line 32: Line 32:
  
 
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>
 
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 (A Clever Explanation You’ll Get Instantly)==
 
==Video Solution (A Clever Explanation You’ll Get Instantly)==

Revision as of 12:03, 31 August 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

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, you can tell that the ratio $g + 2:y - 2$ 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 green side gained $2$ frogs and the yellow side lost $2$ frogs, 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 (A Clever Explanation You’ll Get Instantly)

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

~hsnacademy

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

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

~Math-X

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 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_8_Problems/Problem_21&oldid=226095"