Difference between revisions of "2024 AMC 8 Problems/Problem 21"
(→Solution 2) |
Oliviax2011 (talk | contribs) |
||
(32 intermediate revisions by 18 users not shown) | |||
Line 6: | Line 6: | ||
<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 = | + | 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> frogs: | ||
+ | |||
+ | 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 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>. | |
− | <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 | + | 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:// | + | 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== | ||
Line 32: | Line 54: | ||
==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}} |
Revision as of 06:52, 15 November 2024
Contents
[hide]- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3 (Simple and easy to make sense of)
- 5 Video Solution 1 by Math-X (First fully understand the problem!!!)
- 6 Video Solution (A Clever Explanation You’ll Get Instantly)
- 7 Video Solution by Power Solve (crystal clear)
- 8 Video Solution 2 by OmegaLearn.org
- 9 Video Solution 3 by SpreadTheMathLove
- 10 Video Solution by NiuniuMaths (Easy to understand!)
- 11 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 12 Video Solution by Interstigation
- 13 Video Solution by Dr. David
- 14 Video Solution by WhyMath
- 15 See Also
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 . Then green frogs moved to the sunny side and yellow frogs moved to the shady side. Now the ratio is . What is the difference between the number of green frogs and the number of yellow frogs now?
Solution 1
Let the initial number of green frogs be and the initial number of yellow frogs be . Since the ratio of the number of green frogs to yellow frogs is initially , . Now, green frogs move to the sunny side and yellow frogs move to the shade side, thus the new number of green frogs is and the new number of yellow frogs is . We are given that , so , since , we have , so and . Thus the answer is
-anonchalantdreadhead
Solution 2
Since the original ratio is and the new ratio is , the number of frogs must be a multiple of , the only solutions left are and .
Let's start with frogs:
We must have frogs in the shade and frogs in the sun. After the change, there would be frogs in the shade and frog in the sun, which is not a ratio.
Therefore the answer is: .
-ILoveMath31415926535
Solution 3 (Simple and easy to make sense of)
The ratio of (green) to (yellow) frogs is . When green frogs move to the sunny side and yellow frogs move to the shady side, the ratio becomes which is .
So earlier, , or , of the total frogs were green. Now, , or , of the total frogs are green. When the frogs transferred from the yellow side to the green, the green side gained of the total amount of frogs. So, = , where is the total number of frogs. Solving for we get = .
If the ratio has a total of , then we can multiply each of them by , or , and find that there are green frogs and yellow frogs. Therefore, the difference between the green and yellow frogs is
~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
Video Solution by Power Solve (crystal clear)
https://www.youtube.com/watch?v=HodW9H55ZsE
Video Solution 2 by OmegaLearn.org
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
Video Solution by WhyMath
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.