Difference between revisions of "2024 AMC 8 Problems/Problem 9"
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==Problem== | ==Problem== | ||
− | All the marbles in Maria's collection are red, green, or blue. Maria has half as many red marbles as green marbles and twice as many blue marbles as green marbles. Which of the following could be the total number of marbles in Maria's collection? | + | All the marbles in Maria's collection are red, green, or blue. Maria has half as many red marbles as green marbles, and twice as many blue marbles as green marbles. Which of the following could be the total number of marbles in Maria's collection? |
<math>\textbf{(A) } 24\qquad\textbf{(B) } 25\qquad\textbf{(C) } 26\qquad\textbf{(D) } 27\qquad\textbf{(E) } 28</math> | <math>\textbf{(A) } 24\qquad\textbf{(B) } 25\qquad\textbf{(C) } 26\qquad\textbf{(D) } 27\qquad\textbf{(E) } 28</math> | ||
+ | |||
+ | ~Alice (minor edit) | ||
==Solution 1== | ==Solution 1== | ||
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Since she has half as many red marbles as green, we can call the number of red marbles <math>x</math>, and the number of green marbles <math>2x</math>. | Since she has half as many red marbles as green, we can call the number of red marbles <math>x</math>, and the number of green marbles <math>2x</math>. | ||
Since she has half as many green marbles as blue, we can call the number of blue marbles <math>4x</math>. | Since she has half as many green marbles as blue, we can call the number of blue marbles <math>4x</math>. | ||
− | Adding them up, we have <math>7x</math> marbles. The number of marbles therefore must be a multiple of <math>7</math> | + | Adding them up, we have: <math>7x</math> marbles. The number of marbles therefore must be a multiple of <math>7</math>, as <math>x</math> represents an integer, so the only possible answer is <math>\boxed{\textbf{(E) 28}}.</math> |
+ | ~edited by anabel.disher | ||
==Solution 2== | ==Solution 2== | ||
− | Suppose Maria has <math>g</math> green marbles and | + | Suppose Maria has <math>g</math> green marbles and <math>t</math> total marbles. She then has <math>\frac{g}{2}</math> red marbles and <math>2g</math> blue marbles. Altogether, Maria has |
<cmath>g + \frac{g}{2} + 2g = \frac{7g}{2} = t</cmath> | <cmath>g + \frac{g}{2} + 2g = \frac{7g}{2} = t</cmath> | ||
− | marbles, | + | marbles, so <math>g = \dfrac{2t}{7},</math> so <math>t</math> must be a multiple of <math>7</math>. The only multiple of <math>7</math> in the answer choices is <math>\boxed{\textbf{(E) 28}}.</math> |
+ | |||
+ | -Benedict T (countmath1) and anabel.disher | ||
+ | |||
+ | ==Video Solution by Central Valley Math Circle(Goes through the full thought process)== | ||
+ | https://youtu.be/QZTmQHWAYrI | ||
+ | |||
+ | ~mr_mathman | ||
+ | |||
+ | ==Video Solution by Math-X (First fully understand the problem!!!)== | ||
+ | https://youtu.be/BaE00H2SHQM?si=ORnDevWxiB6tv1jO&t=1912 | ||
− | - | + | ~Math-X |
==Video Solution (A Clever Explanation You’ll Get Instantly)== | ==Video Solution (A Clever Explanation You’ll Get Instantly)== | ||
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==Video Solution 1 (easy to digest) by Power Solve== | ==Video Solution 1 (easy to digest) by Power Solve== | ||
https://youtu.be/16YYti_pDUg?si=FAyIwmaHxE6UdSwP&t=238 | https://youtu.be/16YYti_pDUg?si=FAyIwmaHxE6UdSwP&t=238 | ||
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==Video Solution by NiuniuMaths (Easy to understand!)== | ==Video Solution by NiuniuMaths (Easy to understand!)== | ||
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https://youtu.be/4WSB0osAR2I | https://youtu.be/4WSB0osAR2I | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/_RN-ILHelp4 | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2024|num-b=8|num-a=10}} | {{AMC8 box|year=2024|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:52, 22 December 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Video Solution by Central Valley Math Circle(Goes through the full thought process)
- 5 Video Solution by Math-X (First fully understand the problem!!!)
- 6 Video Solution (A Clever Explanation You’ll Get Instantly)
- 7 Video Solution 1 (easy to digest) by Power Solve
- 8 Video Solution by NiuniuMaths (Easy to understand!)
- 9 Video Solution 2 by SpreadTheMathLove
- 10 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 11 Video Solution by Interstigation
- 12 Video Solution by Daily Dose of Math (Certified, Simple, and Logical)
- 13 Video Solution by Dr. David
- 14 Video Solution by WhyMath
- 15 See Also
Problem
All the marbles in Maria's collection are red, green, or blue. Maria has half as many red marbles as green marbles, and twice as many blue marbles as green marbles. Which of the following could be the total number of marbles in Maria's collection?
~Alice (minor edit)
Solution 1
Since she has half as many red marbles as green, we can call the number of red marbles , and the number of green marbles . Since she has half as many green marbles as blue, we can call the number of blue marbles . Adding them up, we have: marbles. The number of marbles therefore must be a multiple of , as represents an integer, so the only possible answer is
~edited by anabel.disher
Solution 2
Suppose Maria has green marbles and total marbles. She then has red marbles and blue marbles. Altogether, Maria has marbles, so so must be a multiple of . The only multiple of in the answer choices is
-Benedict T (countmath1) and anabel.disher
Video Solution by Central Valley Math Circle(Goes through the full thought process)
~mr_mathman
Video Solution by Math-X (First fully understand the problem!!!)
https://youtu.be/BaE00H2SHQM?si=ORnDevWxiB6tv1jO&t=1912
~Math-X
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/5ZIFnqymdDQ?si=j8wYLXY9iRPR1wis&t=1006
~hsnacademy
Video Solution 1 (easy to digest) by Power Solve
https://youtu.be/16YYti_pDUg?si=FAyIwmaHxE6UdSwP&t=238
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=V-xN8Njd_Lc
~NiuniuMaths
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=L83DxusGkSY
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=zqQTfBWr9T0
Video Solution by Interstigation
https://youtu.be/ktzijuZtDas&t=890
Video Solution by Daily Dose of Math (Certified, Simple, and Logical)
~Thesmartgreekmathdude
Video Solution by Dr. David
Video Solution by WhyMath
See Also
2024 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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.