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Difference between revisions of "2024 AMC 8 Problems/Problem 9"
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==Problem==
 
==Problem==
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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?
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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>
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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>.  
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Adding them up, we have <math>7x</math> marbles. The number of marbles therefore must be a multiple of <math>7</math>. The only possible answer is <math>\boxed{\textbf{(E) 28}}.</math>
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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>
  
 
==Solution 2==
 
==Solution 2==
  
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Suppose Maria has <math>g</math> green marbles and let <math>t</math> be the total number of marbles. She then has <math>\frac{g}{2}</math> red marbles and <math>2g</math> blue marbles. Altogether, Maria has
+
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>
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marbles, implying that <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> is <math>\boxed{\textbf{(E) 28}}.</math>
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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>
  
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-Benedict T (countmath1)
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-Benedict T (countmath1) and anabel.disher
 +
 
 +
 
 +
==Solution 3==
 +
This solution is similar to Solution 1. The most simplest and effective way to solve this problem is equations. We have the following equations:
 +
 
 +
2r=g
 +
and 2g=b
 +
We essentialy want to find out the total number of mables in Maria's collection or in equation terms (r+g+b).
 +
Converting that equation and substituting the values we get r+2r+4r=7r
 +
Now, we know the total is a multiple of 7 leaving our answer as <math>\boxed{\textbf{(E) 28}}.</math> -AADHYA2012
 +
 
 +
==Video by MathTalks 😉==
 +
 
 +
https://youtu.be/9GVWXv9Pg1E?si=lhCKMjJ0wvfc_MfY
 +
 
 +
~rc1219
 +
 
 +
 
 +
 
 +
==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
−
 
−
==Video Solution by Math-X (First fully understand the problem!!!)==
 
−
https://youtu.be/BaE00H2SHQM?si=ORnDevWxiB6tv1jO&t=1912
 
−
 
−
~Math-X
 
  
 
==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
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 +
==Video Solution by WhyMath==
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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}}
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 +
[[Category:Introductory Algebra Problems]]

Latest revision as of 20:25, 7 June 2025

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?

$\textbf{(A) } 24\qquad\textbf{(B) } 25\qquad\textbf{(C) } 26\qquad\textbf{(D) } 27\qquad\textbf{(E) } 28$

Solution 1

Since she has half as many red marbles as green, we can call the number of red marbles $x$, and the number of green marbles $2x$. Since she has half as many green marbles as blue, we can call the number of blue marbles $4x$. Adding them up, we have: $7x$ marbles. The number of marbles therefore must be a multiple of $7$, as $x$ represents an integer, so the only possible answer is $\boxed{\textbf{(E) 28}}.$

Solution 2

Suppose Maria has $g$ green marbles and $t$ total marbles. She then has $\frac{g}{2}$ red marbles and $2g$ blue marbles. Altogether, Maria has \[g + \frac{g}{2} + 2g = \frac{7g}{2} = t\] marbles, so $g = \dfrac{2t}{7},$ so $t$ must be a multiple of $7$. The only multiple of $7$ in the answer choices is $\boxed{\textbf{(E) 28}}.$

-Benedict T (countmath1) and anabel.disher


Solution 3

This solution is similar to Solution 1. The most simplest and effective way to solve this problem is equations. We have the following equations:

2r=g and 2g=b We essentialy want to find out the total number of mables in Maria's collection or in equation terms (r+g+b). Converting that equation and substituting the values we get r+2r+4r=7r Now, we know the total is a multiple of 7 leaving our answer as $\boxed{\textbf{(E) 28}}.$ -AADHYA2012

Video by MathTalks 😉

https://youtu.be/9GVWXv9Pg1E?si=lhCKMjJ0wvfc_MfY 

~rc1219


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)

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)

https://youtu.be/8GHuS5HEoWc

~Thesmartgreekmathdude

Video Solution by Dr. David

https://youtu.be/4WSB0osAR2I

Video Solution by WhyMath

https://youtu.be/_RN-ILHelp4

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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