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Difference between revisions of "2024 AMC 8 Problems/Problem 9"

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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>
  
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~edited by anabel.disher
 
==Solution 2==
 
==Solution 2==
  
 
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 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
 
<cmath>g + \frac{g}{2} + 2g = \frac{7g}{2} = t</cmath>
 
<cmath>g + \frac{g}{2} + 2g = \frac{7g}{2} = t</cmath>
−
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>
+
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> in the answer choices is <math>\boxed{\textbf{(E) 28}}.</math>
  
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-Benedict T (countmath1)
+
-Benedict T (countmath1) and anabel.disher
  
 
==Video Solution (A Clever Explanation You’ll Get Instantly)==
 
==Video Solution (A Clever Explanation You’ll Get Instantly)==

Latest revision as of 12:35, 28 October 2024

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?

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

~Alice (minor edit)

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 $\boxed{\textbf{(E) 28}}.$

~edited by anabel.disher

Solution 2

Suppose Maria has $g$ green marbles and let $t$ be the total number of 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, implying that $g = \dfrac{2t}{7},$ 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

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 Math-X (First fully understand the problem!!!)

https://youtu.be/BaE00H2SHQM?si=ORnDevWxiB6tv1jO&t=1912

~Math-X

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

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. AMC logo.png

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