2024 AMC 8 Problems/Problem 9

1 Problem
2 Solution 1
3 Solution 2
4 Video Solution by Math-X (First fully understand the problem!!!)
5 Video Solution (A Clever Explanation You’ll Get Instantly)
6 Video Solution 1 (easy to digest) by Power Solve
7 Video Solution by NiuniuMaths (Easy to understand!)
8 Video Solution 2 by SpreadTheMathLove
9 Video Solution by CosineMethod [🔥Fast and Easy🔥]
10 Video Solution by Interstigation
11 Video Solution by Daily Dose of Math (Certified, Simple, and Logical)
12 Video Solution by Dr. David
13 Video Solution by WhyMath
14 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?

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

~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 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

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

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