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

Revision as of 13:52, 26 January 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?

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

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 $7$ . The only possible answer is $\boxed{\textbf{(E) 28}}.$

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 $7$ is $\boxed{\textbf{(E) 28}}.$

-Benedict T (countmath1)

Video Solution 1 (easy to digest) by Power Solve

https://youtu.be/16YYti_pDUg?si=FAyIwmaHxE6UdSwP&t=238

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=L83DxusGkSY

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.

@@ Line 23: / Line 23: @@
 ==Video Solution 2 by SpreadTheMathLove==
 https://www.youtube.com/watch?v=L83DxusGkSY
+==See Also==
+{{AMC8 box|year=2024|num-b=8|num-a=10}}
+{{MAA Notice}}

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