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

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==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>.
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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>(E) \boxed{28}</math>.

Revision as of 16:11, 25 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 $(E) \boxed{28}$.

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