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

Revision as of 14:51, 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

@@ Line 8: / Line 8: @@
 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>.
-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{(E) 28.}</math>
+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>
+==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
+<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>
+-Benedict T (countmath1)
 ==Video Solution 1 (easy to digest) by Power Solve==

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