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>. | ||
− | 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> |
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+ | ==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== | ==Video Solution 1 (easy to digest) by Power Solve== |
Revision as of 13:51, 26 January 2024
Contents
[hide]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?
Solution 1
Since she has half as many red marbles as green, we can call the number of red marbles , and the number of green marbles . Since she has half as many green marbles as blue, we can call the number of blue marbles . Adding them up, we have marbles. The number of marbles therefore must be a multiple of . The only possible answer is
Solution 2
Suppose Maria has green marbles and let be the total number of marbles. She then has red marbles and blue marbles. Altogether, Maria has marbles, implying that so must be a multiple of . The only multiple of is
-Benedict T (countmath1)
Video Solution 1 (easy to digest) by Power Solve
https://youtu.be/16YYti_pDUg?si=FAyIwmaHxE6UdSwP&t=238