2017 AMC 8 Problems/Problem 9

Problem

All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Macy could have?

$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5$

Solution 1

The $6$ green marbles and yellow marbles form $1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}$ of the total marbles. Now, suppose the total number of marbles is $x$ . We know the number of yellow marbles is $\frac{5}{12}x - 6$ and a positive integer. Therefore, $12$ must divide $x$ . Trying the smallest multiples of $12$ for $x$ , we see that when $x = 12$ , we get there are $-1$ yellow marbles, which is impossible. However when $x = 24$ , there are $\frac{5}{12} \cdot 24 - 6 = \boxed{\textbf{(D) }4}$ yellow marbles, which must be the smallest possible.

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/ZsGJwbTf4ao

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Video Solution

https://youtu.be/rQUwNC0gqdg?t=770

https://youtu.be/EvmWC1zMHfY

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2017 AMC 8 (Problems • Answer Key • Resources)
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2017 AMC 8 Problems/Problem 9

Contents

Problem

Solution 1

Video Solution (CREATIVE THINKING!!!)

Video Solution

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