2017 AMC 8 Problems/Problem 9

Problem 9

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 Marcy could have?

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

Solution

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 $\frac{5}{12} \cdot 24 - 6 = \textboxed{(D) } 4$ (Error compiling LaTeX. Unknown error_msg) yellow marbles, which must be the smallest possible.

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2017 AMC 8 Problems/Problem 9

Problem 9

Solution

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