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

3

Solution 2

Since $\frac{1}{3}$ of the marbles are blue and $\frac{1}{4}$ are red, it is clear that the total number of marbles must be divisible by $12$ . If there are $12$ marbles, then $4$ are blue, $3$ are red, and $6$ are green, meaning that there are $-1$ yellow marbles. This is impossible. Trying the next multiple of $12$ , $24$ , we find that $8$ are green, $6$ are red, and $6$ are green, meaning that the minimum number of yellow marbles is $\boxed{\textbf{(D) }4}$ .

~cxsmi

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/ZsGJwbTf4ao

~Education, the Study of Everything

Video Solution

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

https://youtu.be/EvmWC1zMHfY

~savannahsolver

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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