2017 AMC 8 Problems/Problem 9

Revision as of 14:27, 22 November 2017 by StellarG (talk | contribs) (Solution)

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 are part of $1 - \frac13 - \frac14 = \frac5{12}$ of the total marbles. If $6 \implies \frac13$ of the total number of marbles, then there would be 18 marbles. Since a fourth of 18 is not a whole number, we cannot have 18 marbles. Then in $6 \implies \frac14$ of the total number of marbles, it works, because 24/4 = 6, and 24/3 = 8. So we have that 24 - 6 - 6 - 8 = 10 - 8 = 4 marbles, or D.

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AJHSME/AMC 8 Problems and Solutions

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