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AMC8 2017 Test Paper
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Sample Question 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?

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




Answer Keys

Question 9: D


Solutions

Question 9
Step 1: Recognize that the green and yellow marbles together form \(1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}\) of the total number of marbles, \(x\).

Step 2: Construct the number of yellow marbles to be \(\frac{5}{12}x - 6\).

Step 3: Realize that 12 must divide \(x\), as \(x\) has thirds and quarters in its division.

Step 4: Test the smallest multiples of \(12\) for \(x\). Try \(x=12\), which gives \(-1\) yellow marbles, unreasonable.

Step 5: On the other hand, when \(x = 24\), there are \(\frac{5}{12} \cdot 24 - 6 = 4\) yellow marbles, which is an acceptable result.

Step 6: From the test, we find that 4 yellow marbles is the smallest possible number, so the answer is \(\textbf{(D) }4\).