2017 AMC 8 Problems/Problem 9

Problem 9

All of Macy'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?

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

Video Solution

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

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.

Solution 2

The 6 green and yellow marbles make up $\(frac{12}{12} \(frac{1}{3} + \frac{1}{4}) = \frac{5}{12}$ (Error compiling LaTeX. Unknown error_msg) of the total marbles. Now we know that there are $\frac{5}{12} - 6$ yellow marbles. Now, because going for the number of yellow marbles right now doesn't work (there would be -1 yellow marbles), we multiply 2 on both sides, to find out there are $\frac{10}{24} - 6$ $\boxed{\textbf{(D) }4}$ yellow marbles.

-elbertpark

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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

Page

Toolbox

Search

2017 AMC 8 Problems/Problem 9

Contents

Problem 9

Video Solution

Solution 1

Solution 2

See Also