Difference between revisions of "2017 AMC 8 Problems/Problem 9"
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<math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5</math> | <math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad\textbf{(E) }5</math> | ||
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| + | ==Solution== | ||
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| + | The 6 green marbles are part of <math>1 - \frac13 - \frac14 = \frac5{12}</math> of the total marbles. If <math>6 = \frac13</math> 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 <math>6 = \frac14</math> of the total number of marbles, it works, because 24/4 = 6, and 24/3 = 8. So we have that 24 - 6 - 8 - 8 = 10 - 8 = 2 marbles, or B. | ||
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| + | ==See Also== | ||
| + | {{AMC8 box|year=2017|num-b=20|num-a=22}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Revision as of 14:13, 22 November 2017
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?
Solution
The 6 green marbles are part of 
of the total marbles. If 
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 
of the total number of marbles, it works, because 24/4 = 6, and 24/3 = 8. So we have that 24 - 6 - 8 - 8 = 10 - 8 = 2 marbles, or B.
See Also
| 2017 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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. 
