Difference between revisions of "2017 AMC 8 Problems/Problem 9"
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==Solution 2== | ==Solution 2== | ||
− | The 6 green and yellow marbles make up <math>\frac{ | + | The 6 green and yellow marbles make up <math>\(frac{12}{12} \(frac{1}{3} + \frac{1}{4}) = \frac{5}{12}</math> of the total marbles. Now we know that there are <math>\frac{5}{12} - 6</math> 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 <math>\frac{10}{24} - 6</math> <math>\boxed{\textbf{(D) }4}</math> yellow marbles. |
-[[User:elbertpark|elbertpark]] | -[[User:elbertpark|elbertpark]] |
Revision as of 01:58, 6 November 2020
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?
Video Solution
https://youtu.be/rQUwNC0gqdg?t=770
Solution 1
The green marbles and yellow marbles form of the total marbles. Now suppose the total number of marbles is . We know the number of yellow marbles is and a positive integer. Therefore, must divide . Trying the smallest multiples of for , we see that when , we get there are yellow marbles, which is impossible. However when , there are 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. ! LaTeX Error: Bad math environment delimiter.) of the total marbles. Now we know that there are 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 yellow marbles.
See Also
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.