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>1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}</math> of the total marbles, just like as in the previous solution. Now we know that there are <math>\frac{5}{12} - 6</math> yellow marbles. Now, because | + | The 6 green and yellow marbles make up <math>1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}</math> of the total marbles, just like as in the previous solution. Now we know that there are <math>\frac{5}{12}(\text{total}) - 6</math> yellow marbles. Now, because 12 marbles for the total doesn't work (there would be -1 yellow marbles), we multiply the 12 by 2, to find out there are <math>\frac{5}{12}(24) - 6=\boxed{\textbf{(D) }4}</math> yellow marbles. |
-[[User:elbertpark|elbertpark]] | -[[User:elbertpark|elbertpark]] |
Revision as of 13:23, 19 January 2021
Problem
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?
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 of the total marbles, just like as in the previous solution. Now we know that there are
yellow marbles. Now, because 12 marbles for the total doesn't work (there would be -1 yellow marbles), we multiply the 12 by 2, to find out there are
yellow marbles.
Video Solution
https://youtu.be/rQUwNC0gqdg?t=770
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.