Difference between revisions of "2017 AMC 8 Problems/Problem 9"
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− | ==Problem | + | ==Problem== |
− | All of | + | 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? |
<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 1== | ==Solution 1== | ||
− | The <math>6</math> green marbles and yellow marbles form <math>1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}</math> of the total marbles. Now suppose the total number of marbles is <math>x</math>. We know the number of yellow marbles is <math>\frac{5}{12}x - 6</math> and a positive integer. Therefore, <math>12</math> must divide <math>x</math>. Trying the smallest multiples of <math>12</math> for <math>x</math>, we see that when <math>x = 12</math>, we get there are <math>-1</math> yellow marbles, which is impossible. However when <math>x = 24</math>, there are <math>\frac{5}{12} \cdot 24 - 6 = \boxed{\textbf{(D) }4}</math> yellow marbles, which must be the smallest possible. | + | The <math>6</math> green marbles and yellow marbles form <math>1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12}</math> of the total marbles. Now, suppose the total number of marbles is <math>x</math>. We know the number of yellow marbles is <math>\frac{5}{12}x - 6</math> and a positive integer. Therefore, <math>12</math> must divide <math>x</math>. Trying the smallest multiples of <math>12</math> for <math>x</math>, we see that when <math>x = 12</math>, we get there are <math>-1</math> yellow marbles, which is impossible. However when <math>x = 24</math>, there are <math>\frac{5}{12} \cdot 24 - 6 = \boxed{\textbf{(D) }4}</math> yellow marbles, which must be the smallest possible. |
==Solution 2== | ==Solution 2== | ||
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− | -[ | + | Since <math>\frac{1}{3}</math> of the marbles are blue and <math>\frac{1}{4}</math> are red, it is clear that the total number of marbles must be divisible by <math>12</math>. If there are <math>12</math> marbles, then <math>4</math> are blue, <math>3</math> are red, and <math>6</math> are green, meaning that there are <math>-1</math> yellow marbles. This is impossible. Trying the next multiple of <math>12</math>, <math>24</math>, we find that <math>8</math> are green, <math>6</math> are red, and <math>6</math> are green, meaning that the minimum number of yellow marbles is <math>\boxed{\textbf{(D) }4}</math>. |
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+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] | ||
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+ | ==Video Solution (CREATIVE THINKING!!!)== | ||
+ | https://youtu.be/ZsGJwbTf4ao | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/rQUwNC0gqdg?t=770 | ||
+ | |||
+ | https://youtu.be/EvmWC1zMHfY | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 10:04, 27 May 2024
Contents
Problem
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 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
Since of the marbles are blue and
are red, it is clear that the total number of marbles must be divisible by
. If there are
marbles, then
are blue,
are red, and
are green, meaning that there are
yellow marbles. This is impossible. Trying the next multiple of
,
, we find that
are green,
are red, and
are green, meaning that the minimum number of yellow marbles is
.
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution
https://youtu.be/rQUwNC0gqdg?t=770
~savannahsolver
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.