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Difference between revisions of "2017 AMC 8 Problems/Problem 9"

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==Problem==
 
==Problem==
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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 Macy could have?
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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==
  
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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.
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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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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.
 
  
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-[[User:s|s]]
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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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==Solution 3==
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~[https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
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Letting <math>y</math> be the number of yellow marbles we get <math>\frac{1}{3} + \frac{1}{4} + \frac{1}{6} + \frac{1}{y} = 1</math>. Solving for <math>y</math> we get <math>y = 4</math>, so there are <math>\boxed{\textbf{(D) }4}</math> yellow marbles.
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==Video Solution (CREATIVE THINKING!!!)==
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https://youtu.be/ZsGJwbTf4ao
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~Education, the Study of Everything
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 09:04, 27 May 2024

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?

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

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

Since $\frac{1}{3}$ of the marbles are blue and $\frac{1}{4}$ are red, it is clear that the total number of marbles must be divisible by $12$. If there are $12$ marbles, then $4$ are blue, $3$ are red, and $6$ are green, meaning that there are $-1$ yellow marbles. This is impossible. Trying the next multiple of $12$, $24$, we find that $8$ are green, $6$ are red, and $6$ are green, meaning that the minimum number of yellow marbles is $\boxed{\textbf{(D) }4}$.

~cxsmi

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

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

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