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Difference between revisions of "2012 AMC 8 Problems/Problem 19"

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<math> \textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}8\qquad\textbf{(C)}\hspace{.05in}9\qquad\textbf{(D)}\hspace{.05in}10\qquad\textbf{(E)}\hspace{.05in}12 </math>
 
<math> \textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}8\qquad\textbf{(C)}\hspace{.05in}9\qquad\textbf{(D)}\hspace{.05in}10\qquad\textbf{(E)}\hspace{.05in}12 </math>
  
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==Solution 1==
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==Solution 1 (Trial and Error)==
  
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6 are blue and green- b+g=6
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<math>6</math> are blue and green - <math>b+g=6</math>
  
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8 are red and blue- r+b=8
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<math>8</math> are red and blue - <math>r+b=8</math>
  
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4 are red and green- r+g=4
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<math>4</math> are red and green- <math>r+g=4</math>
  
  
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We can do trial and error. Let's make blue 5. That makes green 1 and red 3 because 6-5=1 and 8-5=3. To check this let's plug 1 and 3 into r+g=4 and it does work. Now count the number of marbles- 5+3+1=9. So 9 (C) is the answer.
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We can do trial and error. Let's make blue <math>5</math>. That makes green <math>1</math> and red <math>3</math> because <math>6-5=1</math> and <math>8-5=3</math>. To check this, let's plug <math>1</math> and <math>3</math> into <math>r+g=4</math>, which works. Now count the number of marbles - <math>5+3+1=9</math>. So the answer is <math>\boxed{\textbf{(C)}\ 9}.</math>
  
 
==Solution 2==
 
==Solution 2==
  
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We already knew the facts: <math>6</math> are blue and green, meaning <math>b+g=6</math>; <math>8</math> are red and blue, meaning <math>r+b=8</math>; <math>4</math> are red and green, meaning <math>r+g=4</math>. Then we need to add these three equations: <math>b+g+r+b+r+g=2(r+g+b)=6+8+4=19</math>. It gives us all of the marbles are <math>r+g+b = 19/2 = 9</math>. So the answer is <math>\boxed{\textbf{(C)}\ 9}</math>.  ---LarryFlora
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We already knew the facts: <math>6</math> are blue and green, meaning <math>b+g=6</math>; <math>8</math> are red and blue, meaning <math>r+b=8</math>; <math>4</math> are red and green, meaning <math>r+g=4</math>. Then we need to add these three equations: <math>b+g+r+b+r+g=2(r+g+b)=6+8+4=18</math>. It gives us all of the marbles are <math>r+g+b = 18/2 = 9</math>. So the answer is <math>\boxed{\textbf{(C)}\ 9}</math>.   
 +
~LarryFlora
  
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==Solution 3 Venn Diagrams==
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==Solution 3 (Venn Diagrams)==
 
We may draw three Venn diagrams to represent these three cases, respectively.  
 
We may draw three Venn diagrams to represent these three cases, respectively.  
  
 
[[File: Screen_Shot_2021-08-29_at_9.14.51_AM.png]]  
 
[[File: Screen_Shot_2021-08-29_at_9.14.51_AM.png]]  
  
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Let the amount of all the marbles is <math>x</math>, meaning <math>r+b+g = x</math>. The Venn diagrams give us the equation: <math>(x-6)+(x-8)+(x-4) = x</math>. Then <math>3x-18= x</math>. Resolving the equation,<math>x = 18/2 =9</math>, thus, the answer is <math>\boxed{\textbf{(C)}\ 9}</math>.   ---LarryFlora
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Let the amount of all the marbles be <math>x</math>, meaning <math>R+G+B = x</math>.  
 +
 
 +
The Venn diagrams give us the equation: <math>x = (x-6)+(x-8)+(x-4)</math>.
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So <math>x = 3x-18</math>, <math>x = 18/2 =9</math>.
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Thus, the answer is <math>\boxed{\textbf{(C)}\ 9}</math>.   
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~LarryFlora
 +
 
 +
==Solution 4 (Answer Choices)==
 +
Since we know all but <math>8</math> marbles in the jar are green, the jar must have at least <math>9</math> marbles. Then we can just start from <math>C</math> and keep going. If there are <math>9</math> marbles total, there are <math>3</math> red marbles <math>(9-6)</math>, <math>1</math> green marble <math>(9-8)</math>, and <math>5</math> blue marbles <math>(9-4)</math>. Since we assumed there were <math>9</math> marbles and <math>3+1+5=9</math>, the answer is <math>\boxed{\textbf{(C)}\ 9}</math>.
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==Solution 5 (Algebra) ==
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Let <math>x</math> be the number of total marbles. There are <math>x – 6</math> red marbles, <math>x – 8</math> green marbles, and <math>x – 4</math> blue marbles.  
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We can create an equation: <math>(x – 6)+(x – 8)+(x – 4)=x</math>
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Solving, we get <math>x=9</math>, which means the total number of marbles is <math>\boxed{\textbf{(C)}\ 9}</math>.
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-J.L.L
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(Feel free to edit)
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==Solution 6==
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Let <math>x</math> be the number of total marbles, <math>r</math> be the number of red marbles, <math>g</math> be the number of green marbles, and <math>b</math> be the number of blue marbles. Then we have <math>x - r = 6</math>, <math>x - g = 8</math>, <math>x - b = 4</math>, and <math>r + g + b = x</math>. Adding the first three equations together, we get <math>3x - r - g - b = 18</math> or <math>3x - (r + g + b) = 18</math>. Substituting in the fourth equation, we have <math>3x - x = 18</math> <math>\implies</math> <math>\boxed{\textbf{(C)}\ 9}</math>.
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 +
~[https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
 +
 
 +
==Video Solution==
 +
https://youtu.be/mMph7QH1kX0 Soo, DRMS, NM
 +
 
 +
https://youtu.be/-p5qv7DftrU ~savannahsolver
 +
 
 +
==Video Solution by OmegaLearn ==
 +
https://youtu.be/TkZvMa30Juo?t=1316
 +
 
 +
~pi_is_3.14
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2012|num-b=18|num-a=20}}
 
{{AMC8 box|year=2012|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:30, 17 February 2024

Problem

In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?

$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}8\qquad\textbf{(C)}\hspace{.05in}9\qquad\textbf{(D)}\hspace{.05in}10\qquad\textbf{(E)}\hspace{.05in}12$

Solution 1 (Trial and Error)

$6$ are blue and green - $b+g=6$

$8$ are red and blue - $r+b=8$

$4$ are red and green- $r+g=4$


We can do trial and error. Let's make blue $5$. That makes green $1$ and red $3$ because $6-5=1$ and $8-5=3$. To check this, let's plug $1$ and $3$ into $r+g=4$, which works. Now count the number of marbles - $5+3+1=9$. So the answer is $\boxed{\textbf{(C)}\ 9}.$

Solution 2

We already knew the facts: $6$ are blue and green, meaning $b+g=6$; $8$ are red and blue, meaning $r+b=8$; $4$ are red and green, meaning $r+g=4$. Then we need to add these three equations: $b+g+r+b+r+g=2(r+g+b)=6+8+4=18$. It gives us all of the marbles are $r+g+b = 18/2 = 9$. So the answer is $\boxed{\textbf{(C)}\ 9}$. ~LarryFlora

Solution 3 (Venn Diagrams)

We may draw three Venn diagrams to represent these three cases, respectively.

Screen Shot 2021-08-29 at 9.14.51 AM.png

Let the amount of all the marbles be $x$, meaning $R+G+B = x$.

The Venn diagrams give us the equation: $x = (x-6)+(x-8)+(x-4)$. So $x = 3x-18$, $x = 18/2 =9$. Thus, the answer is $\boxed{\textbf{(C)}\ 9}$. ~LarryFlora

Solution 4 (Answer Choices)

Since we know all but $8$ marbles in the jar are green, the jar must have at least $9$ marbles. Then we can just start from $C$ and keep going. If there are $9$ marbles total, there are $3$ red marbles $(9-6)$, $1$ green marble $(9-8)$, and $5$ blue marbles $(9-4)$. Since we assumed there were $9$ marbles and $3+1+5=9$, the answer is $\boxed{\textbf{(C)}\ 9}$.

Solution 5 (Algebra)

Let $x$ be the number of total marbles. There are $x – 6$ red marbles, $x – 8$ green marbles, and $x – 4$ blue marbles. We can create an equation: $(x – 6)+(x – 8)+(x – 4)=x$ Solving, we get $x=9$, which means the total number of marbles is $\boxed{\textbf{(C)}\ 9}$. -J.L.L (Feel free to edit)

Solution 6

Let $x$ be the number of total marbles, $r$ be the number of red marbles, $g$ be the number of green marbles, and $b$ be the number of blue marbles. Then we have $x - r = 6$, $x - g = 8$, $x - b = 4$, and $r + g + b = x$. Adding the first three equations together, we get $3x - r - g - b = 18$ or $3x - (r + g + b) = 18$. Substituting in the fourth equation, we have $3x - x = 18$ $\implies$ $\boxed{\textbf{(C)}\ 9}$.

~cxsmi

Video Solution

https://youtu.be/mMph7QH1kX0 Soo, DRMS, NM

https://youtu.be/-p5qv7DftrU ~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/TkZvMa30Juo?t=1316

~pi_is_3.14

See Also

2012 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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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