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

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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?
 
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?
  
−
<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}18 </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==
+
==Solution 1 (Trial and Error)==
−
Let <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.
 
  
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If "all but 6 are red marbles", that means that the number of green marbles and the number of blue marbles amount to <math>6</math>. Likewise, the number of red marbles and blue marbles amount to <math>8</math>, and the number of red marbles and the number of green marbles amount to <math>4</math>.
+
<math>6</math> are blue and green - <math>b+g=6</math>
  
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We have three equations:
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<math>8</math> are red and blue - <math>r+b=8</math>
  
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<math> g + b = 6 </math>
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<math>4</math> are red and green- <math>r+g=4</math>
  
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<math> r + b = 8 </math>
 
  
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<math> r + g = 4 </math>
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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>
  
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We add all the equations to obtain a fourth equation:
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==Solution 2==
  
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<math> 2r + 2g + 2b = 18 </math>
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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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Now divide by <math> 2 </math> on both sides to find the total number of marbles:
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==Solution 3 (Venn Diagrams)==
 +
We may draw three Venn diagrams to represent these three cases, respectively.
  
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<math> r + g + b = 9 </math>
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[[File: Screen_Shot_2021-08-29_at_9.14.51_AM.png]]
  
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Since the sum of the number of red marbles, green marbles, and blue marbles is the number of marbles in the jar, the total number of marbles in the jar is <math> \boxed{\textbf{(C)}\ 9} </math>.
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Let the amount of all the marbles be <math>x</math>, meaning <math>R+G+B = x</math>.  
  
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Notice how we never knew how many of each color there were (there is 1 green marble, 5 blue marbles, and 3 red marbles).
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The Venn diagrams give us the equation: <math>x = (x-6)+(x-8)+(x-4)</math>.
 +
So <math>x = 3x-18</math>, <math>x = 18/2 =9</math>.
 +
Thus, the answer is <math>\boxed{\textbf{(C)}\ 9}</math>.   
 +
~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>.
 +
 
 +
==Solution 5 (Algebra) ==
 +
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.
 +
We can create an equation: <math>(x – 6)+(x – 8)+(x – 4)=x</math>
 +
Solving, we get <math>x=9</math>, which means the total number of marbles is <math>\boxed{\textbf{(C)}\ 9}</math>.
 +
-J.L.L
 +
(Feel free to edit)
 +
 
 +
==Solution 6==
 +
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>.
 +
 
 +
~[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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