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>
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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>
  
==Solution==
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==Solution 1==
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.
 
  
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>.
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6 are blue and green- b+g=6
  
We have three equations:
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8 are red and blue- r+b=8
  
<math> g + b = 6 </math>
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4 are red and green- r+g=4
  
<math> r + b = 8 </math>
 
  
<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.
  
We add all the equations to obtain a fourth equation:
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==Solution 2==
  
<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=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
  
Now divide by <math> 2 </math> on both sides to find the total number of marbles:
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==Solution 3 Venn Diagrams==
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We may draw three Venn diagrams to represent these three cases, respectively.
  
<math> r + g + b = 9 </math>
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[[File: Screen_Shot_2021-08-29_at_9.14.51_AM.png]]
  
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 is <math>x</math>, meaning <math>R+G+B = x</math>.  
  
Notice how we never knew how many of each color there were (there are 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>.
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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>.    ---LarryFlora
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 +
==Solution 4 Venn Diagrams==
 +
We may draw three Venn diagrams to represent these three cases, respectively.
 +
 
 +
[[File: Screen_Shot_2021-08-29_at_9.14.51_AM.png]]
 +
 
 +
Let the amount of all the marbles is <math>x</math>, meaning <math>R+G+B = x</math>.
 +
 
 +
Adding the three Venn diagrams, it gives us the equation: <math>x+18 = 3x</math>.
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So <math>2x = 18</math>, <math>x = 18/2 =9</math>.
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Thus, the answer is <math>\boxed{\textbf{(C)}\ 9}</math>.   ---LarryFlora
  
 
==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}}

Revision as of 13:55, 3 September 2021

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

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 and it does work. Now count the number of marbles- 5+3+1=9. So 9 (C) is the answer.

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: $b+g+r+b+r+g=2(r+g+b)=6+8+4=19$. It gives us all of the marbles are $r+g+b = 19/2 = 9$. 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 is $x$, 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 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 is $x$, meaning $R+G+B = x$.

Adding the three Venn diagrams, it gives us the equation: $x+18 = 3x$. So $2x = 18$, $x = 18/2 =9$. Thus, the answer is $\boxed{\textbf{(C)}\ 9}$. ---LarryFlora

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
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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