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

## 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.

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.

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

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