# 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 (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.

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

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 $3x - x = 18$ $\implies$ $\boxed{\textbf{(C)}\ 9}$.

## Video Solution

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

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

~pi_is_3.14