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

(Solution 3 Venn Diagram)
(Solution 3 Venn Diagram)
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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
 
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
  
==Solution 3 Venn Diagram==
+
==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.  
  

Revision as of 09:13, 29 August 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 9.09.15 AM.png

Let the amount of all the marbles is $x$. The Venn diagrams give us the equation: $(x-6)+(x-8)+(x-4) = x$. So $3x-18= x$. Then $x = 18/2 =9$. 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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