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

(Solution 3 Venn Diagram)
(Solution 3 Venn Diagram)
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We may draw three Venn diagrams to represent these three cases respectively.  
 
We may draw three Venn diagrams to represent these three cases respectively.  
Let the amount of all the marbles is X , so <math>X+18 = 3X</math>. It gives us the amount of all the marbles is <math>X= 18/2 =9</math>. The answer is <math>\boxed{\textbf{(C)}\ 9}</math>. ---LarryFlora
+
Let the amount of all the marbles is x, so <math>x+6+8+4 = 3x</math>. It gives us <math>x= 18/2 =9</math>. So 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 19:36, 28 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 Diagram

We may draw three Venn diagrams to represent these three cases respectively. Let the amount of all the marbles is x, so $x+6+8+4 = 3x$. It gives us $x= 18/2 =9$. So 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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