Art of Problem Solving
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2012 AMC 8 Problems/Problem 19: Difference between revisions

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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 <math>x</math>. From the Venn diagrams, which gives us the equation: <math>(x-6)+(x-8)+(x-4) = x</math>. So <math>3x-18= x</math>. Then <math>x = 18/2 =9</math>. Thus, the answer is <math>\boxed{\textbf{(C)}\ 9}</math>.    ---LarryFlora
Let the amount of all the marbles is <math>x</math>. The Venn diagrams give us the equation: <math>(x-6)+(x-8)+(x-4) = x</math>. So <math>3x-18= x</math>. Then <math>x = 18/2 =9</math>. 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 08:10, 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 Diagram

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We may draw three Venn diagrams to represent these three cases, respectively. 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination