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

(Solution 3 Venn Diagram)
(Solution 3 Venn Diagram)
Line 20: Line 20:
  
 
==Solution 3 Venn Diagram==
 
==Solution 3 Venn Diagram==
 
+
[[File: Screen_Shot_2021-08-29_at_9.09.15_AM.png]]  
[[File: Screen Shot 2021-08-28 at 7.24.38 PM.png]]  
 
 
 
 
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>. 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
 
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

Revision as of 09:12, 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

Screen Shot 2021-08-29 at 9.09.15 AM.png 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS