Difference between revisions of "2012 AMC 8 Problems/Problem 19"
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==Solution 3 Venn Diagram== | ==Solution 3 Venn Diagram== | ||
| + | We may draw three Venn diagrams to represent these three cases, respectively. | ||
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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 08:12, 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?
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: 
are blue and green, meaning 
; 
are red and blue, meaning 
; 
are red and green, meaning 
. Then we need to add these three equations: 
. It gives us all of the marbles are 
. So the answer is 
. ---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 
. The Venn diagrams give us the equation: 
. So 
. Then 
. Thus, the answer is . ---LarryFlora
See Also
| 2012 AMC 8 (Problems • Answer Key • Resources) | ||
| 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. 
