2012 AMC 8 Problems/Problem 19
Contents
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 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 may 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 . So 9 (C) is the answer.
See Also
2012 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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