Difference between revisions of "2012 AMC 8 Problems/Problem 19"
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==Solution 2== | ==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 <math>19/2 = 9</math>. So the answer is <math>\boxed{\textbf{(C)}\ 9}</math>. ---LarryFlora | + | 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 <math>r+g+b = 19/2 = 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 18:06, 28 August 2021
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 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 
. So 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. 
