Difference between revisions of "2012 AMC 8 Problems/Problem 19"
(→Solution) |
(→Problem) |
||
| Line 4: | Line 4: | ||
<math> \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 </math> | <math> \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 </math> | ||
| − | Please check out my channel Elections Projections. | + | Please check out my channel Elections Projections. The link https://www.youtube.com/channel/UCUf37EvvIHugF9gPiJ1yRzQ |
| + | 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 is the answer | ||
==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 16:02, 8 January 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?
Please check out my channel Elections Projections. The link https://www.youtube.com/channel/UCUf37EvvIHugF9gPiJ1yRzQ 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 is the answer
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. 
