Difference between revisions of "2012 AMC 8 Problems/Problem 19"
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Thus, the answer is <math>\boxed{\textbf{(C)}\ 9}</math>. ---LarryFlora | Thus, the answer is <math>\boxed{\textbf{(C)}\ 9}</math>. ---LarryFlora | ||
| + | ==Solution 4 (Answer Choices)== | ||
| + | Since we know all but <math>8</math> marbles in the jar are green, the jar must have at least <math>9</math> marbles. Then we can just start from <math>C</math> and keep going. If there are <math>9</math> marbles total, there are <math>3</math> red marbles <math>(9-6)</math>, <math>1</math> green marble <math>(9-8)</math>, and <math>5</math> blue marbles <math>(9-4)</math>. Since we assumed there were <math>9</math> marbles and <math>3+1+5=9</math>, the answer is <math>\boxed{\textbf{(C)}\ 9</math>. | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 15:51, 12 January 2023
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: 
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 Diagrams
We may draw three Venn diagrams to represent these three cases, respectively.
Let the amount of all the marbles is 
, meaning 
.
The Venn diagrams give us the equation: 
.
So 
, 
.
Thus, the answer is . ---LarryFlora
Solution 4 Venn Diagrams
We may draw three Venn diagrams to represent these three cases, respectively.
Let the amount of all the marbles is , meaning .
Adding the three Venn diagrams, it gives us the equation: 
.
So 
, .
Thus, the answer is . ---LarryFlora
Solution 4 (Answer Choices)
Since we know all but marbles in the jar are green, the jar must have at least 
marbles. Then we can just start from 
and keep going. If there are marbles total, there are 
red marbles 
, 
green marble 
, and 
blue marbles 
. Since we assumed there were marbles and 
, the answer is $\boxed{\textbf{(C)}\ 9$ (Error compiling LaTeX. Unknown error_msg).
Video Solution
https://youtu.be/mMph7QH1kX0 Soo, DRMS, NM
https://youtu.be/-p5qv7DftrU ~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/TkZvMa30Juo?t=1316
~pi_is_3.14
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. 
