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 (Trial and Error)
are blue and green -
are red and blue -
are red and green-
We can do trial and error. Let's make blue . That makes green and red because and . To check this, let's plug and into , which works. Now count the number of marbles - . So the answer is
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 be , meaning .
The Venn diagrams give 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 .
Solution 5 (Algebra)
Let be the number of total marbles. There are red marbles, green marbles, and blue marbles. We can create an equation: Solving, we get , which means the total number of marbles is . -J.L.L (Feel free to edit)
Solution 6
Let be the number of total marbles, be the number of red marbles, be the number of green marbles, and be the number of blue marbles. Then we have , , , and . Adding the first three equations together, we get or . Substituting in the fourth equation, we have .
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 | |
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All AJHSME/AMC 8 Problems and Solutions |
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