2012 AMC 8 Problems/Problem 19
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
Let be the number of red marbles, be the number of green marbles, and be the number of blue marbles.
If "all but 6 are red marbles", that means that the number of green marbles and the number of blue marbles amount to . Likewise, the number of red marbles and blue marbles amount to , and the number of red marbles and the number of green marbles amount to .
We have three equations:
We add all the equations to obtain a fourth equation:
Now divide by on both sides to find the total number of marbles:
Since the sum of the number of red marbles, green marbles, and blue marbles is the number of marbles in the jar, the total number of marbles in the jar is .
Notice how we never knew how many of each color there were (there are 1 green marble, 5 blue marbles, and 3 red marbles).
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.