2009 AIME II Problems/Problem 1
Contents
Problem
Before starting to paint, Bill had 
ounces of blue paint, 
ounces of red paint, and 
ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When Bill finished, he had equal amounts of blue, red, and white paint left. Find the total number of ounces of paint Bill had left.
Solution
After the pink stripe is drawn, all three colors will be used equally so the pink stripe must bring the amount of red and white paint down to ounces each. Say 
is the fraction of the pink paint that is red paint and 
is the size of each stripe. Then equations can be written: 
and 
. The second equation becomes 
and substituting the first equation into this one we get 
so 
. The amount of each color left over at the end is thus 
and 
.
Solution 2
We know that all the stripes are of equal size. We can then say that 
is the amount of pain per stripe. Then 
will be the amount of blue paint left. Now for the other two stripes. The amount of white paint left after the white stripe and the amount of red paint left after the blue stripe are 
and 
respectively. The pink stripe is also r ounces of paint, but let there be 
ounces of red paint in the mixture and 
ounces of white paint. We now have two equations: 
and 
. Solving yields k = 34 and r = 92. We now see that there will be ounces of paint left in each can. 
ounces * 
cans = \boxed{114}$
See Also
| 2009 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
