2009 AIME II Problems/Problem 1

Revision as of 13:21, 18 April 2009 by Aimesolver (talk | contribs) (Solution 2)

Problem

Before starting to paint, Bill had $130$ ounces of blue paint, $164$ ounces of red paint, and $188$ 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

Solution 1

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 $130$ ounces each. Say $a$ is the fraction of the pink paint that is red paint and $x$ is the size of each stripe. Then equations can be written: $ax = 164 - 130 = 34$ and $(1-a)x = 188 - 130 = 58$. The second equation becomes $x - ax = 58$ and substituting the first equation into this one we get $x - 34 = 58$ so $x = 92$. The amount of each color left over at the end is thus $130 - 92 = 38$ and $38 * 3 = \boxed{114}$.

Solution 2

We know that all the stripes are of equal size. We can then say that $r$ is the amount of paint per stripe. Then $130 - r$ 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 $188 - r$ and $164 - r$ respectively. The pink stripe is also r ounces of paint, but let there be $k$ ounces of red paint in the mixture and $r - k$ ounces of white paint. We now have two equations: $164 - r - k = 188 - r - (r-k)$ and $164 - r -  k = 130 - r$. Solving yields k = 34 and r = 92. We now see that there will be $130 - 92 = 38$ ounces of paint left in each can. $38 * 3 = \boxed{114}$

See Also

2009 AIME II (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
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