Difference between revisions of "2009 AIME II Problems/Problem 1"
Flyhawkeye (talk | contribs) (→Solution: Added solution) |
|||
Line 3: | Line 3: | ||
== Solution == | == Solution == | ||
+ | |||
+ | === Simple Solution === | ||
+ | |||
+ | Let the stripes be <math>b, r, w,</math> and <math>p</math>, respectively. Let the red part of the pink be <math>\frac{r_p}{p}</math> and the white part be <math>\frac{w_p}{p}</math> for <math>\frac{r_p+w_p}{p}=p</math>. | ||
+ | |||
+ | Since the stripes are of equal size, we have <math>b=r=w=p</math>. Since the amounts of paint end equal, we have <math>130-b=164-r-\frac{r_p}{p}=188-w-\frac{w_p}{p}</math>. Thus, we know that | ||
+ | <cmath>130-p=164-p-\frac{r_p}{p}=188-p-\frac{w_p}{p}</cmath> | ||
+ | <cmath>130=164-\frac{r_p}{p}=188-\frac{w_p}{p}</cmath> | ||
+ | <cmath>r_p=34p, w_p=58p</cmath> | ||
+ | <cmath>\frac{r_p+w_p}{p}=92=p=b.</cmath> | ||
+ | Each paint must end with <math>130-92=38</math> oz left, for a total of <math>3 \cdot 38 = \boxed{114}</math> oz. | ||
=== Solution 1 === | === Solution 1 === |
Revision as of 16:27, 16 July 2018
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
Simple Solution
Let the stripes be and , respectively. Let the red part of the pink be and the white part be for .
Since the stripes are of equal size, we have . Since the amounts of paint end equal, we have . Thus, we know that Each paint must end with oz left, for a total of oz.
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 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 paint 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.
Solution 3
Let the amount of paint each stripe painted used be . Also, let the amount of paint of each color left be . 1 stripe is drawn with the blue paint, and 3 stripes are drawn with the red and white paints. Add together the amount of red and white paint, and obtain the following equations : and . Solve to obtain . Therefore is , with three cans of equal amount of paint, the answer is .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.