2011 AIME I Problems/Problem 1
Problem 1
Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C contains one liter of a solution that is acid. From jar C, liters of the solution is added to jar A, and the remainder of the solution in jar C is added to jar B. At the end both jar A and jar B contain solutions that are 50% acid. Given that and are relatively prime positive integers, find .
Solution 1
There are L of acid in Jar A. There are L of acid in Jar B. And there are L of acid in Jar C. After transfering the solutions from jar C, there will be
L of solution in Jar A and L of acid in Jar A.
L of solution in Jar B and $\frac{12}{5}+\frac{k}{100}\cdot \left(1-\frac{m}{n}\right)=\frac{12}{5}+\frac{k}{100}-\frac{mk}{100n}\right$ (Error compiling LaTeX. ! Missing delimiter (. inserted).) of acid in Jar B.
Since the solutions are 50% acid, we can multiply the amount of acid for each jar by 2, then equate them to the amount of solution.
Add the equations to get
Solving gives .
If we substitute back in the original equation we get so . Since and are relatively prime, and . Thus .
Solution 2
One might cleverly change the content of both Jars.
Since the end result of both Jars are acid, we can turn Jar A into a 1 gallon liquid with acid
and Jar B into 1 gallon liquid with acid.
Now, since Jar A and Jar B contain the same amount of liquid, twice as much liquid will be pour into Jar A than Jar B, so of Jar C will be pour into Jar A.
Thus, and .
Solving for yields
So the answer is
Solution 3
One may first combine all three jars in to a single container. That container will have liters of liquid, and it should be acidic. Thus there must be liters of acid.
Jug A contained , or of acid, and jug B or . Solving for the amount of acid in jug C, , or .
Once one knows that the jug C is acid, use solution 1 to figure out m and n.
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
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Followed by Problem 2 | |
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