2011 AIME I Problems/Problem 1
Problem
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
Jar A contains liters of water, and liters of acid; jar B contains liters of water and liters of acid.
The gap between the amount of water and acid in the first jar, , is double that of the gap in the second jar, . Therefore, we must add twice as much of jar C into the jar over jar . So, we must add of jar C into jar , so .
Since jar C contains liter of solution, we are adding of a liter of solution to jar . In order to close the gap between water and acid, there must be more liters of acid than liters of water in these liters of solution. So, in the liters of solution, there are liters of water, and liters of acid. So, 80% of the sample is acid, so overall, in jar C, 80% of the sample is acid.
Therefore, our answer is .
~ ihatemath123
Solution 2
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 transferring 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 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 3
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 4
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.
Jar A contained , or of acid, and jar B or . Solving for the amount of acid in jar C, , or .
Once one knowss that the jar C is acid, use solution 1 to figure out m and n for .
Video Solution
https://www.youtube.com/watch?v=_znugFEst6E&t=919s
~Shreyas S
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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