Difference between revisions of "2011 AIME I Problems/Problem 1"
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Add the equations to get | Add the equations to get | ||
<cmath>\frac{42}{5}+\frac{k}{50}=10</cmath> Solving gives <math>k=80</math>. | <cmath>\frac{42}{5}+\frac{k}{50}=10</cmath> Solving gives <math>k=80</math>. | ||
− | <br>If we substitute back in the original equation we get <math>\frac{m}{n}=\frac{2}{3}</math> so <math>3m=2n</math>. Since <math>m</math> and <math>n</math> are relatively prime, <math>m=2</math> and <math>n=3</math>. Thus <math>k+m+n=80+2+3=\boxed{ | + | <br>If we substitute back in the original equation we get <math>\frac{m}{n}=\frac{2}{3}</math> so <math>3m=2n</math>. Since <math>m</math> and <math>n</math> are relatively prime, <math>m=2</math> and <math>n=3</math>. Thus <math>k+m+n=80+2+3=\boxed{085hi}</math>. |
== Solution 2 == | == Solution 2 == |
Revision as of 21:00, 17 September 2020
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
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 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.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.