Difference between revisions of "2011 AIME I Problems/Problem 1"
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− | == Problem | + | == 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 <math>k\%</math> acid. From jar C, <math>\frac{m}{n}</math> 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 <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>k + m + n</math>. | 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 <math>k\%</math> acid. From jar C, <math>\frac{m}{n}</math> 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 <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>k + m + n</math>. | ||
== Solution 1== | == Solution 1== | ||
− | There are <math>\frac{45}{100}(4)=\frac{9}{5}</math> L of acid in Jar A. There are <math>\frac{48}{100}(5)=\frac{12}{5}</math> L of acid in Jar B. And there are <math>\frac{k}{100}</math> L of acid in Jar C. After | + | There are <math>\frac{45}{100}(4)=\frac{9}{5}</math> L of acid in Jar A. There are <math>\frac{48}{100}(5)=\frac{12}{5}</math> L of acid in Jar B. And there are <math>\frac{k}{100}</math> L of acid in Jar C. After transferring the solutions from jar C, there will be |
<br> <math>4+\frac{m}{n}</math> L of solution in Jar A and <math>\frac{9}{5}+\frac{k}{100}\cdot\frac{m}{n}</math> L of acid in Jar A.<br> | <br> <math>4+\frac{m}{n}</math> L of solution in Jar A and <math>\frac{9}{5}+\frac{k}{100}\cdot\frac{m}{n}</math> L of acid in Jar A.<br> | ||
− | <br> <math>6-\frac{m}{n}</math> L of solution in Jar B and <math>\frac{12}{5}+\frac{k}{100}\cdot \left(1-\frac{m}{n}\right)=\frac{12}{5}+\frac{k}{100}-\frac{mk}{100n} | + | <br> <math>6-\frac{m}{n}</math> L of solution in Jar B and <math>\frac{12}{5}+\frac{k}{100}\cdot \left(1-\frac{m}{n}\right)=\frac{12}{5}+\frac{k}{100}-\frac{mk}{100n}</math> of acid in Jar B. |
<br>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.<br> | <br>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.<br> | ||
<cmath>\frac{18}{5}+\frac{km}{50n}=4+\frac{m}{n}</cmath> | <cmath>\frac{18}{5}+\frac{km}{50n}=4+\frac{m}{n}</cmath> | ||
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Solving for <math>k</math> yields <math>k=80</math> | Solving for <math>k</math> yields <math>k=80</math> | ||
− | So the answer is <math>80+2+3 = \boxed{ | + | So the answer is <math>80+2+3 = \boxed{085}</math> |
== Solution 3 == | == Solution 3 == | ||
One may first combine all three jars in to a single container. That container will have <math>10</math> liters of liquid, and it should be <math>50\%</math> acidic. Thus there must be <math>5</math> liters of acid. | One may first combine all three jars in to a single container. That container will have <math>10</math> liters of liquid, and it should be <math>50\%</math> acidic. Thus there must be <math>5</math> liters of acid. | ||
− | + | Jar A contained <math>45\% \cdot 4L</math>, or <math>1.8L</math> of acid, and jar B <math>48\% \cdot 5L</math> or <math>2.4L</math>. Solving for the amount of acid in jar C, <math>k = (5 - 2.4 - 1.8) = .8</math>, or <math>80\%</math>. | |
− | Once one | + | Once one knowss that the jar C is <math>80\%</math> acid, use solution 1 to figure out m and n for <math>k+m+n=80+2+3=\boxed{085}</math>. |
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=_znugFEst6E&t=919s | ||
+ | |||
+ | ~Shreyas S | ||
== See also == | == See also == | ||
− | {{AIME box|year=2011|n=I|before= | + | {{AIME box|year=2011|n=I|before=First Problem|num-a=2}} |
+ | {{MAA Notice}} |
Latest 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.