Difference between revisions of "2011 AIME I Problems/Problem 1"
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== Solution == | == Solution == | ||
| − | 45/5 | + | 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 transfering 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>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)</math> 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. | ||
Revision as of 12:59, 19 March 2011
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
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 
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.
