Percentage of Alcohol in a SolutionDate: 06/02/99 at 00:28:59 From: Nycole Subject: Algebra Suppose 30 liters of a solution with an unknown percentage of alcohol is mixed with 5 liters of a 90% alcohol solution. If the resulting mixture is a 62% alcohol solution, what is the percentage of alcohol in the first solution? First, I tried to figure out an equation --> (30 + x) + (5 + .90) = .62 (or 62%). I do not know if you should add the percentage of alcohol with the liters, then add that with 5 liters of a 90% alcohol solution. Is that right? Please help! Date: 06/02/99 at 12:05:23 From: Doctor Rob Subject: Re: Algebra Think of the amount of alcohol in each solution. In the first one, it is 30*(x/100) liters. In the second one, it is 5*(90/100). In the last one, it is 35*(62/100) (because the total amount of the mixture is 35 = 30 + 5 liters). When the first two solutions are mixed, the total amount of alcohol is the sum of the amounts in the two ingredients, so you get an equation: 30*(x/100) + 5*(90/100) = 35*(62/100). This equation is pretty easy to solve for x, the percentage you are seeking. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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