How man ounces of a 50% alcohol solution must be mixed with 80 ounces of a 20% alcohol solution to make a 40% alcohol solution?

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- Apr 23rd 2008, 11:32 AMj_drydenAnother mixture one
How man ounces of a 50% alcohol solution must be mixed with 80 ounces of a 20% alcohol solution to make a 40% alcohol solution?

- Apr 23rd 2008, 01:28 PMelizsimca
- Apr 23rd 2008, 01:46 PMicemanfan
I agree with this method for solving the problem. This is equating the amount of alcohol present in each of the partial solutions with the amount of alcohol present in the mixed solution. The $\displaystyle (.40)(80+x)$ represents the total number of ounces present in the mix after combining the two solutions, multiplied by the final percentage of 40% to arrive at the total amount of alcohol present.