SOLUTION: How many liters of 60% alcohol solution and 30% alcohol solution must be mixed to obtain 18 liters of 50% alcohol solution? Complete the table below, then solve.

Question 1095256: How many liters of 60% alcohol solution and 30% alcohol solution must be mixed to obtain 18 liters of 50% alcohol solution?
Complete the table below, then solve.
60% solution 30% solution Final solution
Number of Liters x y
Liters of Alcohol

You need ( ) liters of 60% solution and ( ) liters of 30% solution.

Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39630) (Show Source): You can put this solution on YOUR website!
This solution is being done with just one variable:
x, volume of the 30% alcohol
18-x, volume of the 60% alcohol

, to account for pure alcohol
-

6 liters of 30% alcohol
12 liters of 60% alcohol
Answer by greenestamps(13214) (Show Source): You can put this solution on YOUR website!

For the method you were asked to use, the table might look like this:

Then your two equations (from rows 2 and 3 of the table) are for the total number of liters...

and for the total amount of alcohol...

While that is a good way to learn to solve this kind of problem, the method suggested by the other tutor (using only one variable) is, I think, easier.

And for a much faster way to the answer, notice that the percentage of the final mixture, 50%, is "twice as close" to 60% as it is to 30%; that means you need to use twice as much of the 60% alcohol as the 30% alcohol.

18 liters, using twice as much of the 60% alcohol, means 12 liters of the 60% and 6 liters of the 30%.

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