SOLUTION: Solution A is 50% alcohol. Solution B is 20% alcohol. We want to create a 20 liter mixture of the two solutions that will be 32% alcohol. How many liters of each solution should be

Question 495507: Solution A is 50% alcohol. Solution B is 20% alcohol. We want to create a 20 liter mixture of the two solutions that will be 32% alcohol. How many liters of each solution should be used?
Found 2 solutions by josmiceli, Earlsdon:
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Let = liters of solution A needed
Let = liters of solution B needed
given:
(1)
(2)
---------------------
(2)
(2)
Multiply both sides of (1) by
and subtract (1) from (2)
(2)
(1)

and, since,
(1)
(1)
(1)
8 liters of solution A are needed
12 liters of solution B are needed
check answer:
(2)
(2)
(2)
(2)
OK
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Let x = the required number of liters of solution A (50% alcohol solution), then (20-x) will be the number of liters of solution B (20% alcohol solution).
The sum of these is to be 20 liters of 32% alcohol solution.
This situation can be expressed (after changing the percentages to their decimal equivalents) in the following equation:
Simplify and solve for x.
Combine the x-terms.
Subtract 4 from both sides.
Finally, divide both sides by 0.3
liters and...
liters.
You will need to mix 8 liters of solution A (the 50% alcohol solution) with 12 liters of solution B (the 20% alcohol solution) to obtain 20 liters of 32% alcohol solution.

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