SOLUTION: Pure acid is to be added to a 20 percent acid solution to obtain 48 L of a 40 percent acid solution. What amounts of each should be used?

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Question 968929: Pure acid is to be added to a 20 percent acid solution to obtain 48 L of a 40 percent acid solution. What amounts of each should be used?
Found 2 solutions by josgarithmetic, Alan3354:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Solvable with a single variable.

Let v be volume of the pure acid (ASSUMING it is a liquid), and then 48-v is the volume of the 20% solution liquid.

%28100v%2B20%2848-v%29%29%2F48=40 which is accounting for percents. Numerator is amount of solute, and denominator is amount of solution result.

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Pure acid is to be added to a 20 percent acid solution to obtain 48 L of a 40 percent acid solution. What amounts of each should be used?
=================
T = amount of 20%
P = amount of pure (100%)
---
T + P = 48 (total liquid)
20T + 100P = 48*40 (total acid)
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20T + 100P = 1920
T + 5P = 96
T + P = 48
---------------------- Subtract
4P = 48
etc