SOLUTION: How many liters of a 20% acid solution should be mixed with 30 liters of 50% acid solution in order to obtain a 40% solution.

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: How many liters of a 20% acid solution should be mixed with 30 liters of 50% acid solution in order to obtain a 40% solution.      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 1005187: How many liters of a 20% acid solution should be mixed with 30 liters of 50% acid solution in order to obtain a 40% solution.
Found 2 solutions by Boreal, stanbon:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
x liters of 20%
.20x+30(.50)=(x+30)(.40)
.20x+15=.40x + 12
-.20x= -3
x=15 liters
15*(.20 pure acid)=3 liters
30*(.50 pure acid)=15 liters
That is 18 liters pure acid
That is 45 liters solution *0.45 pure acid=18 liters.
The answer is 15 liters

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
How many liters of a 20% acid solution should be mixed with 30 liters of 50% acid solution in order to obtain a 40% solution.
-----
Equation:
acid + acid = acid
0.20x + 0.50*30 = 0.40(x+30)
-------
20x + 50*30 = 40x + 40*30
----
-20x = -10*30
x = 15 liters (amt. of 20% solution needed)
---------
Cheers,
Stan H.
-----------------