SOLUTION: how many liters of a 60% acid solution must be mixed with a 15% acid solution to get 405 L of a 50% acid solution?

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Question 947427: how many liters of a 60% acid solution must be mixed with a 15% acid solution to get 405 L of a 50% acid solution?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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how many liters of a 60% acid solution must be mixed with a 15% acid solution to get 405 L of a 50% acid solution?
:
let x = amt of 60% solution required
the resulting amt is to be 405L, therefore
(405-x) = amt of 15% solution
:
Use the decimal equiv and write a mixture equation
.60x + .15(405-x) = .50(405)
.60x + 60.75 - .15x = 202.5
.60x - .15x = 202.5 - 60.75
.45x = 141.75
x = 141.75/.45
x = 315L of 60% solution required