<PRE><B>Question 1100802</B>
A 5% solution is not reasonable. I assume you mean a 50% solution
:
A chemical company mixes pure water with their premium antifreeze solution to create an inexpensive antifreeze mixture.
 The premium antifreeze solution contains 80% pure antifreeze.
 The company wants to obtain 320 gallons of a mixture that contains 50% pure antifreeze.
 How many gallons of water and how many gallons of the premium antifreeze solution must be mixed?
:
let w = amt of water required
:
An equation based on the the percent water, an 80% solution is 20% water.
.20(320-w) + w = .50(320)
64 - .2w + w  = 160
.8w = 160 - 64
w = {{{96/.8}}}
w = 120 gal of water to produce 320 gal of 50% antifreeze
:
320 - 120 = 200 gal of 80% solution


 
 Water: _120___ gallons
 Premium Antifreeze: __200___ gallons  </PRE>