<PRE><B>Question 953330</B>
5 gallons of a 20% acid solution is to be mixed with a 12% acid solution to yield an 18% acid solution.
 How many gallons of the 12% acid solution will be needed?
:
let  = amt of 12% acid required
then
(x+5) = resulting amt of the 18% solution
:
A mixture equation in decimal form
.20(5) + .12x = .18(x+5)
1.0 + .12x = .18x + .90
1.0 - .90 = .18x - .12x
.1 = .06x
x - .1/.06
x = 1{{{2/3}}} gal of 12% solution
:
:
Check this on your calc: 
enter .2(5) + .12(1.6667)
then
enter .18(1.6667+5)</PRE>