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A 3-gallon mixture contains one part S and two parts R.
In order to change it to a mixture containing 25% S, how much R should be added?
The the resulting amt will be 75% r, do this in per cent r, ignore s
A 3 gallon mixture 1:2 will have
r in it
let r = no. of gal of r required
the mixture equation
(3) + r = .75(r+3)
2 + r = .75r + 2.25
r - .75r = 2.25 - 2
.25r = .25
r = 1 gallon required to accomplish this