SOLUTION: 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?

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Question 621442: 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?

Answer by ankor@dixie-net.com(15652) About Me  (Show Source):
You can put this solution on YOUR website!
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 2%2F3r in it
:
let r = no. of gal of r required
the mixture equation
2%2F3(3) + r = .75(r+3)
2 + r = .75r + 2.25
r - .75r = 2.25 - 2
.25r = .25
r = 1 gallon required to accomplish this