SOLUTION: A 20-gallon salt-water solution contains 15% pure salt. How much water should be added to make a new solution that is only 12% salt?

Question 635359: A 20-gallon salt-water solution contains 15% pure salt. How much water should be added to make a new solution that is only 12% salt?
Found 2 solutions by mananth, Maths68:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
20 gallons contains 15%
so it contains 3 grams salt. ((15/100) *20)
We need 12 % salt
so we have to add
(3*100)/12
=25 gallons
you have to add 25-20 = 5 gallons of water to make it 12% solution.
Alternative
-------------- percent ---------------- Amount
Salt solution 15 ---------------- 20 gallons
Water 0 ---------------- x gallons
Mixture 12 ---------------- 20 + x gallons

15 * 20 + 0 x = 12 ( 20 + x )
300 + 0 x = 240 + 12 x
0 x -12 x = -300 + 240
-12 x = -60
/ -12
x = 5 gallons Water

m.ananth@hotmail.ca

Answer by Maths68(1474) (Show Source): You can put this solution on YOUR website!
Solution A
Amount=20 gallons
Concentration =15%=0.15

Solution B
Amount=x gallons
Concentration =0%=0 (Pure Water)

Mixture
Amount=(20+x)gallons
Concentration =12%=0.12
(Amount of Solution A)(Concentration of Solution B)+(Amount of Solution B)(Concentration of Solution A)=(Amount of Mixture)(Concentration of Mixture
(20)(0.15)+(x)(0)=(20+x)(0.12)
3+0=2.4+0.12x
3-2.4=0.12x
0.6=0.12x
0.6/0.12=0.12x/0.12
5=x
x=5

A 20-gallon salt-water solution contains 15% pure salt. 5 gallons of water should be added to make a new solution that is only 12% salt.

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