SOLUTION: How much water must be added to a barrel containing 48 gallons of a 10% salt solution to obtain a 6% salt solution?

Question 74741: How much water must be added to a barrel containing 48 gallons of a 10% salt solution to obtain a 6% salt solution?
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!

Let x=amount of water that must be added
Now we know that he amount of pure salt before adding the water must equal the amount of pure salt after adding the water. So our equation to solve is:
0.10(48)=0.06(48+x) get rid of the parens
4.8=2.88+0.06x subtract 2.88 from both sides
4.8-2.88=2.88-2.88+0.06x collect like terms
1.92=0.06x divide both sides by 0.06x
x=32 gal of water must be added

CK
0.10(48)=0.06(48+32)
4.8=0.06(80)
4.8=4.8

Hope this helps----ptaylor

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