SOLUTION: A solution of salt and water contains 18 ounces of pure salt and 36 ounces of water. How many ounces of water must be added to obtain a 24% pure salt solution? I have tried m

Question 203260: A solution of salt and water contains 18 ounces of pure salt and 36 ounces of water. How many ounces of water must be added to obtain a 24% pure salt solution?

I have tried many different solutions and I feel there is something more needed on the right hand side of the equation, but I can't figure it out.
1) x+36
---- = 24%
x+54
2) x+18
---- = 24%
x+24
Found 2 solutions by nerdybill, NaturalBornNerd:
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
A solution of salt and water contains 18 ounces of pure salt and 36 ounces of water. How many ounces of water must be added to obtain a 24% pure salt solution?
.
Let x = amount of water to be added
then
"amount of salt" = 24% of "total volume"
18 = .24(18+36+x)
18 = .24(54+x)
18 = 12.96 + .24x
5.04 = .24x
21 ounces = x (amount of water to be added)

Answer by NaturalBornNerd(5) (Show Source): You can put this solution on YOUR website!
I approached the problem in a different way and got a different answer,
but it seems to check out.
First,18oz. of salt in 36oz. of water is a 50% solution.
I don't think we need to be concerned with the amount of salt as a variable,
but only the amount of water we need to dilute our mixture down to a 24% concentration of salt.
If we let 'w' = water, then what we have is 18/36w = 0.05 (oop's, this is a typo. I mean 0.50) .
What we want is 18/w = 0.24, so...
18/w = 0.24
18 = 0.24w
18/0.24 = w
w = 75. So we need to have 18oz. of salt in 75 ounces of water to have a 24% solution.
To achieve this , we need to add 75oz. - 36oz. = 39oz. of water to our original mixture.
We check our answer by dividing 18oz.(salt)/75oz.(water)= 24%(salt).

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