SOLUTION: I believe this is a mixture problem, but I am stuck at setting it up (formula). Can you please help me! Here is the question: How many ounces of pure water must be added to 60 oz

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Question 5363: I believe this is a mixture problem, but I am stuck at setting it up (formula). Can you please help me! Here is the question: How many ounces of pure water must be added to 60 oz of a solution that is 20% water to make a solution that is 30% water? I know this is not set up right.
Amount . rate = Quantity
Pure H20 x .30 .20x
60 oz. .20
Found 2 solutions by rapaljer, melzaren:
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = amount of pure water at 100% water
60 = solution being mixed at 20% water
x+60 = resulting mixture at 30% water.

From this, the equation is:
1.00(x) + (.20)(60) = (.30)(x+ 60)
1.00x + 1.20 = .30x + 1.80

Appropriate simplification gives you
.70x = .60
x+=.60%2F.70+=+6%2F7 oz.

It usually comes out more even than this, but sometimes life is like that. I think it's right. Anybody see an error?

R^2 at SCC

Answer by melzaren(3) About Me  (Show Source):
You can put this solution on YOUR website!
x * 1.0 + 60 * .20 = .30* (x+60)
1.0x + (60 * .20) = .30*(x + 60)
x + 12 = .30x + 18
x - .30x = 6
.70x = 6

x = 6/ .7
x = 60/ 7