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krisolaw

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Jun 5, 2005
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How much water should be added to 20 ounces of a 15% solution of alcohol to dilute it to a 10% mixture?

I am having some problems setting this up. Could someone please expain it to me? I am trying to get ready for midterms next week. Thanks in advance!
 
(15%)(20)=3, so we have 3oz of alcohol.
So how many oz of solution will we need if (10%)(x)=3?
 
(15%)(20)=3, so we have 3oz of alcohol.
So how many oz of solution will we need if (10%)(x)=3?

Even with this information, I still can't get the correct answer according to the book. The book tells me the answer is 10 ounces of water. What am I missing? Please explain further. Thanks
 
Well (10%)(x)=3oz implies that x=30oz.
So you have to add 10oz to 20oz to get 30oz.
 
Hello, krisolaw!

How much water should be added to 20 ounces of a 15% solution of alcohol to dilute it to a 10% mixture?
"Mixture" problems are always challenging.
This one gives us concentrations of alcohol, but we're adding water.
We should think in terms of the amount of water in the solution.

We have 20 ounces of solution which is 85% water.
. . We have: 0.85 x 20 = 17 ounces of water.

We will add x ounces of pure water.
. . This mixture will contain 17 + x ounces of water.

We will have 20 + x ounces of mixture which is supposed to be 90% water.
. . That is, it will have: 0.90(20 + x) ounces of water.

Note: We just described the final amount of water in two ways.

And there is our equation: . 17 + x .= .0.90(20 + x)
 
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