Question 864372
Let x = amount of pure water to add (in ounces)



We have 50 oz of a 15% saline solution. So we have 50*0.15 = 7.5 oz of pure saline out of 50 oz of the total solution.



So we have this fraction so far: {{{7.5/50}}} and that will equal 0.15 (which is 15%) since we have a 15% solution.



The numerator is the amount of pure saline. The denominator is the total amount of solution (saline + water)



We are adding some x ounces of pure water. So we'll be adding x just in the denominator only. We are NOT adding any pure saline.



So we end up with this fraction {{{7.5/(50+x)}}} and that fraction is forced to be equal to 0.10 since we want a 10% final saline solution. So we get {{{7.5/(50+x) = 0.10}}}



Now solve for x



{{{7.5/(50+x) = 0.10}}}



{{{7.5 = 0.10(50+x)}}}



{{{7.5 = 5+0.10x}}}



{{{7.5-5 = 0.10x}}}



{{{2.5 = 0.10x}}}



{{{2.5/0.10 = x}}}



{{{25 = x}}}



{{{x = 25}}}



So you must add <font size=4 color="red">25 ounces</font> of pure water.