SOLUTION: please help me solve this problem how many ounces of pure water must be added to 5o ounces of 15% saline solution to make a saline solution that is 10% salt?

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Question 864372: please help me solve this problem
how many ounces of pure water must be added to 5o ounces of 15% saline solution to make a saline solution that is 10% salt?
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
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%2F50 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%2F%2850%2Bx%29 and that fraction is forced to be equal to 0.10 since we want a 10% final saline solution. So we get 7.5%2F%2850%2Bx%29+=+0.10


Now solve for x


7.5%2F%2850%2Bx%29+=+0.10


7.5+=+0.10%2850%2Bx%29


7.5+=+5%2B0.10x


7.5-5+=+0.10x


2.5+=+0.10x


2.5%2F0.10+=+x


25+=+x


x+=+25


So you must add 25 ounces of pure water.