document.write( "Question 864372: please help me solve this problem
\n" ); document.write( "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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Algebra.Com's Answer #520946 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
Let x = amount of pure water to add (in ounces)\r
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\n" ); document.write( "\n" ); document.write( "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.\r
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\n" ); document.write( "\n" ); document.write( "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.\r
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\n" ); document.write( "\n" ); document.write( "The numerator is the amount of pure saline. The denominator is the total amount of solution (saline + water)\r
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\n" ); document.write( "\n" ); document.write( "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.\r
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\n" ); document.write( "\n" ); document.write( "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\"\r
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\n" ); document.write( "\n" ); document.write( "Now solve for x\r
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\n" ); document.write( "\n" ); document.write( "\"7.5%2F%2850%2Bx%29+=+0.10\"\r
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\n" ); document.write( "\n" ); document.write( "\"7.5+=+0.10%2850%2Bx%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"7.5+=+5%2B0.10x\"\r
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\n" ); document.write( "\n" ); document.write( "\"7.5-5+=+0.10x\"\r
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\n" ); document.write( "\n" ); document.write( "\"2.5+=+0.10x\"\r
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\n" ); document.write( "\n" ); document.write( "\"2.5%2F0.10+=+x\"\r
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\n" ); document.write( "\n" ); document.write( "\"25+=+x\"\r
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\n" ); document.write( "\n" ); document.write( "\"x+=+25\"\r
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\n" ); document.write( "\n" ); document.write( "So you must add 25 ounces of pure water.
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