document.write( "Question 772776: You have 30 cl of acid solution that is 10% in strenght, how much water would you need to dilute the solution to 7% in strenght.\r
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Algebra.Com's Answer #471159 by Edwin McCravy(20056)\"\" \"About 
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Mixture_Word_Problems/772776 (2013-08-14 12:56:29): You have 30 cl of acid solution that is 10% in strenght, how much water would you need to dilute the solution to 7% in strenght.
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document.write( "10% of 30 is 3. So\r\n" );
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document.write( "30 cl of 10% solution contains 3 cl of pure acid [and the rest (27 cl) is\r\n" );
document.write( "water.  Forget the water, just think about the amt of acid and the amt of\r\n" );
document.write( "liquid].\r\n" );
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document.write( "So you have 30 cl of liquid that contains 3 cl of acid\r\n" );
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document.write( "Now you're going to add x liters of liquid (water).\r\n" );
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document.write( "So now you have 30+x cl of liquid that still contains only 3 cl of acid.\r\n" );
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document.write( "Now those 3 cl of acid must be that 7% of 30+x.\r\n" );
document.write( "          |                 |       |  |\r\n" );
document.write( "So        3                 =     .07  *  (30+x)\r\n" );
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document.write( "                         3 = .07*(30+x)\r\n" );
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document.write( "Multiply both sides by 100 to get rid of the decimal\r\n" );
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document.write( "                       300 = 7*(30+x)\r\n" );
document.write( "                       300 = 210+7x\r\n" );
document.write( "                        90 = 7x\r\n" );
document.write( "                        \"90%2F7\" = x\r\n" );
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document.write( "That works out to about 12.9 cl of water to dilute it down to 7%\r\n" );
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document.write( "Checking.  You'll end up with about 30+12.9 or 42.9 cl. and 7% of\r\n" );
document.write( "that is 42.9(.07) = 3.003, and that's close enough to 3, and we \r\n" );
document.write( "expected it be a little off because the 12.9 was rounded off.\r\n" );
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document.write( "Edwin
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