document.write( "Question 618603: how many liters of 20% saline solution should be added to 40 liters of a 50% saline solution to make a 30% solution? \n" ); document.write( "

Algebra.Com's Answer #389004 by dragonwalker(73) $\"\"$ $\"About$

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You have 40 liters of 50% solution. You want a 30% solution and have a 20% solution to do this with.\r
\n" ); document.write( "\n" ); document.write( "50% is 20% more than the desired concentration
\n" ); document.write( "20% is 10% less than the desired concentration\r
\n" ); document.write( "\n" ); document.write( "so twice as much 20% solution is needed to lower the concentration compared to the 50% solution raising the concentration (10 vs. 20) \r
\n" ); document.write( "\n" ); document.write( "So if there is 40 liters of the 50% which is 20% above the desired concentration twice as much of the 20% is needed to lower it by the same amount.\r
\n" ); document.write( "\n" ); document.write( "So 40 x 2 liters of 20% is required. i.e. 80 liters\r
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