document.write( "Question 400731: How many liters of a 14% alcohol solution must be mixed with 20L of a 50% solution to get a 30% solution? \n" ); document.write( "

Algebra.Com's Answer #283665 by Tatiana_Stebko(1539) $\"\"$ $\"About$

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Let add x liters of a 14% alcohol solution to 20L of a 50% solution, then will have $\"%28x%2B20%29\"$ liters of a 30% solution
\n" ); document.write( "In x liters of a 14% alcohol solution is $\"0.14x\"$ liters of alcohol
\n" ); document.write( "In 20 liters of a 50% alcohol solution is $\"0.5%2A20=10\"$ liters of alcohol
\n" ); document.write( "In $\"%28x%2B20%29\"$ liters of a 30% alcohol solution is $\"0.3%2A%28x%2B20%29\"$ liters of alcohol
\n" ); document.write( "equation
\n" ); document.write( " $\"0.14x%2B10=0.3%2A%28x%2B20%29\"$
\n" ); document.write( " $\"0.14x%2B10=0.3x%2B6\"$
\n" ); document.write( " $\"0.14x-0.3x=6-10\"$
\n" ); document.write( " $\"-0.16x=-4\"$
\n" ); document.write( " $\"x=%28-4%29%2F%28-0.16%29\"$
\n" ); document.write( " $\"x=25\"$ liters of a 14% alcohol solution must be mixed with 20L of a 50% solution to get a 30% solution \n" ); document.write( "

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