document.write( "Question 1074549: One solution contains 20% alcohol and another solution contains 60% alcohol. Some of each of the two solutions is mixed to produce 10 liters of a 50% solution. How many liters of the 60% solution should be used? \n" ); document.write( "
Algebra.Com's Answer #689209 by MathTherapy(10552)\"\" \"About 
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\n" ); document.write( "One solution contains 20% alcohol and another solution contains 60% alcohol. Some of each of the two solutions is mixed to produce 10 liters of a 50% solution. How many liters of the 60% solution should be used?
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Let amount of 60% (needed solution) be S
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document.write( "Then amount of 20% solution = 10 - S
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document.write( "We then get the following MIXTURE equation: .6S + .2(10 - S) = .5(10)
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document.write( ".6S + 2 - .2S = 5
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document.write( ".6S - .2S = 5 - 2
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document.write( ".4S = 3
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document.write( "S, or amount of 60% (needed) solution to mix = \"highlight_green%28matrix%281%2C4%2C+3%2F.4%2C+or%2C+7.5%2C+L%29%29\"
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document.write( "That's how SIMPLE this is.....nothing COMPLEX and/or CONFUSING! \n" );
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