document.write( "Question 129672This question is from textbook Algebra structure and meathod
\n" ); document.write( ": A chemist mixed a 20% acid solution with a 60% acid solution to make 10 L of a 30% solution. How many liters of the 60% solution were used? \n" ); document.write( "
Algebra.Com's Answer #94659 by stanbon(75887)\"\" \"About 
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A chemist mixed a 20% acid solution with a 60% acid solution to make 10 L of a 30% solution. How many liters of the 60% solution were used?
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\n" ); document.write( "20% solution DATA:
\n" ); document.write( "Amount = x liters; active ingredient = 0.20x L
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\n" ); document.write( "60% solution DATA:
\n" ); document.write( "Amount = \"10-x\" liters ; active = 0.60(10-x)= 6-0.6x liters
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\n" ); document.write( "Mixture DATA;
\n" ); document.write( "Amt. = 10 L : active = 0.30*10 = 3 liters
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\n" ); document.write( "EQUATION:
\n" ); document.write( "active + active = active
\n" ); document.write( "0.2x + 6 - 0.6x = 3
\n" ); document.write( "-0.4x = -3
\n" ); document.write( "x = 7.5 liters (amount of 20% solution needed for the mix)
\n" ); document.write( "10-x = 2.5 liters (amount of 60% solution needed for the mix)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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