document.write( "Question 347694: I can't figure this out: Pure acid is to be added to a 10% acid solution to obtain 54 L of a 20% acid solution. What amounts of each should be used? \n" ); document.write( "
Algebra.Com's Answer #248544 by stanbon(75887)\"\" \"About 
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Pure acid is to be added to a 10% acid solution to obtain 54 L of a 20% acid solution. What amounts of each should be used?
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\n" ); document.write( "Equation:
\n" ); document.write( "acid + acid = acid
\n" ); document.write( "x + 0.10(54-x) = 0.20*54
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\n" ); document.write( "Multiply thru by 100 and solve for \"x\":
\n" ); document.write( "100x + 10*54 - 10x = 20*54
\n" ); document.write( "90x = 10*54
\n" ); document.write( "x = 6 liters (amt. of pure acid needed in the mixture)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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