document.write( "Question 30192: Pure acid is to be added to a 10% acid solution to obtain 54L of a 20% acid solution. What amounts of each should be used? \n" ); document.write( "
Algebra.Com's Answer #16890 by longjonsilver(2297)\"\" \"About 
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Let x = amount of pure acid added.\r
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\n" ); document.write( "\n" ); document.write( "So how much of the 10% acid was there? Easy... everything else to make up the 54litres...so (54-x).\r
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\n" ); document.write( "\n" ); document.write( "Now we say BEFORE = AFTER\r
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\n" ); document.write( "\n" ); document.write( "BEFORE is (1.00*x) + (0.10*(54-x))
\n" ); document.write( "AFTER is 0.20*54\r
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\n" ); document.write( "\n" ); document.write( "so (1.00*x) + (0.10*(54-x)) = 0.20*54
\n" ); document.write( "x + 5.4 - 0.1x = 10.8
\n" ); document.write( "0.9x + 5.4 = 10.8
\n" ); document.write( "0.9x = 5.4
\n" ); document.write( "x = 5.4/0.9
\n" ); document.write( "x = 6 litres\r
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\n" ); document.write( "\n" ); document.write( "So we need 6 litres of the pure acid and also 48 litres of the 10% acid solution.\r
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\n" ); document.write( "\n" ); document.write( "jon.
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