document.write( "Question 534337: pure acid is to be added to a 10% acid solution to obtain 42 L of a 40% acid solution. What amounts of each should be used?\r
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document.write( "blank L of pure acid solution and blank L of 10% acid solution \n" );
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Algebra.Com's Answer #351461 by oberobic(2304)![]() ![]() ![]() You can put this solution on YOUR website! x = amount 10% acid \n" ); document.write( "y = amount of pure acid \n" ); document.write( ". \n" ); document.write( "x+y = 42 liters \n" ); document.write( "x = 42-y \n" ); document.write( "y = 42-x \n" ); document.write( ". \n" ); document.write( "The desired solution is 42 liters of a 40% solution. It will contain 42*.40 liters of pure acid. \n" ); document.write( "42*.4 = 16.8 liters of pure acid and 25.2 liters of solvent (probably water). \n" ); document.write( ". \n" ); document.write( "10%*x + 100%*y = 40%*42 \n" ); document.write( ".1*x + y = 16.8 \n" ); document.write( ". \n" ); document.write( "multiply both sides by 10 to eliminate the decimals \n" ); document.write( ". \n" ); document.write( "x + 10y = 168 \n" ); document.write( ". \n" ); document.write( "substitute for x = 42-y \n" ); document.write( ". \n" ); document.write( "42-y +10y = 168 \n" ); document.write( "9y = 126 \n" ); document.write( "y = 126/9 = 42/3 = 14 \n" ); document.write( ". \n" ); document.write( "x = 42-14 \n" ); document.write( "x = 28 \n" ); document.write( ". \n" ); document.write( "Check how much 'pure' acid is in the solution. \n" ); document.write( "28 * 10% acid = 2.8 liters of pure acid \n" ); document.write( "13 * 100% acid = 14 liters of pure acid \n" ); document.write( "2.8 + 14 = 16.8 liters of pure acid in the 42 liters of solution \n" ); document.write( ". \n" ); document.write( "Answer: Mix 14 liters of pure acid and 28 liters 10% acid to make 42 liters of 40% acid. \n" ); document.write( ". \n" ); document.write( "Done. \n" ); document.write( " \n" ); document.write( " |