document.write( "Question 577678: how many liters of a 50% acid solution must be mixed with a 30% acid solution to get 200 L of a 40% acid solution? \n" ); document.write( "
Algebra.Com's Answer #370251 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
percent ---------------- quantity
\n" ); document.write( "Acod solution type I 50.00% ---------------- x liters
\n" ); document.write( "Acid solution type II 30.00% ------ 200 - x liters
\n" ); document.write( "Mixture 40.00% ---------------- 200
\n" ); document.write( "
\n" ); document.write( "50.00% x+30.00%(200-x)=40.00% * 200
\n" ); document.write( "50x+30(200-x)= 8000
\n" ); document.write( "50x+6000-30x=8000
\n" ); document.write( "50x-30x =8000-6000
\n" ); document.write( "20x= 2000
\n" ); document.write( "/ 20
\n" ); document.write( "x= 100 liters 50.00% Acid solution type I
\n" ); document.write( "100 liters 30.00% Acid solution type II 50
\n" ); document.write( "m.ananth@hotmail.ca
\n" ); document.write( " \n" ); document.write( "
\n" );