document.write( "Question 511886: How many liters of a 40% and 80% acids must be mixed in order to make 20 liters of a 55% acid solution?\r
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document.write( "A 12 liter mixture is 60% alcohol. if 8 liters of a 25% alcohol mixture are added, what percent mixture will be formed.
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Algebra.Com's Answer #342566 by Maths68(1474)![]() ![]() You can put this solution on YOUR website! (1) \n" ); document.write( "How many liters of a 40% and 80% acids must be mixed in order to make 20 liters of a 55% acid solution? \r \n" ); document.write( "\n" ); document.write( "Solution A \n" ); document.write( "Amount = x \n" ); document.write( "Concentration =40% = 0.4\r \n" ); document.write( "\n" ); document.write( "Solution B \n" ); document.write( "Amount = 20-x \n" ); document.write( "Concentration =80% = 0.8\r \n" ); document.write( "\n" ); document.write( "Resultant Solution \n" ); document.write( "Amount =20 \n" ); document.write( "Concentration =55%=0.55\r \n" ); document.write( "\n" ); document.write( "[Amount Solution A * Concentration A] + [Amount Solution B * Concentration of B] = Amount of Resultant * Concentration of resultant\r \n" ); document.write( "\n" ); document.write( "(0.4)x+(0.8)(20-x)=(20)(0.55) \n" ); document.write( "0.4x+16-0.8x=11 \n" ); document.write( "0.4x-0.8x=11-16 \n" ); document.write( "-0.4x=-5 \n" ); document.write( "-0.4x/-0.4=-5/-0.4 \n" ); document.write( "x=12.5\r \n" ); document.write( "\n" ); document.write( "Solution A \n" ); document.write( "Amount = x =12.5 liters \n" ); document.write( "Concentration =40% = 0.4\r \n" ); document.write( "\n" ); document.write( "Solution B \n" ); document.write( "Amount = 20-x =20-12.5=7.5 liters \n" ); document.write( "Concentration =80% = 0.8\r \n" ); document.write( "\n" ); document.write( "Result \n" ); document.write( "======= \n" ); document.write( "12.5 liters of a 40% and 7.5 liters of 80% acids must be mixed in order to make 20 liters of a 55% acid solution \n" ); document.write( "================================================================================ \n" ); document.write( "(2) \n" ); document.write( "A 12 liter mixture is 60% alcohol. if 8 liters of a 25% alcohol mixture are added, what percent mixture will be formed.\r \n" ); document.write( "\n" ); document.write( "Solution A \n" ); document.write( "Amount =12 liters \n" ); document.write( "Concentration =60% = 0.6\r \n" ); document.write( "\n" ); document.write( "Solution B \n" ); document.write( "Amount = 8 liters \n" ); document.write( "Concentration =25% = 0.25\r \n" ); document.write( "\n" ); document.write( "Resultant Solution \n" ); document.write( "Amount =(8+12)=20 liters \n" ); document.write( "Concentration =x\r \n" ); document.write( "\n" ); document.write( "[Amount Solution A * Concentration A] + [Amount Solution B * Concentration of B] = Amount of Resultant * Concentration of resultant \n" ); document.write( "(12*0.6)+(8)(0.25)=20x \n" ); document.write( "7.2+2=20x \n" ); document.write( "9.2=20x \n" ); document.write( "9.2/20=20x/20 \n" ); document.write( "0.46=x \n" ); document.write( "x=0.46 \n" ); document.write( "x=46/100= 46%\r \n" ); document.write( "\n" ); document.write( "Result \n" ); document.write( "======= \n" ); document.write( "If we add 12 liter 60% alcohol solution to 8 liters of a 25% alcohol solution, a mixture of 20 liters of 46% alcohol will be formed. \n" ); document.write( " \n" ); document.write( " |