document.write( "Question 1184883: A chemist has three different acid solutions. The first acid solution contains 25 % acid, the second contains 40 % and the third contains 85 % . He wants to use all three solutions to obtain a mixture of 135 liters containing 35 % acid, using 2 times as much of the 85 % solution as the 40 % solution. How many liters of each solution should be used?\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

Algebra.Com's Answer #815546 by MathTherapy(10552) $\"\"$ $\"About$

You can put this solution on YOUR website!

\n" ); document.write( "A chemist has three different acid solutions. The first acid solution contains 25 % acid, the second contains 40 % and the third contains 85 % . He wants to use all three solutions to obtain a mixture of 135 liters containing 35 % acid, using 2 times as much of the 85 % solution as the 40 % solution. How many liters of each solution should be used?
\n" ); document.write( "

Let amount of 40% solution to mix, be F
\n" );
document.write( "Then amount of 85% solution to mix = 2F
\n" );
document.write( "Also, amount of 25% to mix = 135 - F - 2F = 135 - 3F
\n" );
document.write( "We then get: .4F + .85(2F) + .25(135 - 3F) = .35(135)
\n" );
document.write( ".4F + 1.7F + .25(135) - .75F = .35(135)
\n" );
document.write( ".4F + 1.7F - .75F = .35(135) - .25(135)
\n" );
document.write( "1.35F = .1(135)
\n" );
document.write( "Amount of 40% to be mixed, or 
\n" );
document.write( "You should easily be able to calculate the amount of 25% and 85% solutions, to mix. \n" );
document.write( "

\n" );