Answer
120 liters of 25% solution, 60 liters of 40% solution, 20 liters of 50% solution
Work Step by Step
25%, 40%, 50% solutions
Let $x$, $y$, and $z$ be the amount used of the 25%, 40%, and 50% solutions, respectively.
We want 200 liters of a 32% solution. Twice as much of the 25% solution is used as the 40% solution.
$x+y+z=200$
$.25x+.4y+.5z=200*.32$
$.25x+.4y+.5z=64$
$x=2y$
$x+y+z=200$
$x=2y$
$.25x+.4y+.5z=64$
$(.25x+.4y+.5z)*20=64*20$
$5x+8y+10z=1280$
$5x+8y+10z=1280$
$x=2y$
$x+y+z=200$
$5x+8y+10z=1280$
$5x+4*2y+10z=1280$
$5x+4x+10z=1280$
$9x+10z=1280$
$x+y+z=200$
$x+1/2*2*y+z=200$
$x+1/2*(2y)+z=200$
$x+1/2*x+z=200$
$3/2x+z=200$
$3/2x+z-3/2x=200-3/2*x$
$z=200-3/2*x$
$9x+10z=1280$
$9x+10(200-1.5x)=1280$
$9x+2000-15x=1280$
$-6x+2000=1280$
$-6x+2000-1280+6x=1280-1280+6x$
$720=6x$
$720/6=6x/6$
$120=x$
$x=2y$
$120=2y$
$120/2=2y/2$
$60=y$
$x+y+z=200$
$120+60+z=200$
$180+z=200$
$180+z-180=200-180$
$z=20$
