How many liters of each solution should be used?

Question

A chemist has three different acid solutions. The first acid solution contains 25% acid, the second contains 40% and the third contains 85%. They want to use all three solutions to obtain a mixture of 108 liters containing 50% acid, using 2 times as much of the 85% solution as the 40% solution. How many liters of each solution should be used?

Answer 1

.25x+.4y+.85z =.5(108)

z=2y

x+y+z=108

use substitution elimination or matrix algebra to

find x= liters of 25%

y= liters of 40%

z=liters of 85%

x+3y=108

.25(108-3y)+.4y+.85(2y)=.5(108)

(-.75+.4+1.7)y=(.5-.25)(108

1.35y=27

y=28/1.35=about 20.741 liters of 40%

z=41.482 liters of 85%

x=108-3(20.741)=108-62.223

= 45.777 liters of 25%

roughly (x,y,z)=(45.8,20.7,41.5)

45.8+20.7+41.5=108