A chemist has three different acid solutions

Question

A chemist has three different acid solutions. The first acid solution contains 25% acid, the second contains 40% and the third contains 60%. They want to use all three solutions to obtain a mixture of 240 liters containing 35% acid, using 3 times as much of the 60% solution as the 40% solution. How many liters of each solution should be used?

Answer 1

Hi there,

Let x= volume of 25% acid, y=volume of 40% acid and z=volume of 60% acid

Here's the setup:

0.25x+0.4y+0.6z=0.35(240)

x+y+z=240

z=3y

Three equations three unknowns, solve by substitution (preferably) or any other method.

I hope this helps!