Question

A chemist has three different acid solutions. The first acid solution contains
25%
acid, the second contains
40%
and the third contains
60%
. He wants to use all three solutions to obtain a mixture of
60
liters containing
45%
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

Let t, f, and s be the amount of 25%, 40%, and 60% solutions used.

We are told s=3f.  The general equation is:

100(0.25t+0.4f+0.6s)/(t+s+f)=45, and using s=3f  we have:

(0.25t+0.4f+1.8f)/(t+4f)=0.45

(0.25t+2.2f)/(t+4f)=0.45

0.25t+2.2f=0.45t+1.8f  

0.25t+0.4f=0.45t

0.4f=0.2t

t=2f, remembering that s=3f, will allow us to solve for f

t+s+f=60, using t and s from above

2f+3f+f=60

6f=60

f=10 L then

s=3f=30 L  and t=2f=20L

So 20L of 25%, 10L of 40%, and 30L of 60% acid solutions need to be mixed to make 60L of 45% acid solution.