Chemistry meets math.... not in my wheelhouse

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 90 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?

The chemist should use liters of 25% solution, liters of 40% solution, and liters of 60%

solution.

Answer 1

Answers: 30, 15 and 45 Liters of 25%, 40% and 60%

.

Let F= Liters of 25%, S=Liters of 40%, and T =Liters of 60%

F,S, T for First, Second and Third

25F + .4S +.6T = .45(90)

T=3S 3 times as much 60% as 40%

.25F +.4S +.6(3S) = .45(90)

.25F + 2.2S = 40.5

F+S +T = 90

F+S +3S =90

F +4S = 90

.25F + S = 90/4 = 22.5

1.2S = 40.5-22.5 = 18

S = 18/1.2 = 3/.2= 15 Liters of 40%

T = 3(15 = 45 Liters of 60%

F = 90-45-15 = 30 Liters of 25%

to get 90 Liters of 45%

check the solutions:

.25(30) + .4(15) + .6(45)

=7.5 + 6 + 27

= 40.5

= .45(90)

= 40.5