Precalculas problem

Question

You run a laboratory experiment and you need a mixture of 100 liters containing 45% acid to conduct your project. You don’t have that mixture but you have three different acid solutions. The first acid solution contains 10% acid, the second contains 40% acid and third 60% acid. The supply of 40% acid is low so twice as much 10% solution as 40% solution must be used. How many liters of each solution should you use?

Answer 1

Let F = 40% solution amount ("F" for forty)

Let T =10% solution amount ("T for ten)

Let S =60% solution amount

F+T+S=100

.4F + .1T + .6S = .45(100)

T=2F

That's 3 equations 3 unknowns, so you should get a unique solution

F+2F + S = 100

4F + 2F + 6S = 450

6F + 6S = 450

F+S = 4500/6 = 75

T = 100-75=25 liters

T=2F

F=1/2 of T = 12.5 liters

S = 75-F = 75 - 12.5 = 62.5 liters

T=25, F=12.5, S=62.5 That adds to 100 liters

(1/10)T = 2.5

(4/10)F = 5

(6/10)S = 37.5

2.5+5+37.5 = .45 = 45% solution