Solving problems involving mixture

Question

How many liters of a 50% acid solution must be mixed with a 30% acid solution to get 180 L of a 40% acid solution? 

Answer 1

Let A be the liters of the 50% solution.  180 - A is the volume of the 30% solution.

 

.5A + .3(180 - A) = .4(180)

 

Multiply by ten to clear the fractions.

 

5A + 3(180 - A) = 4(180)

 

Simplify.

 

5A + 540 - 3A = 720

 

2A + 540 = 720

 

2A = 180

 

A = 90

 

Ninety liters of the 50% acid solution have to be mixed with 90 liters of the 30% solution to yield 180 liters of 40% solution.