Concentration

Problem:



Solution:

Let the volume of the pure acid be $ x $ in liter.
Let the volume of the 20% acid be $ y $ in liter.

We have two requirements:

$ x + y = 44 $, we ultimately need 44 liters.
$ x + 0.2y = 44 \times 0.4 $, The acid content should be 40%

We simply subtract the second equation from the first and get

$ (x + y) - (x + 0.2y) = 44 - 44 \times 0.4 $, simplifying, get $ y = 33 $.
Now we have $ x + y = 44 $, so we get $ x = 11 $.

As a sanity check, we compute $ 11 + 33 \times 0.2 = 17.6 = 44 \times 0.4 $, so the answer is correct.