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.