$A$ manufacturer has $600 \text{ litres}$ of a $12 \%$ solution of acid. How many litres of a $30 \%$ acid solution must be added to it so that the acid content in the resulting mixture will be more than $15 \%$ but less than $18 \%$?

Let $x$ litres of $30 \%$ acid solution be added to the mixture.
The total volume of the resulting mixture is $(x + 600) \text{ litres}$.
The amount of acid in the mixture is $0.30x + 0.12(600)$.
According to the problem,the acid content must be between $15 \%$ and $18 \%$:
$0.15(x + 600) < 0.30x + 0.12(600) < 0.18(x + 600)$
Solving the first inequality:
$0.15x + 90 < 0.30x + 72$
$18 < 0.15x$
$x > \frac{18}{0.15} = 120$
Solving the second inequality:
$0.30x + 72 < 0.18x + 108$
$0.12x < 36$
$x < \frac{36}{0.12} = 300$
Thus,the number of litres of the $30 \%$ solution must be greater than $120 \text{ litres}$ and less than $300 \text{ litres}$.