$A$ container is filled with $1120 \text{ litres}$ of a solution containing $40 \%$ acid. How many litres of acid must be added so that the resulting mixture contains more than $40 \%$ but less than $50 \%$ water?

Let $x$ be the amount of acid added in litres.
The initial amount of acid is $40 \% \text{ of } 1120 = 0.40 \times 1120 = 448 \text{ litres}$.
The initial amount of water is $1120 - 448 = 672 \text{ litres}$.
After adding $x$ litres of acid,the total volume becomes $(1120 + x) \text{ litres}$.
The amount of water remains $672 \text{ litres}$.
The percentage of water in the new mixture is $\frac{672}{1120 + x} \times 100$.
We are given that $40 < \frac{672}{1120 + x} \times 100 < 50$.
Dividing by $100$: $0.4 < \frac{672}{1120 + x} < 0.5$.
Taking the first inequality: $0.4 < \frac{672}{1120 + x} \implies 1120 + x < \frac{672}{0.4} = 1680 \implies x < 560$.
Taking the second inequality: $\frac{672}{1120 + x} < 0.5 \implies 1120 + x > \frac{672}{0.5} = 1344 \implies x > 224$.
Thus,the amount of acid to be added is between $224 \text{ litres}$ and $560 \text{ litres}$.