How many litres of water will have to be added to $1125$ litres of the $45 \%$ solution of acid so that the resulting mixture will contain more than $25 \%$ but less than $30 \%$ acid content?

(N/A) Let $x$ litres of water be added to the solution.
Total volume of the mixture becomes $(1125 + x)$ litres.
The amount of acid remains constant at $45 \%$ of $1125$ litres,which is $0.45 \times 1125 = 506.25$ litres.
According to the problem,the acid content in the new mixture must be between $25 \%$ and $30 \%$.
So,$25 \% < \frac{506.25}{1125 + x} \times 100 < 30 \%$.
First,consider the inequality $\frac{506.25}{1125 + x} < 0.30$:
$506.25 < 0.30(1125 + x)$
$506.25 < 337.5 + 0.3x$
$168.75 < 0.3x$
$x > \frac{168.75}{0.3} = 562.5$.
Next,consider the inequality $\frac{506.25}{1125 + x} > 0.25$:
$506.25 > 0.25(1125 + x)$
$506.25 > 281.25 + 0.25x$
$225 > 0.25x$
$x < \frac{225}{0.25} = 900$.
Therefore,the amount of water to be added must be between $562.5$ litres and $900$ litres,i.e.,$562.5 < x < 900$.