$A$ solution of $9 \%$ acid is to be diluted by adding $3 \%$ acid solution to it. The resulting mixture is to be more than $5 \%$ but less than $7 \%$ acid. If there is $460 \text{ L}$ of the $9 \%$ solution,how many litres of $3 \%$ solution will have to be added?

(N/A) Let $x \text{ L}$ of $3 \%$ solution be added to $460 \text{ L}$ of $9 \%$ solution of acid.
Total quantity of mixture $= (460 + x) \text{ L}$.
Total acid content in the $(460 + x) \text{ L}$ of mixture $= 460 \times \frac{9}{100} + x \times \frac{3}{100}$.
It is given that acid content in the resulting mixture must be more than $5 \%$ but less than $7 \%$ acid.
$\therefore 5 \% \text{ of } (460 + x) < \left( 460 \times \frac{9}{100} + \frac{3x}{100} \right) < 7 \% \text{ of } (460 + x)$.
$\therefore (460 + x) \frac{5}{100} < \frac{4140 + 3x}{100} < (460 + x) \frac{7}{100}$.
$\therefore 5(460 + x) < 4140 + 3x < 7(460 + x)$.
$\therefore 2300 + 5x < 4140 + 3x < 3220 + 7x$.
Now,solving $2300 + 5x < 4140 + 3x$:
$2x < 1840 \implies x < 920 \quad \dots(i)$.
Solving $4140 + 3x < 3220 + 7x$:
$3x - 7x < 3220 - 4140 \implies -4x < -920 \implies 4x > 920 \implies x > 230 \quad \dots(ii)$.
From $(i)$ and $(ii)$,the number of litres of the $3 \%$ solution must be more than $230 \text{ L}$ and less than $920 \text{ L}$.