Let $x L$ of $3 \%$ solution be added to $460 L$ of $9 \%$ solution of acid.
Total quantity of mixture $=(460+x) L$
Total acid content in the $(460+x) L$ of mixture
$$
=\left(460 \times \frac{9}{100}+x \times \frac{3}{100}\right)
$$
According to question
$5 \%$ of $(460+x) < 460 \times \frac{9}{100}+\frac{3 x}{100} < 7 \%$ of $(460+x)$
$\Rightarrow \frac{5}{100} \times(460+x) < 460 \times \frac{9}{100}+\frac{3}{100} x$
$ < \frac{7}{100} \times(460+x)$
$$
\begin{aligned}
&\Rightarrow 5 \times(460+x) < 460 \times 9+3 x < 7 \times(460+x) \\
&\Rightarrow 2300+5 x < 4140+3 x < 3220+7 x \\
&\text { Consider } 2300+5 x < 4140+3 x \\
&\Rightarrow 2 x < 1840 \Rightarrow x < 920 \\
&\text { Now consider, } 4140+3 x < 3220+7 x \\
&\Rightarrow-4 x < -920 \Rightarrow 4 x>920 \\
&\Rightarrow x>230
\end{aligned}
$$
Hence, the number of litress of the $3 \%$ solution of acid must be more than $230 L$ and less than $920 L$.