According to the question,

Let x litres of 3% solution is to be added to 460 liters of the 9% of solution

Then, we get,

Total solution = \[\left( 460\text{ }+\text{ }x \right)\text{ }litres\]

Total acid content in resulting solution

\[\begin{array}{*{35}{l}}

=\text{ }\left( 460\text{ }\times \text{ }9/100\text{ }+\text{ }x\text{ }\times \text{ }3/100 \right)  \\

=\text{ }\left( 41.4\text{ }+\text{ }0.03x \right)%  \\

\end{array}\]

According to the question, we have,

Resulting mixture should be more than 5% acidic but less than 7% acidic

So we get,

\[\begin{array}{*{35}{l}}

\Rightarrow ~5\text{ }%\text{ }of\text{ }\left( 460\text{ }+\text{ }x \right)\text{ }<\text{ }41.4\text{ }+\text{ }0.03x\text{ }<\text{ }7%\text{ }of\text{ }\left( 460\text{ }+\text{ }x \right)  \\

\Rightarrow ~5/100\text{ }\times \text{ }\left( 460\text{ }+\text{ }x \right)\text{ }<\text{ }41.4\text{ }+\text{ }0.03x\text{ }<\text{ }7/100\text{ }\times \text{ }\left( 460\text{ }+\text{ }x \right)  \\

\Rightarrow ~23\text{ }+\text{ }0.05\text{ }x\text{ }<\text{ }41.4\text{ }+\text{ }0.03x\text{ }<\text{ }32.2\text{ }+\text{ }0.07x  \\

\end{array}\]

Now, we have

\[\begin{array}{*{35}{l}}

\Rightarrow ~23\text{ }+\text{ }0.05x\text{ }<\text{ }41.4\text{ }+\text{ }0.03x\text{ }and\text{ }41.4\text{ }+\text{ }0.03x\text{ }<\text{ }32.2\text{ }+\text{ }0.07x  \\

i.e.,\text{ }0.02x\text{ }<\text{ }18.4\text{ }and\text{ }0.04x\text{ }>\text{ }9.2  \\

\Rightarrow ~2x\text{ }<\text{ }1840\text{ }and\text{ }4x\text{ }>\text{ }920  \\

\Rightarrow ~x\text{ }<\text{ }920\text{ }and\text{ }x\text{ }>\text{ }230  \\

\Rightarrow ~230\text{ }<\text{ }x\text{ }<\text{ }920  \\

\end{array}\]

Hence, solution between 230 l and 920 l should be added.