Step 1:

Let x litres of $2 \%$ boric acid required to be added to 640 litres of $8 \%$ solution.

Then total mixture $=(x+640)$ litres

Percentage of boric acid in the mixture is $2 \% \;of \;x +8 \% of \;640 > 4 \% \;of \; (x+640)$

$=> \large\frac{2}{100} $$x +\large\frac{8}{100} $$ \times 640 > \large\frac{4}{100} $$(x+640)$

Multiplying birth sides by 100

$=> 2x +5120 > 4x+2560$

Subtracting 2560 from both sides

$=> 2x +5120 -2560 > 4x +2560 -2560$

$=> 2x+2560> 4x$

Subtracting $2x$ from both sides

$2560 > 2x$

Dividing by 2 on both sides $1280 > x$ -------(1)

The percentage of boric acid must be less than $ 6 \%$

Hence $2 \% \;of \;x +8 \% of \;640 < 6 \% \;of \; (x+640)$

$=> \large\frac{2}{100} $$x +\large\frac{8}{100} $$ \times 640 < \large\frac{6}{100} $$(x+640)$

Multiplying both sides by 100

$=> 2x +5120 < 6x+3840$

Subtracting 2x from both sides inequality and subtracting 2840 from both sides of inequality

$ 5120 -3840 < 6x -2x$

$=> 1280 < 4x$

$=> 320 < x $ ------(2)

From (1) and (2) the we conculde that $320 < x < 1280$

Thus the number of litres of $2 \%$ bonic acid solution to be added in more than 320 and less than 1280