Explanation

(b)
Let the volume of 0.4 M HCl is $V_{1}$ and that of 0.9 M HCl is $V_{2.}$
We know that,
$\ \ \ \ \ \ \ \ \ \ \ NV\ = \ \ {\ {\ \ \ \ \ \ \ \ \ N}_{1}V}_{1}\ \ \ \ \ \ \ \ + \ \ {\ \ \ N}_{2}\ V_{2}$
(Mixture) (for 0.4 M HCl) ( for 0.9 M HCl)
$\ \ \ \ \ \ \ \ \ 0.7(V_{1} + \ V_{2})\ = \ \ 0.4\ \times \ V_{1} + 0.9 \times \ \ V_{2}$
[$\because$1m HCl = 1N HCl]
$\ \ \ \ 0.7V_{1} + \ {0.7V}_{2}\ = \ \ 0.4\ \ V_{1} + 0.9\ \ V_{2}$
$\ \ \ \ 0.7V_{1} + \ {0.4V}_{1}\ = \ \ 0.9V_{2} + 0.7V_{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.3V_{1} = 0.2V_{2}\$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{V_{1}}{V_{2}} = \frac{0.2}{0.3} = \frac{2}{3}$