29.2% (w/W) HCl stock solution has density of 1.25 g  $m{L}^{-1}$.  The molecular weight of HCl is 36.5 $mo{l}^{-1}$ .  The volume (mL) of stock solution required to prepare a 200 mL solution 0.4 M HCl is  

Mass of HCl in 1.0 mL stock solution
$=1.25\times \frac{29.2}{100}=0.365g$
Mass of HCl required for 200 mL solution of 0.4 M HCl
$=\frac{200}{1000}\times 0.4\times 36.5=0.08\times 36.5g$
$\because$ 0.365 g of HCl is present in 1.0 mL stock solution.
  $\therefore$ 0.08 x 36.5g HCl will be present in 
$\frac{0.08\times 36.5}{0.365}=8.0mL$