Given two mixtures:
I)  $NaOH+N{a}_{2}C{O}_3$
II)  $NaHC{O}_3+N{a}_{2}C{O}_3$
$100ml$  of mixture I required ‘W’ and ‘X’ ml of  $1MHCl$  in separate titrations using phenolphthalein and Methyl orange indicators. While $100ml$ of mixture II required ‘Y’ and ‘Z’ ml of same  $HCl$  solution in separate titration using same indicators.

Column I (Substance)		Column II (Molarity in solution)
(A)	$N{a}_{2}C{O}_3$ in mixture I	(P)	$(2w-x)\times {10}^{-2}$
(B)	$N{a}_{2}C{O}_3$ in mixture II	(Q)	$(z-2y)\times {10}^{-2}$
(C)	$NaOH$ in mixture I	(R)	$y\times {10}^{-2}$
(D)	$NaHC{O}_3$ in mixture II	(S)	$(x-w)\times {10}^{-2}$

Mixture I:
End point with phenolphthalein (disappearance of pink colour) corresponds to the neutralisation of  $NaOH$ and half- neutralisation of  $N{a}_{2}C{O}_3$.
$$\begin{gathered} NaOH+HCl\to NaCl+H_{2}O \\ N{a}_{2}C{O}_3+HCl\to NaHC{O}_3+NaCl \end{gathered}$$
End point with Methyl orange (Appearance of  Red colour) corresponds to the neutralisation of  $NaOH$ and  $N{a}_{2}C{O}_3$.
$NaOH+HCl\to NaCl+H_{2}O$
$N{a}_{2}C{O}_3+2HCl\to 2NaCl+C{O}_2+H_{2}O$
Volume of  $HCl$ required for neutralisation of $N{a}_{2}C{O}_3=2(x-w)$
 Normality of $N{a}_{2}C{O}_3=\frac{1\times 2(x-w)}{100}$
   $2(x-w)\times {10}^{-2}$
Molarity of $N{a}_{2}C{O}_3=(x-w)\times {10}^{-2}$
Volume of  $HCl$  req. for neutralisation  of $NaOH=w-(x-w)=(2w-x)ml$   
Hence, molarity of  $NaOH=\frac{(2w-x)\times 1}{100}=(2w-x)\times {10}^{-2}M$
Mixture II :

 End  point with phenolphthalein corresponds to half- neutralisation  of $N{a}_{2}C{O}_3$ as $N{a}_{2}C{O}_3+HCl\to NaHC{O}_3+NaCl$
 Volume of  $HCl$ req. for complete neutralisation of     $N{a}_{2}C{O}_3=‘2y’ ml$
  $\therefore MolarityofN{a}_{2}C{O}_3=\frac{1}{2}\times \frac{2y}{100}=y\times {10}^{-2}$   
 End point with Methyl orange corresponds to neutralisation of  $NaHC{O}_3$   
 Hence, volume required for neutralisation of  $NaHC{O}_3$  present initially $=(z-2y)ml$
$\therefore$ Molarity of  $NaHC{O}_3=\frac{(z-2y)}{100}=(z-2y)\times {10}^{-2}$