For a given mixture of $NaHC{O}_3$  and $N{a}_{2}C{O}_3$ , volume of a given $HCl$ required is x ml with phenolphthalein indicator and further y ml required with methyl orange indicator. Hence  volume of$HCl$for complete reaction of original $NaHC{O}_3$  is 

Solution: Using phenolphthalein,
$N{a}_{2}C{O}_3+HCl\to NaHC{O}_3+NaCl$ 
        $1M\left(say\right)$            
$NaHC{O}_3+HCl\to$  No reaction
$\therefore$  Milliequivalents of $N{a}_{2}C{O}_3=$  Milliequivalents of $HCl=x$ 
Using methyl orange,
 $NaHC{O}_3+HCl\to NaCl+C{O}_2+H_{2}O$
Milliequivalents of $NaHC{O}_3$  (from $N{a}_{2}C{O}_3$ ) + Milliequivalents of $NaHC{O}_3$  (original mixture) =$(y-x)$
$\therefore$  Volume of $HCl$required =$(y-x) ml.$