$20\text{mL}$ of $0.1M$ solution of compound $N{a}_{2}C{O}_3\cdot NaHC{O}_3\cdot 2H_{2}O$ is titrated against $0.05MHCl$, $x\text{mL}$ of $HCl$ is used when phenolphthalein is used as an indicator and $y\text{mL}$ of $HCl$ is used when methyl orange is the indicator in two separate titrations. Hence $(y-x)$ is:

The correct option is B $80\text{mL}$
When phenolphthalein is used as an indicator:
Only one part of base reacts
By using formula,
$meq.\text{of acid used}=meq.\text{base}$
$V\times M\times n=V\times M\times n$
Where,
$M=MolarityV=Volumen=n-factor$
$0.05x\times 1=20\times 0.1\times 1;x=40\text{mL}$
When methyl orange is used as an indicator:
All parts of base reacts
By using formula,
$meq.\text{of acid used}=meq.\text{base}$
$0.05y\times 1=(20\times 0.1\times 1)+(20\times 0.1\times 2)$
$y=120\text{mL}$
$\therefore y-x=80\text{mL}$