At ${40}^{0}C$ , the vapour pressure in torr of methanol and ethanol solution is $P=119x+135$  where x is the mole fraction of methanol. Hence, correct statement is/are

Total vapor pressure of the methanol and ethanol mixture:

$\mathrm{P}=119\mathrm{x}+135$...(1)

Expression is: 

$\mathrm{P}={\mathrm{P}}_{\mathrm{A}}^0{\mathrm{x}}_{\mathrm{A}}+{\mathrm{P}}_{\mathrm{B}}^0(1-{\mathrm{x}}_{\mathrm{A}})$

$$\begin{gathered} {\mathrm{P}}_{\mathrm{A}}^{°}=\mathrm{vapour} \mathrm{pressure} \mathrm{of} \mathrm{pure} \mathrm{methanol} \\ {\mathrm{x}}_{\mathrm{A}}=\mathrm{mole} \mathrm{fraction} \mathrm{of} \mathrm{methanol} \\ {\mathrm{P}}_{\mathrm{B}}^{°}=\mathrm{vapour} \mathrm{pressure} \mathrm{of} \mathrm{pure} \mathrm{ethanol} \end{gathered}$$

$\mathrm{P}={\mathrm{P}}_{\mathrm{A}}^0{\mathrm{x}}_{\mathrm{A}}+{\mathrm{P}}_{\mathrm{B}}^0-{\mathrm{P}}_{\mathrm{B}}^0{\mathrm{x}}_{\mathrm{A}}$

$\mathrm{P}={\mathrm{P}}_{\mathrm{B}}^0+({\mathrm{P}}_{\mathrm{A}}^0-{\mathrm{P}}_{\mathrm{B}}^0){\mathrm{x}}_{\mathrm{A}}$...(2)

Equating (1) and (2):

${\mathrm{P}}_{\mathrm{B}}^0=135;{\mathrm{P}}_{\mathrm{A}}^0-{\mathrm{P}}_{\mathrm{B}}^0=119$