If the total vapour pressure of the liquid mixture A and B is given by the equation : ${\text{P=180\hspace{0.17em}X}}_{\mathrm{A}}+90$ then the ratio of the vapour pressure of the pure liquids A and B is given by: 

As we know 

$\begin{array}{l}\mathrm{PT}={\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}{\mathrm{X}}_{\mathrm{A}}+{\mathrm{P}}_{\mathrm{B}}^{\mathrm{o}}{\mathrm{X}}_{\mathrm{B}}\\ ={\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}{\mathrm{X}}_{\mathrm{A}}+{\mathrm{P}}_{\mathrm{B}}^{\mathrm{o}}(1-{\mathrm{X}}_{\mathrm{A}})=({\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}-{\mathrm{P}}_{\mathrm{B}}^{\mathrm{o}}){\mathrm{X}}_{\mathrm{A}}+{\mathrm{P}}_{\mathrm{B}}^{\mathrm{o}}.....\left(1\right)\end{array}$

But from question, ${\mathrm{P}}_{\mathrm{T}}=180{\mathrm{X}}_{\mathrm{A}}+90.....\left(2\right)$

Equating equation (1) and (2) 

$\begin{array}{l}\therefore \mathrm{\hspace{0.17em}}{\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}-{\mathrm{P}}_{\mathrm{B}}^{\mathrm{o}}=180\\ {\mathrm{P}}_{\mathrm{B}}^{\mathrm{o}}=90;\mathrm{\hspace{0.17em}}{\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}=180+90=270\\ \frac{{\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}}{{\mathrm{P}}_{\mathrm{B}}^{\mathrm{o}}}=\frac{270}{90}=3:1\end{array}$