Two liquids $\mathrm{X}$ and  $\mathrm{Y}$ form an ideal solution. The mixture has a vapour pressure of $400 \mathrm{mm}$ at $300 \mathrm{K}$ when mixed in the molar ratio of $1:1$ and a vapour pressure of $350 \mathrm{mm}$ when mixed in the molar ratio of $1:2$ at the same temperature. The vapour pressures of the two pure liquids $\mathrm{X}$ and $\mathrm{Y}$ respectively are

We know that 

The molar ratio of $1:1$ means ${\mathrm{x}}_{\mathrm{X}}={\mathrm{x}}_{\mathrm{Y}}=0.5$

And From Raoult's law,

We know that ${\mathrm{P}}_{\mathrm{X}}={\mathrm{p}}_{\mathrm{x}}°{\mathrm{x}}_{\mathrm{x}} ,{\mathrm{P}}_{\mathrm{Y}}={\mathrm{p}}_{\mathrm{Y}}°{\mathrm{x}}_{\mathrm{Y}}$

Now, From Dalton’s Law,

${\mathrm{P}}_{\text{total }}={\mathrm{x}}_{\mathrm{x}}{\mathrm{p}}_{\mathrm{x}}°+{\mathrm{x}}_{\mathrm{Y}}{\mathrm{p}}_{\mathrm{Y}}°$

or ${\mathrm{p}}_{\mathrm{x}}°+{\mathrm{p}}_{\mathrm{y}}{}^{\circ }=800...$(i)

The molar ratio of $1: 2$ means, ${\mathrm{x}}_{\mathrm{x}}=\frac{1}{3},{\mathrm{x}}_{\mathrm{Y}}=\frac{2}{3}$

Therefore,

$350=\frac{1}{3}{\mathrm{p}}_{\mathrm{x}}{}^{\circ }+\frac{2}{3}{\mathrm{p}}_{\mathrm{Y}}°$

or ${\mathrm{p}}_{\mathrm{x}}°+2\mathrm{p}°=1050 \dots$ (ii)

Eqn (ii) - Eqn (i) gives ${\mathrm{p}}_{\mathrm{Y}}{}^{\circ }=250 \mathrm{mm}$

Putting in (i), we get ${\mathrm{p}}_{\mathrm{x}}{}^{\circ }=550 \mathrm{mm}$

Hence, the correct option is (d).