The plot given below shows P - T curves (where, P is the pressure...

From the graph:
For the solvent $X$, 

$∆ {\text{T}}_{bx} = 2$
$∆ {\text{T}}_{bx} = {m}_{NaCl} \times {K}_{b\left(x\right)} . . . \left(i\right)$
For the solvent $Y$,   

$∆ {\text{T}}_{by} = 1$
$∆ {\text{T}}_{b\left(y\right)} = {m}_{NaCl} \times {K}_{b\left(y\right)} . . . \left(ii\right)$
From the equations ($i$) and ($ii$),
$⇒ \dfrac{{K}_{b\left(x\right)}}{{K}_{b\left(y\right)}} = 2$
For the solute $S$,

$i = \left(1 - \dfrac{\text{\alpha}}{2}\right)$
$∆ {\text{T}}_{b\left(x\right)\left(s\right)} = \left(1 - \dfrac{{\text{\alpha}}_{1}}{2}\right){K}_{b\left(x\right)}$
$∆ {\text{T}}_{b\left(y\right)\left(s\right)} = \left(1 - \dfrac{{\text{\alpha}}_{2}}{2}\right){K}_{b\left(y\right)}$
Given: 

$∆ {\text{T}}_{b\left(x\right)\left(s\right)} = 3 ∆ {\text{T}}_{b\left(y\right)\left(s\right)}$
\left(1 - \dfrac{{\text{\alpha}}_{1}}{2}\right){K}_{b\left(x\right)} = 3 \times \left(1 - \dfrac{{\text{\alpha}}_{1}}{2}\right) \times {k}_{b\left(y\right)}$<br >$2\left(1 - \dfrac{{\text{\alpha}}_{1}}{2}\right) = 3\left(1 - \dfrac{{\text{\alpha}}_{2}}{2}\right)$<br >${\text{\alpha}}_{2} = 0 .7$<br >So,$\text{\alpha} = 0 .05