According to the Faraday's law of electrolysis,

$\mathrm{W}=\mathrm{ZQ}=\frac{\mathrm{E}}{96500}\mathrm{Q}$

For the compound $\mathrm{A}$, 

Equivalent mass of $\mathrm{A}$ = $\frac{\mathrm{Atomic} \mathrm{mass}}{\mathrm{Valency}}=\frac{8}{\mathrm{x}}$

$\mathrm{x}$ is the valency factor for compound $\mathrm{A}$.

So, $$\begin{gathered} \mathrm{W}=\frac{\mathrm{E}}{96500}\mathrm{Q}=\frac{{\displaystyle \frac{8}{\mathrm{x}}}}{96500}\mathrm{Q} \\ \mathrm{x}=\frac{{\displaystyle 7}}{2.4\times 96500}\mathrm{Q} \end{gathered}$$

For compound $\mathrm{B}$,

Equivalent mass of $\mathrm{B}$=$\frac{\mathrm{Atomic} \mathrm{mass}}{\mathrm{Valency}}=\frac{18}{\mathrm{y}}$

$\mathrm{y}$ is the valency factor for compound $\mathrm{B}$.

So, $$\begin{gathered} \mathrm{W}=\frac{\mathrm{E}}{96500}\mathrm{Q}=\frac{{\displaystyle \frac{18}{\mathrm{y}}}}{96500}\mathrm{Q} \\ \mathrm{y}=\frac{{\displaystyle 27}}{1.8\times 96500}\mathrm{Q} \end{gathered}$$

For compound $\mathrm{C}$,

Equivalent mass of $\mathrm{C}$ = $\frac{\mathrm{Atomic} \mathrm{mass}}{\mathrm{Valency}}=\frac{50}{\mathrm{z}}$

$\mathrm{z}$ is the valency factor for compound $\mathrm{C}$.

So, $$\begin{gathered} \mathrm{W}=\frac{\mathrm{E}}{96500}\mathrm{Q}=\frac{{\displaystyle \frac{50}{\mathrm{z}}}}{96500}\mathrm{Q} \\ \mathrm{z}=\frac{{\displaystyle 50}}{7.5\times 96500}\mathrm{Q} \end{gathered}$$

Hence, $$\begin{gathered} \mathrm{x}:\mathrm{y}:\mathrm{z}=\raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$2.4$}\right. : \raisebox{1ex}{$18$}\!\left/ \!\raisebox{-1ex}{$1.8$}\right. : \raisebox{1ex}{$50$}\!\left/ \!\raisebox{-1ex}{$7.5$}\right.=3.33 : 10 : 6.66 \\ \mathrm{x}:\mathrm{y}:\mathrm{z} = 1:3:2 \end{gathered}$$

Hence, the valencies of $\mathrm{A},\mathrm{B}$ and $\mathrm{C}$ respectively are $1,3,2$.