For precipitation to take place, it is required that the calculated ionic product exceeds the $K_{s p}$ value.

Before mixing:

$\left[ S ^{2-}\right]=1.0 \times 10^{-19} \,M \,\,\,\,\left[ M ^{2+}\right]=0.04 \,M$

volume $=10\, mL \quad \quad \quad \,\,$ volume $=5 \,mL$

After mixing:

$\left[ S ^{2-}\right]=?$       volume $=(10+5)=15\, mL$

$\left[ M ^{2+}\right]=?$        volume $=15 \,mL$

$\left[ S ^{2-}\right]=\frac{1.0 \times 10^{-19} \times 10}{15}=6.67 \times 10^{-20}\, M$

$\left[ M ^{2+}\right]=\frac{0.04 \times 5}{15}=1.33 \times 10^{-2} \,M$

Ionic product $=\left[ M ^{2+}\right]\left[ S ^{2-}\right]$

$=\left(1.33 \times 10^{-2}\right)\left(6.67 \times 10^{-20}\right)$

$=8.87 \times 10^{-22}$

This ionic product exceeds the $K_{s p}$ of $Z n s$ and $C d S$ Therefore, precipitation will occur in $CdCl _{2}$ and $ZnCl _{2}$ solutions.