A $20kg$ kid is sitting in a $60kg$  boat in a lake. The distance of kid from the bank of lake is $20m$. If the kid moves $8m$ on the boat towards the bank, then the distance of kid from the bank is in meter (The system is initially at rest and there is no friction between boat and water)

Let c.m. of boat is at distance  $x_0$ from bank
C.m. of(boat + kid) (before)
 $x_{c.m.}=\frac{20\times 20+60x_0}{20+60}------\left(1\right)$
When the kid moves towards the bank, by $8m,$ the boat moves away from the bank by x, because here ${\overrightarrow{F}}_{ext}=0.$  Initially ${\overrightarrow{V}}_{c.m.}=0$ and hence location of c.m. is fixed.
c.m. of (boat+kid) (after)
  $x_{c.m.}=\frac{20(20-8+x)+60(x_0+x)}{20+60}$
As the location of c.m. is fixed,
 $\begin{array}{l}{x}_{c.m.}=x_{c.m.}^{\text{'}}\\ x=2m\end{array}$
Distance of kid from the bank = $20-8+x=14m$