Let $j= 4x, k= x$
When $10$ litre is removed, $j$ become $4x- \frac{4}{5} * 10$ = $4x-8$
When $10$ litre is removed, $k$ become $x-\frac{1}{5}*10$ = $x-2$
$10$ litre of $k$ is added, $k$ become= $x-2+10= x+8$
Final ratio of $j$ and $k$ is $\frac{2}{3}$= $\frac{4x-8}{x+8}$=$\frac{2}{3}$
On solving $x= 4$, Initial amount of $j$ is $4x= 4*4 = 16$ litre