Answer: (A)
Each process runs for q period and if there are n process: $p_{1}$, $p_{2}$,$p_{3}$,, ....., $p_{n}$,.
Then $p_1$'s turn comes again when it has completed time quanta for remaining process p2 to pn, i.e, it would take at most $(n-1)q$ time.
So,, each process in round robin gets its turn after $(n-1)q$ time when we don't consider overheads but if we consider overheads then it would be $ns + (n-1)q$
So, we have $ns + (n-1)q \leq t$