ISRO CS 2025 Q.67: solution and answer

Official answer: CSource: ISRO's final answer key of 15 May 2026

1. FCFS serves processes in order of arrival time, not in the order they are listed. The arrival order is P2 (at $0$), P3 (at $3$), P1 (at $6$).

2. Gantt chart: P2 runs from $0$ to $3$, P3 from $3$ to $6$, P1 from $6$ to $30$. The CPU is never idle, because each process arrives no later than the moment the previous one finishes.

3. Turnaround time $=$ completion time $-$ arrival time. P2: $3-0=3$. P3: $6-3=3$. P1: $30-6=24$.

4. Average turnaround time $=\frac{3+3+24}{3}=\frac{30}{3}=10\text{ms}$.

5. Waiting time $=$ turnaround time $-$ execution time. P2: $3-3=0$. P3: $3-3=0$. P1: $24-24=0$. Every process starts the moment it arrives.

6. Average waiting time $=\frac{0+0+0}{3}=0\text{ms}$.

7. The average turnaround time is $10\text{ms}$ and the average waiting time is $0\text{ms}$, which is option (c). Option (a) quotes $30$, the sum of the turnaround times, instead of their mean, and its waiting time does not match the schedule. Options (b) and (d) also do not match the schedule.