Prepare for TCS, round by round.

The idea. A percentage is a ratio out of 100. Most TCS-style problems reduce to one of three moves: convert the percentage to a multiplier, split a total in a given ratio, or compare "before" and "after".

Worked example. A price rises by 20% and then falls by 20%. Is it back where it started?

Multiply, do not add. Rising 20% is x 1.2 and falling 20% is x 0.8, so the net factor is 1.2 x 0.8 = 0.96. The price ends 4% lower, not unchanged. Adding and subtracting percentages directly is the most common mistake here.

Practice (try each one before reading the answer).

1. A class of 40 has boys and girls in the ratio 3 : 2. How many boys?
2. 35% of a number is 140. What is the number?
3. A town of 20,000 grows by 10% in year one and 10% in year two. What is the population after two years?
4. Split 96 in the ratio 5 : 7.
5. A shirt is marked down from 800 to 680. What is the percentage discount?

Answers.

1. 3 + 2 = 5 parts, so each part is 40 / 5 = 8. Boys = 3 x 8 = 24.
2. Number = 140 / 0.35 = 400.
3. 20,000 x 1.1 x 1.1 = 24,200. Growth compounds, so it is not 20% of 20,000.
4. 5 + 7 = 12 parts, so one part is 8. The shares are 40 and 56.
5. Discount = 120, and 120 / 800 = 15%.

The idea. Find the rule that turns one term into the next. Check these in order: constant difference, differences that themselves follow a pattern, multiply-and-add, squares or factorials, and alternating series.

Worked example. 4, 9, 19, 39, ? Each term is the previous one x 2 + 1: 4 x 2 + 1 = 9, 9 x 2 + 1 = 19, 19 x 2 + 1 = 39, so the next is 79.

Practice.

1. 3, 7, 15, 31, 63, ?
2. 2, 6, 12, 20, 30, ?
3. 2, 3, 5, 9, 17, ?
4. 1, 2, 6, 24, 120, ?
5. 100, 96, 88, 72, 40, ?

Answers.

1. Previous x 2 + 1, so 127.
2. Terms are n x (n + 1): 1x2, 2x3, 3x4, 4x5, 5x6, so the next is 6x7 = 42.
3. Differences are 1, 2, 4, 8, so the next difference is 16 and the term is 33.
4. Factorials: 1!, 2!, 3!, 4!, 5!, so the next is 6! = 720.
5. The differences are 4, 8, 16, 32, each double the one before. The next difference is 64, so the term is 40 - 64 = -24. Series can go negative.

What to know.

• Process and thread: a process has its own memory, threads of one process share it (see the process vs thread answer below).
• CPU scheduling: the OS decides which ready process runs next. FCFS runs in arrival order, SJF runs the shortest job first, and Round Robin gives each process a fixed time slice in turn.
• Deadlock: processes wait on each other forever. It needs four conditions together: mutual exclusion, hold and wait, no preemption and circular wait. Break any one and deadlock cannot happen.
• Memory: paging splits memory into fixed-size pages so a program need not sit in one continuous block. Virtual memory lets a program use more memory than is physically free by keeping part of it on disk.

Example (we calculated this). Three jobs arrive together with CPU bursts of 6, 8 and 3.

• FCFS in that order: waiting times 0, 6 and 14, so the average wait is 6.67.
• SJF (3, 6, 8): waiting times 0, 3 and 9, so the average wait is 4.0. Running short jobs first lowers the average wait.
• Round Robin with a time slice of 2 on two jobs of 5 and 3: the 3-unit job finishes at time 7 and the 5-unit job at time 8.

Commonly confused. Starvation is one process never getting the CPU; deadlock is processes blocking each other. Candidates report process, thread and deadlock questions in TCS technical rounds, but TCS does not publish this.

Practice questions.

1. What are the states of a process? New, Ready (waiting for the CPU), Running, Waiting or Blocked (waiting for I/O or an event) and Terminated. A running process moves to Waiting when it asks for I/O, and back to Ready when the I/O finishes.
2. What is a context switch, and why is it costly? Saving the state of the running process and loading the state of the next one. No useful work happens during the switch, and the CPU cache also tends to go cold, so too many switches waste time.
3. Compare FCFS, SJF and Round Robin on bursts 5, 3, 8, 6 (all arrive at time 0). Average waiting time: FCFS 7.25 (waits [0, 5, 8, 16]), SJF 6.25 (waits [3, 0, 14, 8]), Round Robin with a quantum of 3 10 (waits [9, 3, 14, 14]). SJF gives the lowest average wait, but it can starve long jobs and needs the burst time in advance.
4. What is the difference between a mutex and a semaphore? A mutex is a lock owned by the thread that locks it, and only that thread unlocks it, so it protects one resource. A semaphore is a counter that lets up to N threads in, and any thread can signal it, so it also coordinates work between threads.
5. What is a race condition and a critical section? A critical section is code that touches shared data. A race condition is when the result depends on the unlucky timing of two threads in it, for example two threads both reading a balance of 100 and both writing back 110. Locks make the critical section run one thread at a time.
6. How does paging turn an address into a page and an offset? Divide the logical address by the page size. With a 1 KB (1024 byte) page, address 5000 is page 4, offset 904 (5000 = 4 x 1024 + 904).
7. Compare FIFO, LRU and Optimal page replacement on a reference string. We simulated the reference string 7 0 1 2 0 3 0 4 2 3 0 3 2 1 2 0 1 7 0 1 with 3 frames: FIFO gave 15 page faults, LRU 12 and Optimal 9. Optimal is the lowest possible but needs the future, so it is a yardstick, not something an OS can use. Note: FIFO can even get worse with more frames (Belady's anomaly).
8. What is internal and external fragmentation? Internal fragmentation is unused space inside an allocated block. A 10 KB process in 4 KB pages needs 3 pages and wastes 2 KB in the last one. External fragmentation is free memory scattered in pieces too small to use, even though the total is enough.
9. What are user mode, kernel mode and a system call? Kernel mode can run privileged instructions and touch hardware. User mode cannot. A system call (such as reading a file) is how a user program asks the kernel to do something privileged on its behalf.
10. What is thrashing? When the system spends more time swapping pages in and out than running programs, because too many processes compete for too little memory. The fix is fewer running processes, more memory, or better page replacement.

Answer. A process is a running program with its own memory space. A thread is a path of execution inside a process, and all threads of one process share that process's memory.

Why it is asked. It checks whether you understand that sharing memory makes threads cheap to create and fast to communicate, but also means two threads can corrupt shared data unless access is controlled.

Example. A text editor is one process. Its UI, autosave and spell-check can run as separate threads that all read the same document in memory. Two separate editor windows opened as two processes do not share that document.

Answer. A primary key identifies each row of a table and cannot be NULL, and a table has only one. A unique key also prevents duplicate values, but a table can have several, and it can usually hold NULL.

Why it is asked. It checks whether you know how a database enforces uniqueness and identity.

Example. In a users table, id is the primary key (every user has exactly one id). email can be a unique key: no two users share an email. We ran this in SQLite: inserting a second row with the same email failed with UNIQUE constraint failed: users.email, and inserting two rows with a NULL email succeeded. How many NULLs a unique key allows differs between databases, so check the one you use.

How to answer. Keep it to about a minute in three parts: who you are now (degree, college, main skill), what you have done (one or two projects or experiences), and why you want this role.

Example. "I am a final-year B.Tech student in computer science. I built a lab-booking web app in Python and SQL and practise coding every day. I want to start my career as a software developer where I can learn from large real projects."

Avoid. Reading your resume line by line, or starting with your family and hobbies.

TCS says both sections are part of the integrated test and that the Advanced section is mandatory if you want a Digital or Prime offer. Your test performance decides whether you are interviewed for Prime, Digital or Ninja.

190 minutes in total: 75 minutes for the Foundation part and 115 minutes for the Advanced part. It is taken in person at a TCS iON centre, not from home.

TCS's official page does not say. Different prep sites say different things, so rely on the instructions shown on your test screen rather than on a website.

For the 2026 drive, TCS says no pending backlog is permitted at the time you appear for the selection process. Check the notice for the drive you are applying to.

No. Hirenix is independent and not affiliated with TCS. This page helps you prepare, and TCS decides who is selected.