Amrita University B.Tech 2026
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JEE Main Application Date:31 Oct' 25 - 27 Nov' 25
JEE Main 2025 Normalization: How scores are calculated? -With JEE Main being a computer-based test held in multiple sessions, candidates are often confused about the JEE Main 2025 normalization process and the process through which scores are calculated to arrive at the ranks. The National Testing Agency (NTA), under whose aegis the exam will be conducted, announces the relevant information regarding normalization of JEE Main and how the scores and ranks will be calculated. Other important factors like preparation of the rank list, inter-se-merit guidelines have also been announced. Given below is the JEE Main 2025 Normalization process; how scores are calculated to allot the ranks to candidates of various shifts.
Since JEE Main 2025 will be conducted in different sessions & different dates, the question of maintaining equivalence among the different sets of question papers comes. It is likely that some candidates got a tough paper while some may find theirs easy. To ensure that no candidate is at a loss, or even benefits from this, normalisation in JEE Mains will be done for the purpose of ranking. NTA will rank students on the basis of their percentile scores which will be calculated according to a pre-determined formula.
JoSAA opening and closing rank 2025 for BTech artificial intelligence at NITs are given below.
Institute | Quota | Gender | Opening Rank | Closing Rank |
National Institute of Technology Karnataka, Surathkal | HS | Gender-Neutral | 2815 | 3579 |
National Institute of Technology Karnataka, Surathkal | HS | Female-only (including Supernumerary) | 8949 | 10004 |
National Institute of Technology Karnataka, Surathkal | OS | Gender-Neutral | 2060 | 2650 |
National Institute of Technology Karnataka, Surathkal | OS | Female-only (including Supernumerary) | 3987 | 3987 |
National Institute of Technology, Rourkela | HS | Gender-Neutral | 8001 | 9099 |
National Institute of Technology, Rourkela | HS | Female-only (including Supernumerary) | 17226 | 18371 |
National Institute of Technology, Rourkela | OS | Gender-Neutral | 2962 | 4448 |
National Institute of Technology, Rourkela | OS | Female-only (including Supernumerary) | 5770 | 7401 |
Sardar Vallabhbhai National Institute of Technology, Surat | HS | Gender-Neutral | 8112 | 12415 |
Sardar Vallabhbhai National Institute of Technology, Surat | HS | Female-only (including Supernumerary) | 11267 | 18764 |
Sardar Vallabhbhai National Institute of Technology, Surat | OS | Gender-Neutral | 8812 | 10721 |
Sardar Vallabhbhai National Institute of Technology, Surat | OS | Female-only (including Supernumerary) | 14641 | 17523 |
NTA JEE Main 2025 Percentile scores are based on the relative performance of all those candidates who appeared for the examination. It is obtained after transforming the scores into a scale ranging from 100 to 0 for each session of examinees. Percentile score shows the percentage of candidates who scored equal to or below a particular percentile in an exam. It is the normalized score and not the raw score. After normalization, each topper of a session will get the same percentile of 100 which is the desirable one. Also, the marks between the lowest and highest JEE Main scores are also converted to the respective percentiles.
The formula to be used for JEE Main normalisation procedure is:
Formula - (100 x number of candidates appeared in the session with raw score equal to or less than the candidate) / total number of candidates appeared in that session
Note: To avoid bunching effect and reduce tie breaking, the JEE Main percentile scores will be calculated up to 7 decimal places.
Normalisation in JEE Mains 2025 is a procedure implemented to ensure fairness and equity in evaluating the performance of students across different shifts or sessions of the exam. Since JEE Main is conducted over multiple shifts to accommodate a large number of candidates, the difficulty level of the question papers may vary slightly between different shifts. JEE Main normalisation procedure helps account for these variations and ensures that no candidate is unfairly advantaged or disadvantaged due to the difficulty level of their particular shift.
The JEE Main normalisation procedure takes into consideration the principle that the overall ability of a candidate remains constant, regardless of the shift's difficulty level. It aims to create a level playing field for all candidates, regardless of the specific set of questions they face.
Also Check:
Since the calculation is session wise, all the highest raw scores will have a JEE Main normalisation procedure based on percentile score of 100 for their respective sessions.
How to calculate JEE Main Percentile Score?
Let's take a look at how the JEE Main percentile is calculated. The JEE Main percentile score will be calculated using the following percentile formula:
Percentile Score of a Candidate = 100 x (Number of candidates who secured a raw score (or actual score) equal or less than the candidate)/(Total number of candidates who appeared in that session)
Step 1: Raw Scores Preparation
The marks obtained by each candidate in JEE Main eac session 2025 for all correct answers in each subject (Maths, Physics and Chemistry) shall be summed up to arrive at the raw scores.
Step 2: Preparation of JEE Main Percentile Scores for each of the subject and total percentile score
Total Percentile (T1P): | (100 x No. of candidates from the session with raw score equal to or less than T1 score) / Total No. of candidates appeared in the session | |
Mathematics Percentile (M1P) | (100 x No. of candidates appeared from the session with raw score equal to or less than than M1 score in Mathematics) / Total No. of candidates appeared in the session | |
Physics Percentile (P1P) | (100 x No. of candidates appeared from the session with raw score equal to or less than P1 score in Physics) / Total No. of candidates appeared in the session | |
Chemistry Percentile (C1P): | (100 x No. of candidates appeared from the session with raw score equal to or less than C1 score in Chemistry) / Total No. of candidates appeared in the session | |
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Inter-se-merit to resolve tie:
When two or more candidates obtain equal percentile, then ranking will be as follows
Candidates who obtain Percentile Score in Mathematics will be ranked higher
Candidates who obtain Percentile Score in Physics will be ranked higher
Candidates who obtain Percentile Score in Chemistry will be ranked higher
Date of Birth – the older candidate will be ranked higher
JEE Main 2025 Rank list
The authorities shall release the final JEE Main rank list 2025 after conclusion of both the sessions of the exam. This JEE Main normalisation procedure is considered to create the final merit list. This shall be used for admissions into JEE Main participating institutes.
First attempt compilation of NTA Scores:
After the conclusion of JEE Main January 2025 session, JEE Mains results will be displayed with the four NTA scores – for each of the three subjects (Mathematics, Physics, and Chemistry) and the total NTA score for that attempt. JEE Main NTA scores will be the percentile scores calculated as mentioned above.
Second attempt compilation of NTA Scores:
Similar to the January attempt, the result of the JEE Main 2025 April session will be declared after that. The JEE Main 2025 scores will be calculated for the total and that of the three subjects individually as mentioned for the first attempt.
Compilation of NTA Score for JEE Main Result and Merit List/ Ranking for Paper 1
The ranking or merit list for JEE Main 2025 will be compiled as follows
In case of candidates who will appear for both the attempts the best of the two NTA JEE Main scores will be used. Here the total NTA score will be used for ranking purpose and not those of the subjects.
In case of candidates who will appear for only one exam, the four JEE Main 2025 NTA scores of that attempt will be used for the preparation of the merit list
The final NTA scores will be used to rank the candidates and for the declaration of the JEE Main 2025 result. In case of any tie, the inter-se-merit guidelines mentioned above in this article will be used to resolve it.
JEE Main 2025 Rank Declaration:
The ranks derived through the method explained above will be used for the purpose of admissions to NITs, IIITs and GFTIs.
The JEE Main 2025 All India rank and All India Category Rank will be declared once the NTA scores are compiled by the authorities for April session.
How will the JEE Main 2025 scores for Paper 2 be calculated?
Raw marks for the first attempt will be announced in February 2025 for students who appeared for the exam held in January.
The actual marks for the April exam will be announced soon after the conclusion of exam.
The JEE Main Paper 2 marks will not be normalized as the exam was held in a single shift so there will be no question of equivalence of the question papers and variation in difficulty levels
Rank List preparation for JEE Main 2025 Paper II
The raw marks for both attempts will be considered for candidates who will give both the attempts and the better of the marks considered.
For candidates who appeared for only one attempt, the raw marks obtained for the same will be used for preparing JEE Main rank list.
The NTA JEE Main 2025 Rank lists will be prepared as per the marks obtained as mentioned above
Inter-se-Merit will be resolved as follows:
On Question asked by student community
Hello,
If you’re appearing for Class 12 (HSC) in 2026 and don’t have your board admit card or roll number yet, you can leave the “Registration No./Enrollment No./Roll No.” field blank or enter “NA” (Not Applicable) if the form allows.
Do not enter your school GR number, as it’s not recognized by the exam board or NTA.
Once your board issues the admit card or registration number, it can be updated later during JEE Main form correction or at the time of result verification.
So, for now, safely enter “NA” or leave blank — not your GR number.
Hope you understand.
Hello,
For JEE Main EWS certificate, your parent's income certificate is required, not yours.
The EWS category is based on the family’s annual income, which must be below 8 lakhs from all sources.
The certificate should be issued by a competent government authority and must be valid for the current financial year.
Hope you understand.
Hello,
To secure admission to the University of Hyderabad for the Integrated BTech + MTech in Computer Science and Engineering (CSE) course with an EWS certificate and home state quota, you would likely need a JEE Main percentile of approximately 95-97 percentile or higher. Also, you can check the JEE Main Cutoff , to know more details.
I hope it will clear your query!!
Hello,
Here’s how you can get JEE Main Previous Year by following these steps :
Go to the official JEE Main website.
Click on the “Previous Year Question Papers” or “Downloads” section.
Choose the year and session you want (for example, 2024, 2023, 2022, etc.).
Download the paper in PDF format.
You can also find solved papers and answer keys for the last 5 years on Careers360, where they are available year-wise and subject-wise for free. Click on this link to get papers : JEE Main Question Papers with solution
Hope it helps !
You can get the most repeated and comprehensive questions in the JEE MAINS for the JEE MAINS 2026 EXAMINATION. You can get their pdf from the article link given below by careers360 and download the topic wise questions from every subject to strengthen the preparation.
link- Most Repeated Questions in JEE Mains: Comprehensive Analysis
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