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.
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.
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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
The approximate annual cost for 11th/12th PCM and JEE coaching is around 1.5 to 3.5 lakhs for institutes, excluding hostel and mess fees. The total fees including hostel as well as mess fees can rise upto 4.5 to 6.5 lakhs and above, depending on location and institute quality.
Hi dear candidate,
You can anytime visit our official website to find the previous 10 years JEE Mains question papers with solutions. Kindly refer to the link attached below to download them in PDF format:
JEE Main Last 10 Years Question Papers with Solutions (2025 to 2015)
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Hello,
For JEE Main and JEE Advanced , the cut-offs are lower for ST category students. Here is a simple idea based on recent trends:
JEE Main qualification for ST : Around 50–60 marks is usually enough to qualify for JEE Advanced.
JEE Advanced qualification for ST : You just need to clear the JEE Main cut-off, then appear for Advanced.
To get good NITs or IITs , you will need higher marks.
For NITs (Hyderabad or good branches), try for 120+ marks in JEE Main .
For IITs , even with ST quota, you should aim for at least 80–100+ marks in JEE Advanced for decent branches.
Since you are from ST category and Hyderabad , you don’t need 300 marks in JEE Main. Try to score as high as possible to get better branches, but even moderate marks can qualify you.
Hope it helps !
Yes, you can. If you do 11th from CBSE in 2025 and 12th from NIOS in 2026, then you are allowed to attempt JEE. NIOS is a valid board, so you can give JEE in 2026 and again in 2027. Just take admission on time and pass your 12th exams.
For the JEE 2026, priority-wise, high weightage chapters include Integral
1. Calculus, Coordinate Geometry, Vectors, 3D Geometry, Complex Numbers, and Quadratic Equations in Maths
2. Semiconductors, Current Electricity, Work, Power and Energy, and Gravitation in Physics
3. Organic Chemistry Containing Oxygen, Hydrocarbons, and P-Block Elements in Chemistry.
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