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JEE Main 2025 Normalization: How scores are calculated?
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.
What is JEE Main Percentile Score?
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.
How will the JEE Main 2025 scores for Paper 1 be calculated?
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.
What is normalisation in JEE Mains 2025?
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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What will be the Highest Raw Score and Percentile Score in JEE Main 2025?
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)
Preparation of JEE Main 2025 Result for each 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 | |
JEE Main 2026: Preparation Tips & Study Plan
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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
- In case the tie is still unresolved, the same rank will be given to the candidates
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:
- Candidates who scores more marks in Aptitude Test will be ranked higher
- In case the tie is unresolved, Candidates who scores more marks in Drawing Test will be given the higher rank
- Lastly the age of the candidates i.e. the older candidate will be awarded the higher rank
- If the tie is still unresolved, the candidates will be given the same rank
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Have a question related to JEE Main ?
Here’s a plan for JEE Mains 2026 in 4 months:
1. Divide time: 2 months for Class 12 syllabus, 1 month for Class 11, 1 month for full revision & mock tests.
2. Daily schedule: 6–7 hours study; 50% for theory & problem-solving, 50% for practice & revision.
3. Topic-wise focus: Prioritize high-weightage chapters and weak areas first.
4. Daily problem practice: Solve previous year questions and chapter-wise exercises.
5. Weekly tests: Take 1 full-length test weekly, analyze mistakes, and revise weak concepts.
6. Consistency: Avoid skipping days; maintain notes and formula sheets for quick revision.
If you want to crack JEE exam you read to dedicatedly prepared for that from the scratch to the advance focus on high weightage topic and prepare question in the time based and continuously practice the previous question this will help to know the pattern of JEE exam questions
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With your JEE Main CRL 476199 and OBC rank 173160, admission to Civil Engineering at Indira Gandhi Institute of Technology (IGIT), Sarang under JoSAA counseling is highly unlikely, since even civil branch closing ranks in IGIT usually remain within 50k–80k CRL (general) and much lower for OBC. However, you may try through state counseling (Odisha JEE) where cutoffs are sometimes more flexible, but chances are still very low. You should keep other private/state colleges as backup.
Hey! With a JEE Main rank of around 1 lakh, you still have a good chance to get admission in several engineering colleges that provide quality education at a reasonable fee. For government colleges, you can aim for NITs or GFTIs in branches like Civil, Mechanical, or Chemical Engineering, as these usually have slightly higher closing ranks. For example, colleges like NIT Goa, NIT Puducherry, and NIT Srinagar might have seats available in certain branches.
Apart from NITs, state government engineering colleges are also a good option, especially under home state quota. They often have lower tuition fees and good faculty, so you can get a solid education without spending much.
If you are open to private colleges, there are options like VIT Vellore, KIIT Bhubaneswar, or Amity University that provide decent infrastructure and placements. However, the fees may be higher than government colleges.
My advice is to focus on branches that are less competitive, check all counseling rounds, and make use of state quotas if applicable. With proper planning, a rank of 1 lakh can still help you get a good college and start your engineering career success
fully.
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