JEE Mains 2025 for Qualifying Percentile: Candidates can access the qualifying percentile data for JEE Main through this page. The National Testing Agency will release the JEE Mains 2025 cutoff on the official website, jeemain.nta.nic.in. To qualify for JEE Advance participants need to meet the minimum percentile specified by JEE Mains 2025. The cutoff marks to qualify in JEE Main 2025 varies every year based on the factors like difficulty level of exam, number of candidates and seats available. The authority conducted the JEE Main entrance exam on January 22 to 30, 2025.
NTA permits candidates to take the Joint Entrance Examination (JEE) Main up to six times over a three-year period.
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JEE Mains 2025 qualifying percentile is calculated using the normalisation process for all the exam shifts. Securing the JEE Main qualifying percentile allows students to appear for JEE advance, but to secure admission into the NITs, IIIT’s and GFTI’s, candidates must score the respective institute cutoff and marks. In case of General category candidates, the cutoff percentile ranges from 88 to 92. In case of OBC, it is around 72 to 76, and for SC/ST the cutoff falls between 30 to 55. For more details regarding what is the qualifying percentile for JEE Mains, candidates can refer to the article given below.
The National Testing Agency evaluates JEE Main 2025 scores for percentile across three distinct subject categories (Physics, Chemistry, Mathematics) and a combined overall scheme using this normalization formula.
Total Percentile (T1P) | (100 x No. of candidates from the session with a raw score equal to or less than T1 score) / Total No. of candidates who appeared in the session |
Mathematics Percentile (M1P) | (100 x No. of candidates appeared from the session with a raw score equal to or less than than M1 score in Mathematics) / Total No. of candidates who appeared in the session |
Chemistry Percentile (C1P): | (100 x No. of candidates appeared from the session with a raw score equal to or less than C1 score in Chemistry) / Total No. of candidates who 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 who appeared in the session |

Candidates can check the Top 10 NTA Score and Overall Rank List in the table below.
RAW SCORE | NTA SCORE (Percentile) | RANK | ||||||
Math | PHY | CHE | Total | Math | PHY | CHE | Total | |
120 | 110 | 101 | 331 | 100.0000 | 100.000 | 99.9201471 | 100.0000 | 1 |
120 | 101 | 111 | 332 | 100.0000000 | 99.9778342 | 99.9926114 | 100.0000000 | 2 |
115 | 115 | 105 | 335 | 99.9964301 | 99.9750107 | 99.9571612 | 100.0000000 | 3 |
115 | 116 | 115 | 346 | 99.9692695 | 99.9938539 | 99.9938539 | 100.0000000 | 4 |
115 | 95 | 111 | 321 | 99.9903209 | 99.9080482 | 99.9975802 | 99.9975802 | 5 |
106 | 110 | 105 | 321 | 99.9007888 | 100.0000000 | 99.9661230 | 22051995 | 6 |
120 | 110 | 100 | 330 | 99.9975371 | 100.0000000 | 100.0000000 | 99.8916336 | 7 |
115 | 110 | 120 | 345 | 99.9969270 | 99.9692695 | 99.9600504 | 100.0000000 | 8 |
106 | 115 | 111 | 332 | 99.9964301 | 99.9500214 | 99.9750107 | 99.9750107 | 9 |
120 | 100 | 100 | 320 | 99.9950743 | 100.0000000 | 99.9704455 | 99.8916336 | 10 |
Compilation and display of Result for multi session Papers of JEE(Main) For examinations conducted twice (First and Second attempt) before admissions to the next academic session.
1. Compilation of Result for First attempt:
Since, the first attempt will be conducted in multi-sessions, NTA scores will be calculated corresponding to the raw marks obtained by a candidate in each session as per above procedure. The calculated NTA scores for all the sessions will be merged for declaration of result. The result shall comprise the four NTA scores for each of the three subjects (Mathematics, Physics, and Chemistry) and the total for the first attempt.
2. Compilation of Result for Second attempt:
Similarly, the second attempt will be conducted in multi-sessions, NTA scores will be calculated corresponding to the raw marks obtained by a candidate in each session as per above procedure. The calculated NTA scores for all the sessions will be merged for declaration of result.The result shall comprise the four NTA scores for each of the three subjects (Mathematics, Physics, and Chemistry) and the total for the second attempt.
3. Compilation of Result and Preparation of Merit List / Ranking:
The four NTA scores for each of the candidates for the first attempt as well as for the second attempt will be merged for compilation of result and preparation of overall Merit List / Ranking.Those appeared in both the attempts; their best of the two NTA scores will be considered further for preparation of Merit List /Rankings as explained in the following example:
Roll No | First Attempt – NTA Score | Second Attempt-NTA Score | Final NTA Score for compilation of result (best of the two in Total) | |
C20045511 | Total | 99.8620723 | 99.8740051 | 99.8740051 |
Mathematics | 99.6249335 | 100.0000000 | 100.0000000 | |
Physics | 99.7338237 | 99.2102271 | 99.2102271 | |
Chemistry | 99.8572327 | 99.7203528 | 99.7203528 | |
C20045512 | Total | Did not Appear | 99.8620723 | 99.8620723 |
Mathematics | 99.6249335 | 99.6249335 | ||
Physics | 99.7338237 | 99.7338237 | ||
Chemistry | 99.8572327 | 99.8572327 | ||
C20045513 | Total | 99.8740051 | Did not Appear | 99.8740051 |
Mathematics | 100.0000000 | 100.0000000 | ||
Physics | 99.2102271 | 99.2102271 | ||
Chemistry | 99.7203528 | 99.7203528 | ||
C20045514 | Total | 99.8740051 | 99.8620723 | 99.8740051 |
Mathematics | 100.0000000 | 99.6249335 | 100.0000000 | |
Physics | 99.2102271 | 99.7338237 | 99.2102271 | |
Chemistry | 99.7203528 | 99.8572327 | 99.7203528 |
IIT JEE aspirants can refer to the JEE Mains Marks vs Percentile vs Rank for 2025, which will be provided here. According to the JEE Main 2025 marks vs percentile vs rank data, candidates need to score above 250 marks to achieve a percentile greater than 90.
Marks Out of 300 | Percentile | Overall Rank |
|---|---|---|
290- 280 | 99.99908943 - 99.99745041 | 13 - 36 |
280 - 250 | 99.99745041 - 99.96976913 | 36 - 428 |
250 - 240 | 99.96976913 - 99.94664069 | 428 - 755 |
240 - 230 | 99.94664069 - 99.91595453 | 755 - 1189 |
230 - 220 | 99.91595453 - 99.86623749 | 1189 - 1893 |
220 - 210 | 99.86623749 - 99.80777899 | 1893 - 2720 |
210 - 200 | 99.80777899 - 99.73129123 | 2720 - 3803 |
200 - 190 | 99.73129123 - 99.62402626 | 3803 - 5320 |
190 - 180 | 99.62402626 - 99.48033855 | 5320 - 7354 |
180 - 170 | 99.48033855 - 99.2955842 | 7354 - 9968 |
170 - 160 | 99.2955842 - 99.06985426 | 9968 - 13163 |
160 - 150 | 99.06985426 - 98.77819917 | 13163 - 17290 |
150 - 140 | 98.77819917 - 98.40768884 | 17290 - 22533 |
140 - 130 | 98.40768884 - 97.94047614 | 22533 - 29145 |
130 - 120 | 97.94047614 - 97.35425213 | 29145 - 37440 |
120 - 110 | 97.35425213 - 96.60949814 | 37440 - 47979 |
110 - 100 | 96.60949814 - 95.64338495 | 47979 - 61651 |
100 - 90 | 95.64338495 - 94.39636137 | 61651 - 79298 |
90 - 80 | 94.39636137 - 92.76234617 | 79298 - 102421 |
80 - 70 | 92.76234617 - 90.4109851 | 102421 - 135695 |
70 - 60 | 90.4109851 - 87.06073037 | 135695 - 183105 |
60 - 50 | 87.06073037 - 81.57582987 | 183105 - 260722 |
50 - 40 | 81.57582987 - 73.08140938 | 260722 - 380928 |
40 - 30 | 73.08140938 - 59.84001311 | 380928 - 568308 |
30 - 20 | 59.84001311 - 40.3469266 | 568308 - 844157 |
20 - 10 | 40.3469266 - 20.95045141 | 844157 - 1118638 |
10 - 0 | 20.95045141 - 6.599800585 | 1118638 - 1321716 |
The National Testing Agency will release the JEE Main 2025 cutoff on the official website. Candidates can check the expected cutoff for various categories: 95+ for General, 83+ for OBC, 85+ for EWS, 65+ for SC, and 50+ for ST.
Frequently Asked Questions (FAQs)
To calculate the JEE Mains percentile for 2025, candidates should use the following formula: Percentile = (100 × Number of candidates in the session with a raw score equal to or lower than the T1 score) / Total number of candidates who appeared in the session.
A score that reaches beyond 95% of all observed data points belongs to the 5th percentile group. The top 5 percentile indicates your test score surpasses what 95% of test-takers achieved. Becoming part of the top 5 means your ranking surpasses 95% of potential candidates thus sitting at the 95th percentile position.
| Category | Expected cut off |
|---|---|
| General | 91-95 |
| EWS | 77-85 |
| OBC NCL | 77-85 |
| SC | 58-65 |
| ST | 44-50 |
On Question asked by student community
Hello,
You should select Jharkhand for the JEE Mains application form because you will be reappearing in the exam as a private candidate in Jharkhand. The state you will be appearing in and your passing year are what you should fill in your JEE Mains application form, but not your 12th class state.
Hope it helps!!!
Hello Aspirant
If you want to get into IIT for AI Engineering, then you should definitely choose Pure Mathematics in Class 11–12. Applied Maths won’t help in JEE because IIT entrance exams use advanced math concepts. Pure Maths will build your problem-solving skills and cover everything needed for JEE Main and Advanced. It also prepares you better for AI, Data Science, and Coding in college. So for your dream of AI engineering at IIT, Pure Maths is the right and safest choice.
That's an excellent resource for targeted, efficient practice!
The JEE Main Chapter-Wise Previous Year Questions (PYQs) is one of the most effective study tools because it allows you to:
Identify High-Weightage Areas: Pinpoint chapters frequently tested in previous exams.
Understand Pattern: Master the specific style and depth of questions asked by NTA.
Improve Speed & Accuracy: Practice time management by solving questions sequentially.
You can download the PDF containing these essential chapter-wise questions (organized by Physics, Chemistry, and Mathematics) right here: https://engineering.careers360.com/download/ebooks/jee-main-chapter-wise-pyqs . Start integrating these into your daily study plan
Hi there,
You should take pure mathematics if you want to pursue AI engineering at IIT after JEE, because the syllabus of JEE Advanced is based on pure mathematics, so it will help you to have a strong base foundation in the concepts. You can perform well in your exam.
Hope it helps!!!
Hi SHOURYA
There is no problem if you upload a mismatched photo in the JEE Mains registration form, which does not match your newly updated Aadhar card photo.
Here are some steps which you can take -
I hope this information helps you. If you know more about the JEE Mains exam, you can visit our official website, careers360.
Here is the link - https://engineering.careers360.com/articles/jee-main-application-form-correction
Best Wishes.
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