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.
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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 aspirant,
If your syllabus is completed with theory , use the next 30 days only for smart revision . Make short notes and revise formulas daily for Physics , Chemistry and Maths . Solve previous year JEE Main questions topic-wise and then full mock tests every 3-4 days. Analyse mistakes properly and revise weak areas again . Avoid new topics and focus on accuracy , speed and confidence building during revision.
FOR REFERENCE : https://engineering.careers360.com/articles/jee-main-revision-strategy
Hope the details will help you.
THANK YOU
Preparing for the JEE Main in just 30 days is a challenging but achievable task if you follow a highly disciplined and strategic approach. According to the Careers360 30-day study plan , the key is to shift your focus from learning everything to mastering high-weightage topics and practicing rigorously.
During the first 15 days, prioritize topics that frequently appear in the exam.
Physics: Modern Physics, Heat & Thermodynamics, Optics, and Current Electricity.
Chemistry: GOC (General Organic Chemistry), Chemical Bonding, p-Block elements, and Solutions.
Maths: Matrices & Determinants, Sequences & Series, Coordinate Geometry, and Vector & 3D Geometry.
Study Strategy: Use NCERT for Chemistry and simplified notes for Physics/Maths. Spend 3-4 hours on each subject daily.
Short Notes: Go through the short notes you made during the first two weeks.
Flashcards: Use flashcards for inorganic chemistry reactions and physics formulas.
Mock Tests: Start giving one full-length mock test every alternate day. Analyze your mistakes immediately to avoid repeating them.
Previous Year Papers (PYQs): Solve the last 3-5 years of JEE Main papers in the actual exam time slot (9 AM–12 PM or 3 PM–6 PM) to sync your body clock.
No New Topics: Stop picking up new chapters. Focus solely on what you already know to build confidence.
Accuracy over Speed: Focus on getting the questions right rather than attempting all of them, as negative marking can significantly lower your percentile.
You can download the comprehensive day-by-day schedule, which includes specific topics to cover each morning and evening, by visiting the link : https://engineering.careers360.com/download/ebooks/jee-main-study-plan-30-days
Hello
I think your question sounds like this: "Can a candidate who passed Class 12 in 2025 and filled JEE Main Session 1 apply for Session 2 as 'Appearing' in 2026 as a fresh board candidate, and will both sessions have the same details? "
So yes, you can fill the JEE Main Session 2 as "Appearing". The information filled in Session 1 will remain exactly as it was and will not change.
Session 1 and Session 2 are treated as separate applications. There is no issue if the qualifying status is different in both sessions. During counselling, the board marks that meet the 75 rule will be considered.
Hello,
If you filled the JEE Main January form with Class 12 passed in 2025 and are planning to appear again for the Class 12 exam through HOS, there is usually no serious issue. You were eligible to apply since you had already passed Class 12. Reappearing through HOS for improvement or requalification is allowed, provided HOS is a recognized board. During counselling, your latest valid Class 12 result will be considered. Make sure you meet the 75% marks or top 20 percentile requirement where applicable. If a correction window opens, update details if needed.
Hope this has solved your query. Thank You.
Good Evening,
Yes, you are eligible for both JEE Mains and Advanced, as you completed your 12th with physics, chemistry and biology. Moreover, you passed mathematics in 2025, which makes you fit the eligibility criteria of both exams.
Aspirant, I would like to inform you that Careers360 recently launched a free mock test series for JEE students. The last date of registration is 8th January, 2026. Enroll and solve chapter wise question papers and improve you concept and assess your learning. The link to the mock test series is attached herewith. https://learn.careers360.com/test-series-jee-main-free-mock-test/
Best regards.
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