Amity University Noida B.Tech Admissions 2025
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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.
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.
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 |
Hello navineesh,
with a 73.45% JEE score, you can target reputed private engineering colleges in Tamil Nadu like SRM Institute of Science and Technology, VIT Vellore, and Satyabama Institute of Science and Technology. Government colleges through JEE Main in Tamil Nadu are limited, as most use TNEA. consider applying through management or entrance quota in private universities.
Hello,
SRMJEE Rank: 15,190 and VITEEE Rank: 1,25,900
You wish to study Computer Science Engineering. Here are your choices:
SRM University (Rank: 15,190):
You have a fair chance of securing CSE (core) at SRM Ramapuram, SRM Vadapalani, or SRM NCR (Delhi) campus.
You may also get CSE specializations like AI/ML or Data Science at SRM Main Campus (Kattankulathur) under Fee Category 2 or 3.
Fees:
Fee Category 1: Rs 2.5L/year
Fee Category 2 or 3: Rs 3 to 3.5L/year
VIT (Rank: 1,25,900):
You are eligible for Bhopal or AP campus but not for Vellore or Chennai campus for CSE.
You can anticipate CSE core or its specializations at VIT-AP or VIT-Bhopal under Category 1 or 2 fees, possibly.
Fees:
Around Rs 1.76L to Rs 2.95L per year based on category and campus
Comparison Summary:
SRM provides older, settled campuses such as Ramapuram and NCR with good placements (average Rs 4–5 LPA).
VIT Bhopal and AP campuses are fresher but carry stronger brand power, higher-quality infrastructure, and marginally superior average placements (Rs 5–6 LPA) of late.
Verdict:
Go with SRM if you like the setup at an already established location.
Go with VIT (Bhopal/AP) if better branding appeals and you're comfortable with fresh campuses.
Hi aspirant,
Based on last year's data and with your VJNT-D status, earning a Computer Science seat at K.K. Wagh Institute of Engineering Education and Research with a 96.24 percentile in JEE Mains appears to be a good possibility.
However, cutoffs change from year to year, so for the most up-to-date information, consult the official admission lists and actively participate in the counseling process.
All the best!
Based on your 987 marks in intermediate and 84.6 percentile in JEE, you may have a chance to secure a seat in the Computer Science and Engineering (CSE) branch at SASTRA University. The university considers both your 12th marks and JEE scores for admission. In previous years, the cutoff for CSE was around 987 marks in intermediate, and your score is close but slightly below this threshold. While admission is not guaranteed, there may still be opportunities, especially if you belong to a reserved category. It's best to check the official SASTRA University website for updates on the admission process and cutoffs.
With an EWS rank of 6328 and a percentile of 97.01 in JEE Mains 2025, you have a good chance of admission in VNIT, though it depends on the branch and the overall competition. Given your HSC score of 72.12%, consider applying for branches with comparatively lower cutoff ranks.
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