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 Ishan,
With an SC rank of 4763 in NCHMCT JEE 2025, it is going to be near impossible to get admission in IHM Kolkata in the 1st round, as based on in previous years, the cutoff ranks for the SC category in top tier IHMs like Kolkata seem to close much earlier (most often under 1000–1500 in rank). However, you could possibly get in different rounds/spot rounds if there are empty seats.
With an EWS rank of 11,895 in JEE Mains 2025 , getting admission into a top NIT (like NIT Trichy, NIT Warangal, NIT Surathkal, etc.) through the General EWS category is quite difficult, especially in high-demand branches like CSE, ECE, or IT. These branches usually close under EWS rank 2,000–5,000 for top NITs.
However, you still have chances at some lower-tier NITs or less competitive branches such as:
NIT Mizoram
NIT Nagaland
NIT Meghalaya
NIT Arunachal Pradesh
NIT Puducherry (for home state candidates)
NIT Agartala
Hello Anushka,
With a JEE Main Paper II rank of 3300 in the OBC category, you have a good chance of getting admission in several reputed colleges offering B.Arch courses.
Key points:
Top government and private colleges usually have closing ranks for OBC candidates around or above your rank.
Institutes like National Institutes of Technology (NITs), some State Government colleges, and well-known private universities may offer seats at this rank.
Admission depends on factors like state quota, counseling rounds, and seat availability.
It is important to participate in the JoSAA counseling and state-level counseling processes to maximize your options.
You should explore colleges such as NITs with architecture programs, IIIT Hyderabad (if available), and prominent private universities offering B.Arch. Checking official counseling details and cutoffs each year will give clearer guidance.
I would recommend you to use the JEE Mains B.Arch College Predictor Tool by Careers360 to get the list of possible colleges you can get with your rank.
I hope this answer helps you. If you have more queries, feel free to share your questions with us, and we will be happy to assist you.
Thank you, and I wish you all the best in your bright future.
Dear Candidate ,
Yes , you can apply to IET Lucknow for B. Tech with your 87 % in Class 12 and CUET UG score even if you have not given JEE mains . While JEE main scores are usually preferred , IET also considers CUET UG scores for admission through UPTAC counselling , especially if seats remain after JEE candidates are admitted . Make sure you meet the eligibility criteria, have all necessary documents ready , and participate in the counselling process . Keep an eye on official dates for application and counseling, typically starting around May - June .
Hello,
Yes, you can apply to IET Lucknow without giving JEE Main if you have appeared in CUET UG and scored well.
With 87% in class 12, you meet the basic eligibility, but admission will depend on your CUET UG score and the counselling process through UPTAC . You need to register for UPTAC counselling, choose IET Lucknow, and participate in seat allotment.
Hope it helps !
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