JEE Main expected cut-offs 2025 for General, OBC, SC, ST, EWS; how many qualify for IIT JEE Adv?
Anu Parthiban | February 10, 2025 | 04:51 PM IST | 2 mins read
The NTA will prepare the JEE Main results 2025 by converting the raw score into percentiles. The JEE Main final answer key 2025 is out at the official website, jeemain.nta.nic.in.
Check your college admission chances based on your JEE Main percentile with the JEE Main 2026 College Predictor.
Try NowNEW DELHI: The National Testing Agency (NTA) will declare the Joint Entrance Examination Main (JEE Main 2025) results on the official website, jeemain.nta.nic.in, for admission to BE, BTech in IITs, NITs and other participating top engineering colleges. Candidates will have to obtain a minimum qualifying JEE Main cut-off to be eligible for BTech admission and JEE Advanced 2025. JEE Main 2025 Result LIVE
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Now that the JEE Main final answer key 2025 is out for session 1 exam held between January 22 to 29 for BE, BTech paper 1, the NTA is expected to announce the results on the official website, jeemain.nta.nic.in 2025. As per past trends, the NTA hosts the JEE Main score card link on the same day it publishes the final answer key.
JEE Main marks vs percentile vs rank
The NTA will prepare the JEE Main results 2025 by converting the NTA raw score into percentile scores. The marks obtained by the candidates are converted into JEE Main percentile to ensure that no candidate is benefitted from the varying difficulty level of question paper in multiple shifts.
“In the examination held in two shifts, if the 40% marks correspond to a percentile score of 78 in shift 1 and 79 in shift 2, then all those equal to or above 78 percentiles (percentile score of 100 to 78) in both shifts will become eligible in the general category. A similar method will be adopted for the other categories to determine eligibility cut-offs. In case the examination is held in more shifts the same principle shall apply,” the NTA explained the JEE Main score calculation.
As per the JEE Main percentile calculator , the total percentile is calculated by dividing the number of candidates appeared in the session with a raw score equal to or less than the score of the candidate by total number of candidates who appeared in the session and multiplying it with 100. Based on the normalised marks, the NTA JEE Main rank list is prepared.
JEE Main Cut-off 2025: Category-wise cut-offs
Candidates can check the category-wise JEE Main cut-off percentile and number of candidates who obtained the minimum qualifying marks in the exam last year.
|
Category |
JEE Main cut-off percentile |
Number of candidates |
|
UR-ALL |
100.0000000 - 93.2362181 |
97,351 |
|
UR-PwD |
93.2041331 - 0.0018700 |
3,973 |
|
EWS-ALL |
93.2041331 - 81.3266412 |
25,029 |
|
OBC-ALL |
93.2041331 - 79.6757881 |
67,570 |
|
SC-ALL |
93.2041331- 60.0923182 |
37,581 |
|
ST-ALL |
93.2041331 - 46.6975840 |
18,780 |
Also read Are you eligible for JEE Advanced 2025? IIT Kanpur answers FAQs on number of attempts, eligibility
JEE Advanced 2025 cut-off
Candidates who qualify the JEE Mains by obtaining the minimum marks will be eligible for admission to National Institutes of Technology (NITs), Centrally-funded Technical Institutes (CFTIs), and other participating colleges. However, those aspiring to pursue BE, BTech in IITs will have to secure a rank in the top 2.5 lakh candidates in JEE Main results 2025.
IIT Kanpur has announced the category-wise distribution of top 2,50,000 candidates based on the performance in BE, BTech paper 1 of JEE Main 2025. The number of candidates may slightly be greater than 2.5 lakh in case of more number of candidates obtaining same JEE Main rank.
|
Category |
Number of candidates |
|
|
OPEN |
96187 |
101250 |
|
OPEN-PwD |
5063 |
|
|
GEN-EWS |
23750 |
25000 |
|
GEN-EWS-PwD |
1250 |
|
|
OBC-NCL |
64125 |
67500 |
|
OBC-NCL-PwD |
3375 |
|
|
SC |
35625 |
37500 |
|
SC-PwD |
1875 |
|
|
ST |
17812 |
18750 |
|
ST-PwD |
938 |
|
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