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I am an OBC NCL candidate and have cleared cutoff from my category percentile. But I am not in top 250000 rank so can I write JEE advance? As written on website only top 250000 students can give exam. Please answer me


Artemis 9th Aug, 2022
Answers (2)
Piyush 9th Aug, 2022

Hello aspirant,

It does not matter whether you have secured a rank of below 2.5 lakhs in JEE mains, what matters is whether you have cleared the JEE mains 2022 cutoff for your category or not, in this case OBC. If you have cleared the cutoff for OBC category, then you are eligible to give JEE advance exam 2022.

EWS - 63.1114141

OBC - 67.0090297

SC - 43.0820954

ST -26.7771328

Further, to know more about the JEE advance eligibility criteria, you can follow these two below mentioned links :

https://jeeadv.ac.in/eligibility.html

https://engineering.careers360.com/articles/jee-advanced-eligibility-criteria

I hope that this will help.

Anurag Khanna 9th Aug, 2022

Hello Aspirant,

See your rank does not define whether you are eligible for JEE-Advance or not but percentile does so if you are qualified for JEE-Advance as per the percentile then you are eligible for JEE-Advance although your rank is more than 2.5 Lakhs.

The cut-offs are released along with the JEE-Main 2022 result so now you can check whether you have qualified for JEE-Advance 2022 or not, JEE-Main 2022 qualifying cut-offs are enlisted below :-

==>> Common Rank List (CRL) -  88.4121383 percentile

==>> GEN-EWS - 63.1114141 Percentile

==>> Other Backward Classes (OBC-NCL) - 67.0090297  Percentile

==>> Scheduled Caste (SC) - 43.0820954 Percentile

==>> Scheduled Tribe (ST) - 26.7771328 Percentile

==>> PwD - 0.0031029 Percentile


You can check out previous year's cut-off trends from the link given below :-

https://engineering.careers360.com/articles/jee-main-cutoff



To have a list of all those colleges which you can get based upon your JEE-Main score or rank you can use JEE-Main College Predictor:-

https://engineering.careers360.com/jee-main-college-predictor?icn=QnA&ici=qna_answer


I hope this information helps you.

Good Luck!!

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