My expected marks of Jee advanced 2020 are 137.how much rank will I get in General & OBC ncl ?
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Answer (1)
4th Oct, 2020
Dear student,
As you are getting 137 marks in JEE advanced then congratulations you have successfully cleared the cutoff of JEE advanced and definitely you will get an IIT.
As the cut off set by IIT Delhi this year - for general candidates the individual subject cut off was set as 10% and the overall aggregate was 35% of 396 marks, but for OBC candidates there was a slight change in cutoff and it was around 9% of 132 marks in each subject and around 31.5% aggregate of 396 marks.
From your general category you have cleared your cutoff but your rank is not upto the mark.
You will get around 8000 to 9000 rank, but from your OBC category you can get under 2000 rank in jee advanced and you can definitely get an IIT.
You can also use our JEE advanced college predictor tool to get an appropriate idea of the colleges that you can get on the basis of your JEE advanced score or rank :-
https://engineering.careers360.com/jee-advanced-college-predictor?icn=QnA&ici=qna_answer
As you are getting 137 marks in JEE advanced then congratulations you have successfully cleared the cutoff of JEE advanced and definitely you will get an IIT.
As the cut off set by IIT Delhi this year - for general candidates the individual subject cut off was set as 10% and the overall aggregate was 35% of 396 marks, but for OBC candidates there was a slight change in cutoff and it was around 9% of 132 marks in each subject and around 31.5% aggregate of 396 marks.
From your general category you have cleared your cutoff but your rank is not upto the mark.
You will get around 8000 to 9000 rank, but from your OBC category you can get under 2000 rank in jee advanced and you can definitely get an IIT.
You can also use our JEE advanced college predictor tool to get an appropriate idea of the colleges that you can get on the basis of your JEE advanced score or rank :-
https://engineering.careers360.com/jee-advanced-college-predictor?icn=QnA&ici=qna_answer
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