I have 92.23 percentage marks in jee mains 2019 i am in obc category what is the rank of me plzz reply
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Your rank will be 66045 as per our rank predictor.
Your rank range will be in the range 60000-70000 .
Our rank predictor formula =( (100 - your percentile)/100)*total number of candidate.
This year 850000 candidate given exam in Jan Jee.
This rank is good to get good college with help of state quota for NIT.
As you are from OBC category you have little more advantage than general candidate.
If any queries remain than comment below.
Good luck...
If you have state quota of this state than it is absolutely sure as per previous cutoff .
If you not have state quota than difficult to say about.
Good luck...
Previous year cutoff for NIT CALICUT
For general candidate
Home state rank
Opening rank :4500
Closing rank:16000
Previous year
For OBC category rank:2591 opening rank
4753 closing rank
If any queries remain than comment below.
Good luck...
You require little extra percentile to get into NIT Calicut for mechanical engineering.
You should to model question paper to sharp your performance in limited time period with little more effort in Jee April you can clearly get into NIT CALICUT.
Good luck...
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