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Hello aspirant ,you can get your predicted rank and category rank using the rank predictor in this site
This is the link
https://engineering.careers360.com/jee-main-rank-predictor
Your approx rank can be found using the formula (100-nta score)*(874469/100)
So based on your percentile, you will have a gen rank of about 79k . If you belong to gen category, your chances for getting a decent branch in top nits are quite slim. You may get sits in the new nits like silchar,puducherry but your chances of getting good branches are still slim.
I hope this helps you.
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Hello Lata
Congrats for scoring 90.99 percentile in JEE mains 2019
We can't say that you are qualified or not because NTA has not released any Cutt off marks or rank list to say that you are qualified or not .NTA will display this list after the second attempt in april.
But we can predict the rank with this percentile .
There is a pedicted formula from our team to predict rank
Total No of students = 874469
Formula
Probable Rank= ( 100 - NTA percentile ) x 874469 / 100
So your predicted rank = 78789 approx
Our Team predicted that the qualify cutt off would be above 85 percentile .But it just a prediction .So pleae wait till april.
List of NIT's in which you can try are
National Institute of Technology, Jamshedpur, Jharkhand
National Institute of Technology, Kurukshetra, Haryana
Visvesvaraya National Institute of Technology, Nagpur, Maharashtra
National Institute of Technology, Patna, Bihar
National Institute of Technology, Rourkela, Orissa
National Institute of Technology, Srinagar, Jammu & Kashmir
Sardar Vallabhbhai National Institute of Technology, Surat, Gujarat
National Institute of Technology, Raipur, Chhattisgarh
National Institute of Technology, Agartala, Tripura
Indian Institute of Information Technology, Allahabad, Uttar Pradesh
I hope you get my point.
Thank You and GOOD LUCK!!!
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