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I got 90.99 percentile in jee mains... Obc... Bihar state.... Cs branch... Chances?... Anyone


Vaibhav Kumar 18th Jan, 2019
Answers (2)
Nandini Gupta 27th Feb, 2019

Hii Vaibhav,

your rank will be approximately 78789 according to the formula ,rank= (100- NTA percentile score ) X 874469 /100.

vaibhav last year closing rank was 4454 for Computer Science Engneering in Motilal Nehru University.

so at this rank it is very difficult to get admission in this college.

for other courses you can see cutoff using this link-

https://www.careers360.com/university/motilal-nehru-national-institute-of-technology-allahabad-prayagraj/cut-off


although you can get other college for CSE.

first predict your rank using the link given below-

https://engineering.careers360.com/jee-main-rank-predictor

B.Tech/B.Arch Admissions OPEN

Apply to Amrita University B.Tech 2025


and then predict the college you can get at your rank-

https://engineering.careers360.com/jee-main-college-predictor

hope you understad

thank you.


Shahid Khan 27th Feb, 2019
Hello aspirant.



As per your percentile I.e.,90.99 and u belong to OBC category u do have the chances of getting new NITs.



Hope this information is relevant to your question.



Thank you:)





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