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top colleges available for 94.90 percentile in JEE main 2022


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kpdarshangowda451 11th Jul, 2022
Answers (2)
ADITYA KUMAR Student Expert 11th Jul, 2022

Hi

Assuming you from general category.

With   94.90 percentile

  • Your rank  will be roughly around 52173 - 58650


Now,

Making  prediction is difficult as  cutoff  depends upon various factors like :-

-----------) number of students/ candidates  that appeared in the  jee main exam,

-----------) number of available seats,

-----------) jee main examination difficulty level.

-----------) the branch you want,

-----------) the category you belong etc.

-----------) your rank in jee main


But yes,

When we go through previous year trend then


>>>>>>>>> For your  rank

B.Tech/B.Arch Admissions OPEN

Apply to VIT - VITEEE 2026

You have chances in following  which are listed below along with their closing rank in 6th round of josaa counselling 2021 under other state/ all india quota   :-



---------------------)National Institute of Technology Sikkim:- civil engineering:-54730


---------------------)National Institute of Technology Agartala:- btech biotechnology:- 56064



---------------------) National Institute of Technology, Rourkela:- Life Science (5 Years, Integrated Master of Science):- 57878


---------------------)School of Engineering, Tezpur University, Napaam, Tezpur:- electrical engineering:-76357


And a few more

You can check the closing ranks  at official website of josaa by going through the link given below

JOSAA 2021:- https://josaa.nic.in/Counseling/seatallotmentresult/currentorcr.aspx

Or

For previous year cut offs of top colleges like NITs, IIITs , you can also refer our page at :-

NITs :- https://engineering.careers360.com/articles/nits-cutoff-for-top-btech-branches

IIITs:- https://engineering.careers360.com/articles/jee-main-cutoff-for-top-iiits

Above prediction has been made  considering  past year cut off data but  closing rank for colleges varies every year depending upon many factors like jee main question paper difficulty, number of students writing the exam etc so

To have a personalised report on your chances of getting a college you can go through our college predictor, it will give you complete list of colleges which you can get on the basis of the category you belong to,  your home state, your gender etc, the link for  jee main college predictor is provided below

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


To help you  further,

To give an idea of jee main 2022 qualifying cut off ,

Minimum percentile in jee main 2021 to qualify it to be eligible for jee advanced is given below :-

--------) general :-87.8992241 percentile

--------) EWS:- 66.2214845 percentile

--------) OBC-NCL :- 68.0234447 percentile

--------) SC :- 46.8825338 percentile

--------) ST:- 34.6728999 percentile

--------) PWD :- 0.0096375 percentile

You can check the same at our page the link is provided below

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


Thankyou

Anurag Khanna 11th Jul, 2022

Hello Aspirant,

Considering your category as General, since you have not mentioned it.

See If we go through the analysis of marks Vs Percentile Vs rank for JEE-Main 2021 then according to the percentile mentioned by you in the question your CRL rank is expected to be in the range of 52173 - 58650 .


Note that the rank which is predicted is only the expected rank on the basis of previous year's data so your actual rank might vary depending upon current factors.


Based on such rank and category you can go for:-


  • NIT Mizoram - National Institute of Technology
  • Guru Ghasidas Vishwavidyalaya
  • ICT Mumbai - Institute of Chemical Technology, Mumbai
  • Sant Longowal Institute of Engineering and Technology
  • North Eastern Hill University, Shillong
  • Chandigarh College of Engineering and Technology


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