203 Views

I have got 92 percentile in JEE main 2019,how can i know my rank,i want to opt for aerospace engineering,I belong to the general category


JEE Main 2025: Rank Predictor | College Predictor | Marks vs Rank vs Percentile

JEE Main 2025: Sample Papers | Syllabus | Mock Tests | PYQsHigh Scoring Topics

Apply to TOP B.Tech/BE Entrance exams: VITEEE | MET | AEEE | BITSAT

Radhika 2nd Feb, 2019
Answers (3)
Nandini Gupta 2nd Feb, 2019

Dear Radhika.

first of all congratulations.

you got very good marks in JEE main.

Radhika you can use the link given below to know your rank,

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

For it you will have to give your sone information like

  • Your JEE amin Application number
  • DOB
  • Your JEE amin NTA persentile
  • Category
  • Date of Exam
  • 12 th passing year

by giving these information you can see the prediction.


also Radhika in India there are only few colleges which have Aerospace branch.like top IIts.

thank you.

Unnam sreekanth 2nd Feb, 2019

Hello aspirant

I can help you with better information.

Good opt aerospace engineering is one of the prestigious and demanded course in the current generation

This is your approximate rank 69,957.

For more information go through the below  career 360 predictor link.

Hope above information is helpful to you

All the best

Thank you

JEE Main 2025 College Predictor

Know your college admission chances in NITs, IIITs and CFTIs, many States/ Institutes based on your JEE Main result by using JEE Main 2025 College Predictor.

Try Now
Ayush Dudhe 2nd Feb, 2019

To start with,

Calulating the percentage  through NAT percentile is not possible. You should wait for the official results to be out.

But there is a simple formula through which we can calculate the approximate rank using NAT percentile.

JEE Main 2019 rank (probable) = (100- NTA percentile score ) X 874469 /100

If NTA percentile score is 90.70, JEE Main rank will be (100-90.70 ) X 874469/100 = 81325

Consider the NTA score of 80.60 then JEE Main 2019 rank = (100-80.60) X 874469 /100. The rank will be 169646.

** This formula would not be for a 100 percentile score as those are the top ranks.

Know More About

Related Questions

Amity University Noida B.Tech...
Apply
Among Top 30 National Universities for Engineering (NIRF 2024) | 30+ Specializations | AI Powered Learning & State-of-the-Art Facilities
Chandigarh University Admissi...
Apply
Ranked #1 Among all Private Indian Universities in QS Asia Rankings 2025 | Scholarships worth 210 CR
UPES B.Tech Admissions 2025
Apply
Ranked #42 among Engineering colleges in India by NIRF | Highest CTC 50 LPA , 100% Placements
Amrita Vishwa Vidyapeetham | ...
Apply
Recognized as Institute of Eminence by Govt. of India | NAAC ‘A++’ Grade | Upto 75% Scholarships
Anant National University - B...
Apply
B.Arch/B.Tech Admissions 2025 open
Manav Rachna-B.Tech Admission...
Apply
NAAC A++ Grade | NBA Accredited B.Tech programs | 41000+ Alumni network | Students from over 20 countries
View All Application Forms

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books