My percentile in jee main 2020 is 98.7122 what will be my rank in general and obc category. ? please tell
New: Direct link to apply for JEE Main 2025 registration for session 1
Also Check: Crack JEE Main 2025 - Join Our Free Crash Course Now!
JEE Main 2025: Sample Papers | Syllabus | Mock Tests | PYQs | Video Lectures
JEE Main 2025: Preparation Guide | High Scoring Topics | Study Plan 100 Days
Hi Ashutosh,
If you have secured 98.7122 percentile in JEE Main paper, you rank would be approximately in between 13000 to 14000. Also, Above is the range of your possible rank. The predicted rank may vary due to JEE Main tie breaker rules for the same score.
Also, you can check your expected rank using the formula:
(100-your total percentile) x 869010/100
So with 98.71 percentile, your rank would be around 13000.
Also, if the number of students increases to 12 lakh in the April session, the formula for rank calculation would change to:
(100-your total percentile) x 1200000/100
Hence your expected rank in this case would be around 180000.
If you have secured this rank you have a chance to get theses colleges:
- NIT Hamirpur
- NIT Patna
- NIT Trichy
- NIT Calicut
- NIT Rourkela
- VNIT Nagpur
- NIT Uttarakhand
- NIT Kurukshetra
These are the college you will get if you have scored these marks.
Sir can you plz tell me where can I get the CSE branch on this rank
Related Questions
Know More about
Joint Entrance Examination (Main)
Eligibility | Application | Preparation Tips | Question Paper | Admit Card | Answer Key | Result | Accepting Colleges
Get Updates BrochureYour Joint Entrance Examination (Main) brochure has been successfully mailed to your registered email id “”.