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I'm Arjun.S.P. ,I've got 81 percentile in JEE B.arch. ,what would be my expected ranking!


Arjun S Pramod 23rd Dec, 2018
Answers (2)
Kiara Sen 1st Feb, 2019

Hello Arjun,


The JEE Main Paper 2 results is annou nced on 31 st January. NTA has provided a criteria of percentage score this year. You can take help from the previous year cutoff marks

MNIT Jaipur – Rank 77 to Rank 1738

MNIT Bhopal – Rank 74 to Rank 1928

MNIT Calicut – Rank 50 to Rank 1834

MNIT Hamirpur – Rank 91 to Rank 2325

NIT Patna –Rank 78 to Rank 4153

NIT Rourkela – Rank 28 to 1712

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Apply to Amrita University B.Tech 2026

NIT Trichy – Rank 10 to 1931

VNIT Nagpur – Rank 45 to 1952


Hope this helps.

Kowshik mogali 1st Feb, 2019

Hello arjun,

Congratulations for your performance 

Total number of students appeared for jee main 2 paper id 180000.

So you can calculate your rank by this method:

180000×(100- obtained percentile)/100

By this your rank would be around 34000.

Good luck

2 Comments
Comments (2)
1st Feb, 2019
Thank you brother! 
Reply
1st Feb, 2019
1,45,386 is the number of students who appeared for examination this year !!So the Rank Formula for this year is (100-your total marks)x145386/100.Hope this will help !!Thank you!!
Reply

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