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I got 76 percentile in jee mains 2019 what is the best possible rank I can get?


Polana Viswanadh2 20th Jan, 2019
Answers (3)
kashishjuneja2000_8032841 Student Expert 20th Jan, 2019

According to your percentile your rank is 209873. But this is a rough idea. The actual rank can be concluded only after the second JEE mains exam is held in april. This is because the percentile is on the basis of the total number of students appearing for the exam. And the total number of students will also include the ones who will give exam in april.

Mayank kumar singh 20th Jan, 2019
Hello,

You can calculate the jee mains rank from the below given formula:

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

For more information about this, visit the below given link:

https://engineering.careers360.com/articles/how-calculate-jee-main-2019-rank-based-on-nta-percentile-score

Good luck.




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Vikas gehlot 20th Jan, 2019

Hello

Greetings

Your percentile is 76 It means 76% student of the Total student appeared for the Exam are Behind You Therefore 24% students of the Total students appeared for the exam are Ahead of You, Therefore Your calculated rank is 209873

Your Rank would change Either Your Percentile Score is Increases in the next attempt or Number of Student increases in the Next attempt if the Number of Student Increases to 1400000 and Your percentile remains same then Your New rank would be 336000

Your rank is Not Good and You should Try in the Next attempt of JEE Main which is JEE Main April and Try to perform well

If Your Percentile Score Increases in the Next attempt Then Your Rank would Be calculated By the Following Formula

(100 - your total percentile score (Higher of two)) X Student appeared in JEE Main/100

Hope this Helps All the Best

1 Comment
Comments (1)
20th Jan, 2019
I got 44 percentile in the january mains...I'm looking forward to write mains in April how many marks obtained would get me a good percentile..I want to know the minimum percentile required to qualify in jee mains and how many bits needed to be attempted to secure the marks.
Reply

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