Hello aspirant

The total number of students attended for JEE paper 2 is 6.35 lakh.

Your percentile is 93 by calculating using the formula

((100-your percentile) x 6,35,000) / 100 expected all India rank will be around 44,450.

Where 8,39,010 is the number of candidates registered for JEE mains paper 2 2020.

93 percentile is quite good score but it's difficult to say wheather you will be able to get IIT college or not. It depends on cut off of the colleges and cut off depends on various factors such as number of candidates appeared and qualified in exam, difficulty level of exam, previous year's cut off etc.,

2019 cut off for qualifying JEE mains was 89+ score,but cut off score for IIT s is very high.

For securing CSE in top 5 NIT colleges they gave a cutt off with 99.4 percentile that means a rank below 5000 and marks more than 190 out of 300.

You can check the cut off in colleges around your location like IIT pune, College of engineering pune, NIT pune.

For college prediction following link may help you

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

Hello,

Well, You can first calculate your rank by rank=(100-percentile)*145386/100.With this calculated rank, you can now compare previous year's cutoff for different colleges and can predict some of the colleges you may be getting. This is not the final rank.

This is the college predicator 2020 for JEE-examinations just go there enter your details because along with your rank a lot of other factor i.e Home state,reservations.

College predictor for JEE Paper