Hi,

This year around 9.65 lakh students appeared for JEE Mains. Hence your rank would be around be: JEE Main 2019 rank (probable) = (100 - NTA percentile score ) X 965000 /100 = 1,66,945. Around 2,24,000 students are selected for JEE Advanced. To be eligible for JEE Advanced, a general category student can expect a cutoff of 75-78 percentile. Also, you will need at least 75% marks in class 12th or have to be in the top 20 percentile in your respective board.

To get admission in an NIT, you will need a percentile of 97+. So, only hope for you is your home state quota but the chances are negligible.

Hope this helps.

Cheers!

Hi, lets check your rank first. With the given percentile and knowing that around 10 lakh student sat in the examination, your rank is around 1,83,000. This rank is in the borderline of eligibility of JEE Mains and no guareented comments can be made until the combined results are out. Now coming to NITs, sorry to say but this rank would not fetch any NIT.