Hello manasi

Well, it's very difficult as of now,to predict the cutoff for JEE mains paper 2 based on percentile.You can first calculate your rank by rank=(100-percentile)*145386/100.With this rank,you can compare previous year's cutoff for different colleges and can predict some of the colleges you may be getting.But remember this

Hi Muhammed, 

According to your percentile, your general probable rank can be translated to be 2,24,114. However, this rank is subject to increase or decrease as and when the NTA announces the final percentiles after normalizing the scores for both the sessions of JEE Mains. 

And to qualify for JEE

Hello! Saikat

Your percentile is awesome.Your rank will be enough to get NITs.I can give you rough idea about your rank under general quota  using this formula:{100-percentile score}×145386/100.where 145386 total number of aspirants that apoera in 1st phase of jee mains paper 2 examination.Your rank will be around 1.7k.Usually with

Hello Richa,

 I would like to suggest you to give second attempt so that you can improve your score.

Use this formula to predict the rank by your percentile.

Rank Formula = (100 - your total score) X 145386 /100.

On the basis of the rank only the cutoff for

Hello Kajal

To give 2019 advance first you have to be eligible for JEE mains 2019 and then you have to crack JEE mains paper then only then you are eligible for JEE advance.

JEE mains eligibility Criteria :

• There is no age limit to be eligible for JEE Main

Hello,

I would like to tell you that the Diploma holders are not permitted to take JEE Mains as Diploma program is for 3 years.

There are many other eligibility criteria for JEE like:

• For Btech courses via JEE Main, candidates need to be from Scienece stream only with PCM

• The mode of the exam is computer based online mode only.
• Also the registrations will be opened tentatively from first week of may to second week of may.
• both for paper