Hello Aspirant,

As your JEE score is 48. And you are of SC category. So your Percentile based on category is:88.47 - 88.67  . Your Overall Percentile is:79.6 - 79.8  . Your Rank based on category is:9508 - 11158  . Your Predicted Rank is:185917 - 218067 .

So use this

Hi Shweta

As you have qualified JEE advanced and you want to go for CS branch, I would suggest take the jammu IIT college with your preferred branch because dropping that only because it is not top of IIT would be not appropriate.

As if you have passion in particular

Hello Aspirant, hope you are doing good.

So, I'm really sorry to tell you that with your rank,  it's not really possible to get a seat in most  IITs, because most of them have a really tight cutoff, and the number of seats are also very less.

You can check

Hello Aspirant,

As you have not mentioned your reservation category so it is quite difficult to guide you with an accurate information but if you belongs to OBC and you have mentioned your OBC-NCL rank then you can get the following IITs :-

• IIT Dhanbad
• IIT Hyderabad
• IIT Patna
• IIT

Hello Chatur Varma , If you want to know about the needed rank in IIT JEE Mains exam for pursue B.tech in Computer Science and Engineering in IITs then you can see the cutoff on careers360 page here:

https://engineering.careers360.com/articles/iits-cutoff-for-b-tech-computer-science-and-engineering

Hope it will be helpful for you :)

Hello
mudhapakamanohar , May be you do not do that thing because doing engineering in two colleges simultaneously is not possible but you can follow that:

Study Engineering & Economics Together as UGC Approves Dual Degrees. Now, students can simultaneously study subjects such as Economics and Engineering together as UGC

Hello Aspirant,

Hope you are doing well.

Just now JEE Advanced results are declared, you can check your rank in official website jee.adv.nic.in.

According to JEE Advanced result 2020 - last year's cutoff, for your score

• IIT (ISM) Dhanbad - closing rank will be 6,408 for Electrical Engineering.

This is