Hi Dhananjay,

Congratulation on getting 87.28 percentile in first attempt of jee mains as this performance of yours proves your caliber. As there is no offical notice about the cutoff and all, based on your percentile your general rank is between 137,027-157,027,on my opinion you should work more harder to confirm your seat in your dream nit by appearing in JEE mains april attempt by scoring good marks.

Hello. It is difficult to exactly predict your rank or college now because it may change after the April session of Jee Mains. But for now as per your current percentile your probable category rank can be around 44420 - 48420. As per your rank it will be difficult for you to get CSE branch in top NITs. But you might get into NIT Manipur under home state quota.

You can check previous years cutoff for different NITs & IIITs from the below link.

https://www.google.com/amp/s/engineering.careers360.com/articles/jee-main-cutoff/amp

Hope this helps. Good luck!

The exact prediction would be difficult about the qualifying score or rnk as JEE Main is being conducted twice by NTA for the first time. Based on this score your probable rank will be 137,027 - 157,027 and your probable category rank range will be 44420 - 48420, With this rank you may be eligible for JEE Advance but the chances are less to get a seat in NITs.

I would suggest you to appear for JEE Main April session again &  score well to get a seat in NITs.

NOTE: Rank Predicted is on the basis of the expected total number of students i.e. 1152650 appearing in the JEE Mains April Session.

Good Luck!

Dear aspirant,
Based on your score your rank will be in the range 137,027 - 157,027.The chance of getting seat in an NIT is less.Any way Let's hope for the best. I recommend you to appear for jee main April session to have a better chance to get into an NIT.There is crash courses by Entrance360 for this.Go to the link for more details
https://www.entrance360.com/engineering/jee-main/#courses

Please tell me your marks that secured in JEE Main and also your rank. Then I am able to give the answer of your question.

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