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sir I got 91.50 percentile in jee mains 2020 and got 26480 obc rank i have any chance of getting any branch in hbtu and already 2 round of counselling are over. I m scaring about admission in hbtu


ANURAG JAISWAL 1st Nov, 2020
Answers (2)
Bxhb 1st Nov, 2020

Hey aspirant !!


Congratulations you've score great !! Ofcourse you can get top branches in HBTU. The previous year cutoff of CSE branch for OBC is 14038-23874. So wait you will definitely get good branch in HBTU. From my prediction you will get IT or EE branch in HBTU.


To know your admission chances in JOSSA & State counseling you can use JEE Main college Predictor tool.

https://engineering.careers360.com/jee-main-college-predictor?icn=QnA&ici=qna_answer



Hope this helps you !!

ARYAN SAGAR 1st Nov, 2020
Dear student,

Hbtu kanpur is one of the oldest and reputed engineering colleges in India. The admision in undergraduate course (B.tech) is only given on the basis of jee mains rank.

Now, as you have secured 26480 obc category rank in jee mains then your crl rank must be above 80000. And with this much huge crl rank you cannot get hbtu kanpur as per seeing the last year cutoff trends for obc category candidates.

Here are the last year cutoff closing ranks for round 4 or spot round counselling for obc category candidates on the basis of their CRL ranks :-

Bio-chemical Engineering - 71450
Chemical Engineering - 59321
Civil Engineering - 53819
Computer Science and Engineering - 31889
Electrical Engineering - 53560
Electronics Engineering - 52255
Food Technology - 79524
Information Technology - 37098
Leather Technology - 79625
Mechanical Engineering - 56638
Oil Technology - 74514
Paint Technology - 72440
Plastic Technology - 77439

So, as you can see, with this crl rank you can't get HBTU kanpur.
So, better you try for some other colleges.

Also check out our jee mains college predictor tool to get an appropriate idea of colleges that you can get on basis of your jee mains rank :-
https://engineering.careers360.com/jee-main-college-predictor?icn=QnA&ici=qna_answer
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