I've got 25,867 OBC category rank in jee mains 2019. Can I get CSE in HBTU Kanpur in the second round of counseling?
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Hello Aakash, with 25,867 OBC rank you can easily get CSE in HBTU Kanpur since normally a rank of 20-30K is required for it.
For general, a minimum of 97-98 percentile is required to get admission in B.Tech programme in HBTU, Kanpur. The cutoff rank for CSE is around 21,000 which is around 97.5 percentile. The last ranks for other branches are almost EE - 31,000; ME - 29,000; CE - 35,000 etc.
Also feel free to use this JEE Main college predictor to be assured of which college you might have a chance of getting into
https://engineering.careers360.com/jee-main-college-predictor
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