KCET Cut-off

Hello Afityh, I must say that you have got very reasonable rank in CET though it may not be the best, but still good. With this I am sure you will get into a good college for ECE or EEE. However to help you I am attaching a link that will predict the college and the branch based on the ranking in KCET. https://www.shiksha.com/engineering/kcet-college-predictor
Best wishes to you.

Visit for details: http://www.btechadmission.co.in/
Aerospace engineering has become a very sought after branch in a small span of time. There is an immense future job prospects and the college is number one. Aerospace has become top 5 branches of RVCE in recent times. CSE&get;ISE&get;ECE&get;MECH&get;AERO
so the cutoff which you should be looking at must be around 2000 in first round and most probably around 2500 in second round. I will also advise you to keep checking the COMEDK official website as they have not uploaded while I am writing this. I hope this will help. Thanks. Visit for details: http://www.btechadmission.co.in/.

KCET cut off is below 2000 ranking for computer science engg. It also depends upon the people choose for CS in our college. But the ranking has to be minimum for CS because NMIT has great reputation for CS since they provide 90% + campus placements and well equipped lecturers.

Here you are able to read reviews on B.Tech IT course given by the bona fide students of this college
https://www.shiksha.com/b-tech/course/b-tech-in-information-science-technology-presidency-university-Bangalore-yelahanaka-362003.

Hi, Visit this link to check colleges based upon your rank:
https://www.shiksha.com/b-tech/resources/kcet-college-predictor
Good Luck. Hope it helps.

Hi, Visit this link to check colleges based upon your rank:
https://www.shiksha.com/b-tech/resources/kcet-college-predictor
KCET Cut off - General - 1478
SC - 7615
ST - 3610
Good Luck. Hope it helps.

Please visit this link hope this will help you
https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.shiksha.com/b-tech/amp/kcet-exam&ved=2ahUKEwjezev45KriAhXEheYKHSx0AosQFjAAegQIBhAB&usg=AOvVaw0G-4-6zyFxSNU3C6nAUN-0&cf=1