College of Engineering and Technology, Bikaner Eligibility

BE COMPUTER ENGINEERING ELIGIBILITY CRITERIA:
A candidate must hold a graduate/post graduate degree with a minimum of 10+2 years of formal education with Mathematics as one of the subjects from a recognized university/institute.
Candidates, who are in the final year, are also eligible to apply provided they complete their formal education.
The candidates must be a minimum of 17 years of age or above. However, the maximum age is set to be 23 years.
Candidates should have passed 12th standard or equivalent examination recognized by AICTE and State Government in Physics and Mathematics subjects along with Chemistry / Electronics / Biology / Biotechnology / Computer Science as optional subjects and English as one of the languages of study.
Students must have obtained a minimum of 45% marks in aggregate (40% for SC/ST/OBC).
The colleges which accept admission through JEE Main follow the eligibility criteria set by the JEE authority.
For being eligible for JEE Main, a candidate must secure minimum 75% aggregate marks at the 10+2 level.
For the other colleges, which may not consider JEE Main Score, the minimum required aggregate marks ranges between 50% - 65%. However, one must note that the eligibility criteria for BE Computer Engineering varies from college to college.
BE MECHANICAL ENGINEERING ELIGIBILITY CRITERA:
To apply for any of the engineering colleges, the candidates should fulfill the minimum eligibility criteria.
The candidate should have completed class 12th from a recognized board.
The candidate should have studied Physics, Chemistry, and Mathematics in class 12th.
They should have scored at least 60% marks in class 12th. Some relaxation is provided to the reserved categories usually by government institutes.
The candidates who have done a 3-year diploma in the related field, from a recognized university can apply for lateral entry. HOPE THIS HELPS! WISH YOU LUCK FOR YOUR FUTURE ENDEAVORS!

Visit for details: http://www.btechadmission.co.in/
I've provided enough details regarding cutoff to save you plenty of time looking around. Assuming you already know enough about the two exams, I'll just mention important bits to know the difference between the two exams better.
Number of applicants in COMEDK'18: 72,152 (against 20,000 seats, out of which 18733 are reserved for engineering)
Number of applicants in KCET'18: 1,98,639 (against 19942 seats, out of which 16942 are reserved for engineering)
Both exams are different in the following ways:
More competition in KCET, compared to COMEDK.
Tough COMEDK, easier to get a good rank, whereas the opposite is true for KCET.
COMEDK colleges more expensive than KCET (this is because COMEDK has more private affiliation) [53,460 (KCET seat) and 1,83,600 (COMEDK seat)]
Reservations for SC/ST/OBC high in KCET, more seats for GEN in COMEDK.
Outsiders not allowed to write KCET, whereas COMEDK is all-India EE.
Keeping the above points in mind, let's look at the cutoffs.
Generally, the closing rank in KCET for RVCE:
Civil: 3354
CSE: 573
ECE: 798
EE: 622
ME: 584
Now, COMEDK for RVCE:
Civil: 3199
CSE: 180
ECE: 390
EE: 732
ME: 829
From the above tables, we can suppose that KCET which see a number of applicants higher than that of COMEDK, but yet latter has a similar cut off as KCET but for less number of candidates. Although one can't exactly tell what should be your rank to be able to get a seat in RV college of engineering as the last year cutoffs can't exactly predict your chances as the cutoffs are based on so many factors so it isn't ever constant.
But to be on safe side, its best if you have:
Civil: 2500 (COMEDK), 2700 (KCET)
CSE: 150 (COMEDK), 300 (KCET)
ECE: 300 (COMEDK), 300 (KCET)
EE: 600 (COMEDK), 500 (KCET)
ME: 700 (COMEDK), 500 (KCET)
Visit for details: http://www.btechadmission.co.in/