SMEAT Eligibility

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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)
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Thapar Institute of Engineering and Technology admits students in B.E. Courses through the following two modes. Mode - 1- (Based on JEE Main 2020 score) for 70% of seats : 70% of the total BE/B.Tech seats shall be filled on the basis of percentile based merit of JEE Main 2020 examination. Eligibility criteria : The candidate has passed 10+2 or equivalent examination with at least 60% marks (55% for
SC/ST candidates) in aggregate of three subjects, namely, Physics, Mathematics and any one subject out of Chemistry, Biology, Biotechnology and Technical Vocational subject or Minimum 60% (55% for SC/ST
candidates) marks in a Diploma recognised by AICTE or a state board of technical education of at least 3-year duration and also has appeared in JEE (Main)-2020. Mode - 2- (Based on 10+2 marks in PCM) for 30% of seats : 30% of the total BE/B.Tech seats shall be filled on the basis of merit of marks secured in 10+2 or equivalent examination in aggregate of three subjects, namely, Physics, Mathematics and any one subject out of Chemistry, Biology, Biotechnology and Technical Vocational subject. Eligibility criteria : The candidate has passed 10+2 or equivalent examination with at least 75% marks (70% for SC/ST candidates) in aggregate of three subjects, namely, Physics, Mathematics and any one subject out of Chemistry, Biology, Biotechnology and Technical Vocational subject and also has appeared in JEE (Main)-2020. So you can apply in TIET through the above mentioned two modes as per your eligibility. Further if you wants to go through there Mode-1, than keeping the last year of trends in consideration, you have to score above 75 percentile in JEE Main for asked engineering branch.