How can I get into CVR College of Engineering for UG/PG courses? Am I eligible for admission?
| Courses | Eligibility |
|---|---|
| M.E./M.Tech | Candidate must have completed B.E./B.Tech. /AMIE in CSE/ CSIT/ Electronics & Computers Engg. /IT & Computer Science and Systems Engineering or equivalent from a UGC recognised University/ Institute. Preference will be given to GATE qualified candidates over TS PGECET qualified candidates. Candidate must have secured 50% marks in group subjects taken together in the qualifying examination (45% for reserved category). |
| B.E. / B.Tech | Candidates who have passed or appeared at the final year of the Diploma examination in Engineering conducted by the State Board of Technical Education and Training, Telangana / Andhra Pradesh or any other examination recognised as equivalent thereto by the State Board of Technical Education and Training, Telangana / Andhra Pradesh. Candidates should have completed 16 years of age, as on 31st December of the year of admission. There is no upper age limit. |
Upvote 
Similar Questions for you
The TS EAMCET maximum cutoff rank required for CSE admission at CVR College of Engineering was 6590 for the students belonging to the Home State quota. Therefore, candidates must target a lower cutoff rank to get CSE admission at this college. Candidates can refer to the table below to view the CSE
CVR College of Engineering TS EAMCET cutoff 2025 vary by rounds, specialisations and categories. For BTech admission, the last round cutoff ranks ranged between 6781 and 28422, with Computer Science and Engineering being the most competitive for admission. Refer to the table below to view the last
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else.
On Shiksha, get access to
Learn more about...
Most viewed courses
Didn't find the answer you were looking for?
Search from Shiksha's 1 lakh+ Topics
Ask Current Students, Alumni & our Experts
Have a question related to your career & education?
See what others like you are asking & answering