JNTUACE Pulivendula All Questions

Hi Ramu,

You need to provide me your category as well because there are many categories in the AP EAMCET examination.

I can provide you the AP EAMCET 2016, 2017 and 2018 cutoff ranks of JNTUA, Pulivendula for Open Category (Boys) candidates, so if you belong to a specific category,

• CSE : 9066
• ECE : 8547
• EEE : 13703
• MECH : 14186
• CIV : 13712

hello aspirant,

as per your rank you have chances in below mention colleges for computer science engineering,

SR Engineering College - [SREC] - Warangal

Sreenidhi Institute of Science and Technology - [SNIST] - Hyderabad

CMR Engineering College - [CMREC] - Hyderabad

Anurag Group of Institutions - Ghatkesar

JNTUH College of

Hello Ramakrishna, hard luck this time bro. Sorry to say but with 21615 rank in EAMCET, it is very difficult to get admission in JNTUA college of Engineering. As the cut-off rank for this college is very high and with this rank despite being from reserved category, it is very

Hi Suleman,

AP EAMCET 2019 cutoff is hard to predict.

But, I can provide you the AP EAMCET 2016, 2017 and 2018 cutoff ranks of JNTUA, Pulivendula for CSE branch.

These cutoff ranks were for BC-B (Boys) category, so you can get a clear idea.

AP EAMCET 2016-

Closing Rank

hi aspirsnt

as per the rank you secured you are not elligible to get in to the university level.so you have to go with the other affiliated colleges.

in andhra pradesh you have good colleges like gmr institute of technology,rvr college of engineering,srkr engineerinfg college,gitam,pullareddy college of engineering anmd few

• CSE - 4424
• CIV - 8973
• MEC - 11228
• CHE - 17423
• EEE - 7279
• ECE - 4854

9785 rank in OC category may not bring you the streams of your choice, but there is a possibility that you get a chance into jntu for other courses. Just wish that most of the rankers get good Central Universities and they join there so that you could fulfill your

• CIVIL ( CR 24544)
• CSE ( CR 13800)
• ECE ( CR 12885)
• MECHANICAL ( CR 21152)
• EEE( CR 16100)