my name is BALARAJU i got emcet in 47 marks with rank 99000 in which clg i got seat in any group in Chittoor

Hello,

You haven't mentioned your category assuming that you belong to General category here is the list of institutes in which you have chances to get seat in Chittor

• SRINIVASA INSTITUTE OF TECHNOLOGY AND MANG STUDIES: EEE
• SRINIVASA INSTITUTE OF TECHNOLOGY AND MANG STUDIES: Mechanical
• SRI VENKATESWARA COLLEGE OF ENGG. AND TECHNOLOGY: Civil
• VEMU INSTITUTE OF TECHNOLOGY: Mechanical

Cut off is subjected to change it may increase or decrease depending on the total number of candidates applied, difficulty level of examination and total number of seats available etc. To check the cut off click use the link given below

https://engineering.careers360.com/articles/ap-eamcet-cutoff

To get the list of more institutes in which you may get seat you may use our AP EAPCET college predictor tool

https://www.careers360.com/ap-eamcet-college-predictor?utm_source=qna&utm_medium=ini-cet_cp

Hello Aspirant,

It is not possible to provide you an exact list of colleges, but we can predict some chances of your admissions in some colleges based on your rank because cut-off keeps on changing depending upon various factors such as :

** Number of candidates appeared for examination.

** Top score for that particular year.

** Total number of seats available.

Based on previous years (2020) allotment of seats, on such rank you have good chances in Chittoor in following colleges :-

SRINIVASA INSTITUTE OF TECHNOLOGY AND MANG STUDIES

SRI VENKATESWARA COLLEGE OF ENGG. AND TECHNOLOGY

VEMU INSTITUTE OF TECHNOLOGY

You can check out previous year's cut-off from the link given below :-

AP EAMCET Cutoff 2022 (Soon) - Check Previous Year Cut Off Here (careers360.com)

You can use the AP EAMCET 2022 College Predictor tool to predict the colleges that you can get based upon your AP EAMCET 2022 rank.

You can use AP EAMCET College predictor from the link given below :-

https://www.careers360.com/ap-eamcet-college-predictor?icn=QnA&ici=qna_answer

I hope this information helps you.

Good Luck!!