Hi Sonu,Sastra UniversityEligibility:A pass in the +2 (or) its equivalent examination with Mathematics & Physics as compulsory subjects with either Chemistry / Bio-Technology / Computer Science / Biology and recognized by any State Board / Central Board or any other accepted by SASTRA as equivalent to 10+2 (or) its equivalent examination, with a minimum aggregate of 60% marks in Mathematics, Physics and Chemistry / Bio-Technology / Computer Science / Biology obtained in a single sitting.Selection Process:*70% of the candidates will be selected for Sastra University B.Tech Admissions 2014 on the basis of their performance in JEE Main 2014 and ...
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Hi Sonu,
Sastra University
Eligibility:
A pass in the +2 (or) its equivalent examination with Mathematics & Physics as compulsory subjects with either Chemistry / Bio-Technology / Computer Science / Biology and recognized by any State Board / Central Board or any other accepted by SASTRA as equivalent to 10+2 (or) its equivalent examination, with a minimum aggregate of 60% marks in Mathematics, Physics and Chemistry / Bio-Technology / Computer Science / Biology obtained in a single sitting.
Selection Process:
*70% of the candidates will be selected for Sastra University B.Tech Admissions 2014 on the basis of their performance in JEE Main 2014 and +2 examinations conducted by various boards across the country.
*For the remaining 30% of admissions, the normalised aggregate +2 marks will be considered.
The ranks of aspirants will be calculated on the basis of 75% weightage to the normalized aggregate +2 marks and 25% weightage to the JEE Main scores.
Normalization of +2 (or its equivalent) total marks: To bring all candidates from different examining authorities/ boards in the same scale of comparison and to create a list in the order of merit, the +2 or its equivalent total marks will be normalized. A first rank student of each board is considered to have obtained
100% mark and aggregate mark of all other students of the board is normalized with reference to that of the first rank student.
For example, if the first rank is 97% and an applicant’s aggregate is 90% then the
applicant’s normalized +2 marks is 92.78 (90/97x100).
Tuition Fees:
For all B. Tech. Programmes
Tuition Fees Rs 45,000 per semester
Development Fees Rs 10,000 (one-time during admission); Other fees Rs 7,000 (one-time during Admission for Personality Development, Sports, Unit Tests, Marks verification, Medical Test, ID Card, etc.).
For any other detail kindly refer below links:
Link:
http://www.sastra.edu/Link:
http://www.sastra.edu/index.php/2014-01-29-07-16-11/2014-04-14-06-49-48/advertisement-for-admissions-2014-15Hope it helps.Please feel free to revert. Click on thumbs up or select it as best answer if you find this useful.
Wish You Luck!
Hi Sonu,
Sastra University
Eligibility:
A pass in the +2 (or) its equivalent examination with Mathematics & Physics as compulsory subjects with either Chemistry / Bio-Technology / Computer Science / Biology and recognized by any State Board / Central Board or any other accepted by SASTRA as equivalent to 10+2 (or) its equivalent examination, with a minimum aggregate of 60% marks in Mathematics, Physics and Chemistry / Bio-Technology / Computer Science / Biology obtained in a single sitting.
Selection Process:
*70% of the candidates will be selected for Sastra University B.Tech Admissions 2014 on the basis of their performance in JEE Main 2014 and +2 examinations conducted by various boards across the country.
*For the remaining 30% of admissions, the normalised aggregate +2 marks will be considered.
The ranks of aspirants will be calculated on the basis of 75% weightage to the normalized aggregate +2 marks and 25% weightage to the JEE Main scores.
Normalization of +2 (or its equivalent) total marks: To bring all candidates from different examining authorities/ boards in the same scale of comparison and to create a list in the order of merit, the +2 or its equivalent total marks will be normalized. A first rank student of each board is considered to have obtained
100% mark and aggregate mark of all other students of the board is normalized with reference to that of the first rank student.
For example, if the first rank is 97% and an applicant’s aggregate is 90% then the
applicant’s normalized +2 marks is 92.78 (90/97x100).
Tuition Fees:
For all B. Tech. Programmes
Tuition Fees Rs 45,000 per semester
Development Fees Rs 10,000 (one-time during admission); Other fees Rs 7,000 (one-time during Admission for Personality Development, Sports, Unit Tests, Marks verification, Medical Test, ID Card, etc.).
For any other detail kindly refer below links:
Link: http://www.sastra.edu/
Link: http://www.sastra.edu/index.php/2014-01-29-07-16-11/2014-04-14-06-49-48/advertisement-for-admissions-2014-15
Hope it helps.Please feel free to revert. Click on thumbs up or select it as best answer if you find this useful.
Wish You Luck!
Hi Sonu,Sastra UniversityEligibility:A pass in the +2 (or) its equivalent examination with Mathematics & Physics as compulsory subjects with either Chemistry / Bio-Technology / Computer Science / Biology and recognized by any State Board / Central Board or any other accepted by SASTRA as equivalent to 10+2 (or) its equivalent examination, with a minimum aggregate of 60% marks in Mathematics, Physics and Chemistry / Bio-Technology / Computer Science / Biology obtained in a single sitting.Selection Process:*70% of the candidates will be selected for Sastra University B.Tech Admissions 2014 on the basis of their performance in JEE Main 2014 and ... view more