Hello

Admission will be offered under two streams.

Stream1: Based on JEE Main scores and +2 Marks for 70% of the seats

Candidates have to apply and appear for JEE Main and participate in the counselling process.

Admission will be based on JEE Main percentile score and normalized +2 marks with a weightage of 25% & 75% respectively.

Based on their JEE Main scores, they have to further register and apply online for admission to the university.

SASTRA University will allot ranks to the candidates on the basis of their scores in JEE Main and class 12th marks.

Further, candidates have to participate in an institute level counselling and get their documents verified and pay the fee.

Stream2: Based on +2 marks for 30% of seats

Under this channel, the candidates can apply for direct admission based on their class 12th marks.

A merit list will be prepared on the basis of +2 marks by the university.

In order to have the same scale of comparison, the +2 or its equivalent aggregate marks will be normalized.

Further, candidates have to participate in an institute level counselling and get their documents verified and pay the fee.

Final admission will be based on the merit list, counselling, fee payment and document verification.

For more details do visit the link given below :-

https://www.careers360.com/university/shanmugha-arts-science-technology-research-and-academy-thanjavur

I hope this information helps you.

Good Luck!!

Hello dear

Sastra University for Btech admission , 70% of the total seats are filled on the basis of the JEE Mains rank and rest 30% based on class 12 merits. So your mains score was quite good rank to secure seat in any branch of  Sastra University.

All the best!!

Basith, There’s nothing like minimum mark in JEE for admission in SASTRA. The weightage for JEE Main scores and +2 marks will be 25% & 75% respectively. The aggregate JEE Main score and normalized +2 aggregate will be taken together as per the weightage and applicants will be ranked based on the resultant combined score.

For example: If a student secures a JEE Main Score of 185 and normalized +2 aggregate is 95%, the combined score of the student will be (0.25*185) + (0.75*95) = 117.5

The combined score of all students will be used to determine the SASTRA Rank and admission will be based on the SASTRA Rank secured.

Ps: you need to have 100+ of JEE score and 95+ % in 12th as minimum to get into SASTRA under stream 1