Filter IconAll FiltersClear All
Thanjavur
Information Technology
B.E. / B.Tech
Full Time
LocationUp Arrow Icon
Search IconClose Icon
Total FeesUp Arrow Icon
  • (2)
  • (1)
  • (1)
RatingUp Arrow Icon
  • (1)
  • (1)
CourseUp Arrow Icon
11 results

B.Tech Information Technology Colleges in Thanjavur

Sort By:Popularity
4.72 L - 6.8 L
– / –
– / –
    – / –
– / –
– / –
– / –
– / –
– / –
Q&A Icon
Commonly asked questions
On B.Tech Information Technology Colleges in Thanjavur
Q:   What is the placement record of SASTRA University?
A: 

The table given below depicts the course-wise total strength and the number of students placed over the past three years of SASTRA University placements:

CoursePlacement Statistics (2021)Placement Statistics (2022)Placement Statistics (2023)
BTech

- Total students: 1,730

- Students placed: 1,504

- Total students: 1,751

- Students placed: 1,548

- Total students: 1,806

- Students placed: 1,617

MTech

- Total students: 94

- Students placed: 50

- Total students: 84

- Students placed: 50

- Total students: 124

- Students placed: 79

BA LLB/ BCom LLB/ BBA LLB

- Total students: 132

- Students placed: 112

- Total students: 132

- Students placed: 107

- Total students: 175

- Students placed: 135

NOTE: The above details are fetched from the NIRF report 2024.

Q:   How are B Tech placements at Shanmugha Arts Science Technology & Research Academy?
A: 

Shanmugha Arts Science Technology and Research Academy (SASTRA) is a deemed university.It offers 4 years B.Tech course (8 semesters) and 5 years integrated M.Tech course (10 semesters). The admission is strictly based on merit and it follows two methods:

1. 70% of seats are filled based on JEE Main score - 25% and normalised 12th mark - 75%

For example: JEE Main score is 185 and normalised 12th score is 95% then  the score of the student is (0.25×185) + (0.75×95) = 117.5

2. 30% of seats are filled based on normalised aggregate 12th mark or its equivalent marks.

For example:The first rank is 97% and the aggregate percentage is 90%, then the normalised percentage is (90/97) × 100 = 92.78

– / –
– / –
    – / –
– / –
– / –
2 L
– / –
    – / –
– / –
– / –
1.67 L - 2 L
– / –
– / –
– / –