What is the JEE Mains cut-off 2019 for NITs?

The I2IT Pune JEE Main cutoff 2025 was released in the form of ranks for the All India categories. For the General AI category, the BE in the Computer Engineering branch had the overall cutoff ranging from 42136 in the first round to 52834 in the last round. Hence, General AI students needed to

The selection process for the All India quota students is based on their JEE Main ranks. Students who want admission to I2IT Pune need to appear in the CAP counselling process to avail a seat for any BE course. 

The minimum rank required for admission was 68635 for the General AI category. Hence

The minimum rank required to get admission at I2IT Pune would depend on the category you belong to. The I2IT Pune JEE Main cutoff 2025 for the General AI category says that a rank of at least 61370 is required to be eligible for admission to the BTech in Information Technology department. This

Students who attempted the JEE Main  should start preparing for the JEE Advanced entrance examination if they want admission at IITs. Those who score above the 98^th percentile in JEE Main session 2 should start to study for the JEE Advanced exam.

Students who get around the 95^th  and 98^th &

As of now, the easiest shift of JEE Main 2026 April session was the April 2 shift 1 exam.