If your JEE Main percentile score is around 88%, your rank would be between 1,20,000 and 1,80,000. This rank may not be sufficient to secure a seat in the top NITs or IIITs, but you can still consider applying to reputed private engineering colleges that accept a percentile score within this range.
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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
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