JEE Advanced Cutoff 2025: Qualifying Marks and IIT Branch Ranks

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Understanding JEE Advanced cutoffs and IIT closing ranks helps you set realistic targets and make strategic choices during JoSAA counselling. Here is a complete analysis.

JEE Advanced 2024 Qualifying Cutoff

Minimum marks required to appear in IIT rank list (approximate): General: 25% in each subject AND 40% aggregate. OBC-NCL: 22.5%/36%. SC: 12.5%/20%. ST: 12.5%/20%. In raw marks (360 total): General needs ~90 marks with no subject below 30.

IIT CSE Closing Ranks (2024)

IIT Bombay CSE: ~100 (General). IIT Delhi CSE: ~120. IIT Madras CSE: ~200. IIT Kharagpur CSE: ~300. IIT Kanpur CSE: ~250. IIT Roorkee CSE: ~400. IIT Guwahati CSE: ~1000. These change ±15% year to year.

Branches vs Rank Chart

Rank 1-500: IIT B/D/M/K/KGP/R top branches (CSE, EE). Rank 500-1500: IITs older branches (Mechanical, Chemical, Civil) + newer IITs CSE. Rank 1500-5000: Middle IITs good branches. Rank 5000-12000: All IITs — consider any branch vs NIT CSE.

OBC-NCL and SC/ST Advantages

Category-wise cutoffs are significantly lower. OBC-NCL students can qualify JEE Advanced with ~70 marks where General needs 90. SC/ST students with 50+ marks in each subject often qualify. Category candidates should target IIT Bombay/Delhi level — seats are available at lower ranks.

JEE Advanced Score Target Setting

Top 5 IITs best branches: 200+ marks (top 500 rank). Old IITs any branch: 150+ marks (top 2000 rank). All IITs: 120+ marks (top 5000 rank). NIT + IIT combination: 90-120 marks may qualify for IITs but secure NIT via JEE Main.

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