Hi The minimum marks required to get admission to NIT Delhi in computer science engineering is the cutoff, which the institute releases for all category candidates. The cutoff is released with opening and closing ranks for candidates belonging to home state that means Delhi, in this case and other state, candidates from outside Delhi. The opening rank means that admission starts with it and closing rank means admission ends with it. You will not get admission in NIT Delhi if you have marks less than its closing rank for your particular category and state. The cutoff for NIT Delhi 2017 is mentioned below : Home State General Category Opening Rank - 3287 Closing Rank - 11564 OBC-NCL Category Opening Rank - 2076 Closing Rank - 3822 SC Category Opening Rank - 491 Closing Rank - 3403 ST Category Opening Rank - 818 Closing Rank - 4014 Other State General Category Opening Rank - 3421 Closing Rank - 7816 OBC-NCL Category Opening Rank - 1764 Closing Rank - 6742 SC Category Opening Rank - 785 Closing Rank - 1530 ST Category Opening Rank - 342 Closing Rank - 716 You can use this cutoff and estimate it for this year. All the best!

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