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!

NIT Delhi

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