JEE Main 2020 Results: NTA Announces Cut-off for JEE Advanced

While the cut-off for Common Rank List is 90.3765335, for Economically Weaker Section it is 70.2435518.

Indian Institute of Technology (IIT) Roorkee declared IIT JEE Advance 2019 examination result today.

(Photo: Erum Gaur/The Quint)

Following declaration of the JEE Main 2020 results, the National Testing Agency has released the category-wise cut-offs for JEE Advanced.

While the cut-off for the Common Rank List is 90.3765335, for Economically Weaker Section it is 70.2435518 .

For Other Backward Classes, the cut-off stands at 72.8887969, while Scheduled Caste, Scheduled Tribe and and People with disability, the cut-off stands at 50.1760245, 39.0696101 and 0.0618524 respectively.

While JEE Main is the basis for admissions to NITs, IIITs and other non-IIT Centrally-funded technical Institutes, only the top 2,50,000 scorers from this test will be eligible for the JEE Advanced – which is the entrance test for admission to the various IITs.

The cut-offs for admissions to NITs, IIITs and other Centrally Funded Technical Institutes will be decided by the Joint Seat Allocation Authority at a later date.

While ranking students, NTA has taken into consideration the best score for those who have appeared for JEE Main in both January and September. NTA Score for Paper-II (B.Arch. &B. Planning) will be declared later on.

Out of the 8.41 lakh students registered for the JEE Main in April/September, only 6.35 lakh students had appeared for the exam conducted across two shifts between 1 to 6 September.

Combining January, a total of 11.74 lakh candidates had registered for JEE Main, out of which 10.23 lakh appeared for the exam.