Hello Aspirant,

JEE Main cut off for Advances exam was 74 for General, 45 for OBC, 29 for SC, 24 for ST and -35 for PwD Categories. Other than this, there is also an opening and closing cut off of JEE Main. It is the maximum or minimum rank at which candidates admission in NITs, IIITs, and CFTIs.

We all know that cut off is the minimum marks required to qualify the test. However, in JEE Main 2019, there are no minimum marks required to qualify it. To make it simple, it is like a candidate with a negative score also gets a rank in the JEE Main exam, but he/she may not be eligible to appear for JEE Advanced 2019. Hence, to qualify JEE Main 2019 there is not cut off but the candidates need to obtain marks to qualify for JEE Advanced 2019 .

Admission to NITs, IIITs and CFTIs participating through Central Seat Allocation Board (CSAB) is done on the basis of AIR. To be considered for admission in NIITS, IITs, etc, the candidate should have secured at least 75% marks in the class 12th examination (65% in case of SC/ST), or be in the top 20 percentile in the 12th class examination conducted by the respective Boards.

All the best!

Hello!

ITs hard to predict cutoff for jee mains but almost similar to last year cutoff range vary from +5 to -5.

The last year cutoff of jee mains are

For general category :-74

For obc :-45

For sc:-29

For st:-24

For pwd:-  -35 for more abount jee mains you may visit below link

Https://engineering.careers360.com/articles/jee-main-cutoff

All the best!

Hello
The cutoff percentile will be declared by NTA on 30th April 2019. So you will have to wait for now.
However, as per our experts candidates having 85 %ile or more will qualify the JEE Mains and for JEE Advanced level. This is just a predicted range on the previous years analysis and so it may vary.

Go through the below given link to get idea on previous years cutoffs:

https://engineering.careers360.com/articles/jee-main-cutoff

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