JEE Main Cut off 2020 for January session will be released with the announcement of result by NTA. The cut off mark is the minimum qualifying mark which must be obtained in order to be eligible for B.Tech/B.E. and B.Arch/B.Plan admissions in NITs, IIITs and other top engineering colleges accepting JEE Main score.
Latest: JEE Main Result – April 2019 has been announced on April 30. Download Now
JEE Main Cut Off April 2019
|Reservation Category||Cut off for Paper 1|
|Common Rank List||89.7548849|
|General – EWS||78.2174869|
|Other Backward Classes (OBC-NCL)||74.3166557|
|Scheduled Caste (SC)||54.0128155|
|Scheduled Tribe (ST)||44.3345172|
JEE Main cut off is the closing rank at which admission to a particular B.Tech program in a particular institute is closed.
The final cut off will be released by NTA soon. For the first time, NTA will use percentile score and Normalization Procedure to rank the candidates after compiling the result of both January and April sessions.
With the announcement of the qualifying marks, NTA also releases the All India Ranks (AIR) for JEE Main which will be considered in seat allotment process. Check JEE Main Counselling Process
JEE Main Previous Years Cut off
JEE Main cut off is also available institution-wise and branch-wise and it varies from candidate to candidate as per the institution/ programme/ category. The JEE Main 2018 Cut Off to qualify for JEE Advanced was 74, 45, 29, 24, -35 for CRL, OBC-NCL, SC, ST, PwD categories respectively. Read the following article to know more about JEE Main cut off, qualifying marks, previous years’ trends and factors affecting cut off.
Factors Affecting JEE Main Cut Off
There are numerous factors that affect JEE Main cut off. Joint Admission Board has already announced that class 12th marks will no more be considered for calculating AIRs in JEE Main exam. This simply hints that applicants need to perform even better in the entrance for securing a good rank in JEE Main 2019. Check JEE Main 2019 April Paper Analysis
The following factors play a vital role in determining the JEE Main cut off every year:
Latest News Updates
- Chapters for JEE Main 2020 Last Minute Preparation
- NTA Security Measures Expected for JEE Main 2020 (January Session)
- JEE Main 2020: Rise in Number of Female Aspirants for January Session
- 9.65 Lakh Candidates Applied for the JEE Main 2020 January Exam
- JEE Main 2020: NTA Releases Official Notification for CBT Exam
- Total number of candidates appeared for the exam
- Level of difficulty of exam
- Number of available seats
- Overall performance of the candidate
JEE Main Cut off Trends
The JEE Main 2019 cutoff will be updated here as and when announced meanwhile candidates can take a look at the previous year cutoff. JEE Main 2018 Cutoff is the lowest in the past 6 years.
|Category||JEE Main 2018 Cut Off||JEE Main 2017 Cut Off||JEE Main 2016 Cut Off||JEE Main 2015 Cut Off||JEE Main 2014 Cut Off||JEE Main 2013 Cut Off|
JEE Main Qualifying Marks
The qualifying marks for JEE Main 2020 can be predicted from the following table which depicts the JEE Main 2019 Cutoff for JEE Advanced.
|Category||Qualified Candidates||Maximum Marks||Minimum Marks|
|Common Rank List||111275||350||74|
JEE Main Counselling
JoSAA and CSAB holds the seat allotment process for admission in 100 participating institutes (23 IITs, 31 NITs, 23 IIITs and 23 GFTIs). Candidates will be allocated seats based on the merit, availability of seat and choice preference (at the time of registration for counselling). Prior to the actual allotment of seats, mock allotment will be held in 2 rounds. This way, candidates will be able to know and choose better preferences for themselves in the actual rounds of seat allotment.
Know more about JEE Main Counselling
JEE Main Marking Scheme
JEE Main result also determines whether the candidates have qualified for JEE Advanced or not. The marking scheme followed to calculate JEE Main scores and ranks are as follows:
- Maximum score for paper 1 is 360 marks.
- Maximum score for paper 2 is 390 marks.
Marks obtained in Paper 1 = Number of correct answers x 4 – Number of incorrect answers x 1.
Marks obtained in Paper 2 = Number of correct answers in Maths and Aptitude Test x 4 – Number of incorrect answers in Maths and Aptitude Test x 1 + marks obtained in Drawing Test.
Calculation of Ranks
Based on obtained percentile in JEE Main through normalization procedure, candidates will be allocated the respective ranks. In case 2 or more candidates obtain equal marks then inter-se-merit is applied to determine the ranks and break the tie.
Tie-break criteria (other than IITs)
Higher percentile in Mathematics > Higher percentile in Physics > Higher percentile in Chemistry > candidate elder in age
Note: In case the tie is not broken, the candidates will be given same ranks while declaring JEE Main Result 2020.