National Testing Authority (NTA), the examination conducting authority responsible for JEE will be releasing the JEE Main 2019 cut off. Since JEE Main 2019 will be held in two sessions which will be conducted in January 2019 and April 2019. The JEE Main 2019 cut off will be declared twice, both for the January 2019 session along with the April 2019 session. CBSE will no longer be conducting the exams or release the results and cut-offs.
When JEE Main results will be announced the All India Rank (AIR) will also be out. NTA will also publish the JEE Main 2019 cut off which is required to appear for JEE Advanced.
| Session | Registration | Exam dates |
| January 2019 | 1st -30th September | 6th-20th January 2019 |
| April 2019 | 8th February-7th March | 6th -20th April 2019 |
The JEE Main cutoffs for NITs is based on JEE Main qualifying marks and All India Ranks (AIR). The cut off varies based on the category and the different programs offered. The allotment of B.Tech seats in NITs are purely based on JEE Main cut off. During the counselling session, each NIT in the country releases its own cut off list in the form of Opening and Closing Ranks after each round.
Here are the expected JEE Main 2019 cutoffs for NITs:
It is based on calculations, last year’s JEE Main cutoffs, question difficulty level along with the number of qualifying students in the recent years, the number of seats available in NITs, the total number of applicants this year and many more factors.
Note: The actual JEE Main 2019 cutoffs for NITs will be released after the declaration of the JEE Mains 2019 results and the All India Rank (AIR).
The JEE Main 2019 cutoffs are as follows:
| Category | Expected Qualifying Marks |
| General | 115 |
| SC | 60 |
| ST | 50 |
| OBC-NCL | 80 |
Now, that we have understood the expected JEE Main cutoff score. Let us take a look at the JEE Main cutoffs for the previous years to establish a pattern which the aspiring students will understand better. It will help them to have all the information under one roof, allowing them to make an informed choice and to be prepared well for the JEE Main in 2019.
The JEE Main cutoffs are the minimum qualifying marks an aspiring student requires to gain eligibility to get admission into B.Tech/B.E. and B.Arch/B. Plan programs in various NITs, IIITs, and other top engineering colleges.
While the JEE Main 2018 cut offs to qualify for JEE advanced is as follows:
| General- 74 |
| ST-24 |
| SC-29 |
| OBC-NCL-45 |
| PwD-35 |
(This list indicates the minimum marks to qualify to JEE Advanced)
Many students feel that the cut off mentioned in the scorecard is the cut off in the exam. That is not true. It is actually the minimum qualifying marks to advance to JEE Advanced. Based on the opening and closing ranks the cut-offs are counselled for seats in NITs.
Here is a detailed division of JEE Main 2018 cut off marks to go for JEE Advanced.
| Category | Qualified Candidates | Maximum Marks | Minimum Marks |
| Common Rank List | 111275 | 350 | 74 |
| SC | 34425 | 73 | 29 |
| ST | 17256 | 73 | 24 |
| PwD | 2755 | 73 | 35 |
| OBC-NCL | 65313 | 73 | 45 |
Now, after such a detailed table analysis, let us study the factors which might influence the JEE Mains 2019 cutoffs:
- Total number of candidates appearing for the paper
- The difficulty level of the exam
- Number of available seats in NITs
- The overall performance of the aspiring candidate
It is very important for the aspiring candidates to understand these factors and then go on to the expected JEE Mains cutoffs. So, study hard and keep these cutoffs in your mind to help you prepare for your upcoming JEE Main exams in 2019.
Keep your mind sharp and aim for NITs.
