JEE MAIN 2017 COUNSELLING
NATIONAL LEVEL ONLINE/OFFLINE TEST
JEE Main 2017 (Joint Entrance Examination) is organized by CBSE for engineering aspirants who wish to pursue various UG programs offered by different engineering colleges across India. The JEE Mains Results is expected to be declared by April 27, 2017 and successful candidates are allowed to participate in JEE Main Counselling as per their order of merit. The candidates are offered admission in their choice of course and college, subject to availability and on the basis of All India Rank of a candidate.
The JEE Main 2017 Counselling Process is tentatively scheduled to start from June 24, 2017. The admissions to the seats offered by Participating Institutes through JEE Main 2017 and JEE Advanced 2017 take place through three counselling process mentioned below and the JEE Main Seat Allotment Process.
The tentative JEE Main 2017 Counselling schedule is being mentioned below. Candidates can check detailed schedule to plan their participation accordingly:
|Online Registration and Choice Filling||June 24, 2017 - June 29, 2017|
|Display of mock seat allocation||June 27, 2017|
|Seat Allotment (First round)||June 30, 2017|
|Physical Reporting (First round)||July 1, 2017 - July 5, 2017|
|Display of filled / available seats||July 6, 2017 (10:00 am)|
|Seat Allotment (Second round)||July 6, 2017 (05:00 pm)|
|Physical Reporting (Second round)||July 7, 2017 – July 9, 2017|
For selection through JEE Main 2017, CBSE would incorporate a two stage process. In this process, your performance in the Class 12 or equivalent qualifying examination and in JEE Main examination is considered with 60% weightage is given to the candidate's JEE Main performance and 40% weightage is given to Class 12th or equivalent qualifying examination performance
So lets discuss one by one, the role of qualifying marks in the selection-
Qualifying marks have been divided in three categories:
Add Z number of students in the group with total marks less than the student whose percentile is required to be estimated.
Number of students who have registered for exam: 14198
Number of students who have appeared for exam: 13711
Case 1: A candidate has scored 45% marks in board & 2328 students have scored below him – Percentile score for 45% marks in board: 2328*100/13711=16.09
Case 2: A candidate has scored 50% marks in board & 6885 students have scored below him – Percentile score for 50% marks in board: 6885*100/13711=50.21.
Important points to note-
JEE Main score- The JEE Main score will be determined on the basis of your performance in Paper-1 & Paper-2 and 60% weightage will be given to your performance while preparing the rank list. So, it is very important to perform well in it. Moreover, your JEE Advanced Exam eligibility will be determined on the basis of the JEE Main 2017 Score.
|Performance in JEE Main||60|
Performance with respect to above parameters gives you the chance for choice locking for NIT’s & for other universities too. Admission in NIT’s is solely the decision of NIT authorities only. If there is any student who have cleared JEE Advanced then preference will be given to him/her over the one who have applied through JEE main score only.There are almost 25000-30000 seats available in IIT’s, NIT’s & IIIT’s in total. Up to rank of 70000 you can expect last NIT & in extreme cases up to 80000.
|JEE MAIN 2017||JEE Mains Result||JEE Main Syllabus||JEE Main Cut off|
Colleges other than IIT’s have some different selection process which we have mentioned here:
You can understand this with an example. Let us say a student has scored total 232 & other one scored 212 marks out of 360, & subject wise weightage is given below:
|Subject||Students (1) weightage of marks||Student (2) weightage of marks||Questions attempted||Negative marking|
|Physics||70||82||Total attempt- 67||3|
|Wrong attempt- 3|
|Chemistry||70||76||Total attempt-79 Wrong attempt-26||26|
Student 2 would have been ranked higher than student 1 in the merit list & will be called first for choice locking. Student with higher absolute value will be ranked on top in the merit list.
Note: Absolute value is the ratio of positive & negative marks obtained by a student. After adopting this process if there is no any chance of further resolution then students will be allotted same rank.
Student 1 has solved only 64 questions in total, & most of them are correct. In case of student 2 who has solved 79 questions, attempts 26 incorrect questions. Conclusion is that if 2 has not solved too many questions then he has more chances of getting good score & rank in merit list too. So try to avoid temptation of solving questions with probable answer.JEE Main 2017 Admission Process in B.Arch/B.Planning courses in colleges other than IIT’s:
Chances of Selection:-Possibilities of selection will entirely depend on certain factors given below:
Note: If a student who has received call regarding short listing, but have not attended the count selling proceedings then in place of that other student with lower rank will get call for choice licking. We strongly advise you to not to miss the counselling.
The process used for admission in various NIT’s, IIIT’s & CFTI’s has been adopted in JEE Main 2017. This process has been carried out on class 12TH marks only.
This is the way of indexing a large amount of ambiguous data in an unambiguous way. It does not include any type of redundancy in the date but presents a clear picture of figures.
To identify which subjects have been considered for all calculations, you can go through the eligibility criteria of JEE Main. In case of alternative subject best marks will be considered.
Normalized Score would be compiled as (Bnor): 0.5* (b1+b2)
Normalization formula is slightly different for (CFTI’s). Here performance in JEE (Main) & Normalized board performance matters the most. It has been considered in the ratio of 60: 40.
Apply this formula, C= 0.6*A0+ 0.4* B nor.
Note: Parameter B 0 has been decided only to perform basic calculations for normalization.
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