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JEE MAIN 2017 RESULT & COUNSELLING

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JEE Main 2017 Result

CBSE (Central Board of Secondary Education) is reponsible for organizing. The JEE Mains Result would be announced by the board in final week of April-2017. All the engineering aspirants can view their results by entering their Roll Number and Date of Birth. All the successfull candidates would be allotted rank on the basis of their score secured in the written test. 

The top 2 lakh JEE Main Rank holders are shortlisted for participating in JEE Advanced 2017. JEE Advanced is scheduled to take place on May 21, 2017 for admission to engineering programmes offered by IITs, ISM Dhanbad and IISC Bangalore. All the candidates would be issued JEE Score Card after the announcement of All India Ranks (AIRs) which is expected to be announced by June 30, 2017.

The shortlisted candidates would be invited to participate in the JEE Main Counselling. The Counselling process for admission to various engineering seats across different institutes would be conducted by JoSAA (Joint Seat Allocation Authority), CSAB (CSAB SFTI and CSAB NEUT) and respective state counselling authorities.


JEE Main 2017 Result Highlights

  • JEE Main 2017 result will be in the form of JEE Main 2017 score card and All India Ranks (AIR). The score card would be available after the declaration of JEE Main AIR ranks in the final week of June-2017.
  • JEE Main 2017 Score will be used to prepare the JEE Main2017 rank list on basis of which admissions in NITs, IIITs and other Central and State Government funded institutes will be done
  •  As per CBSE 1,98,228 candidates have been shortlisted to appear in JEE Advanced 2017 based on JEE Main 2017 score
  • Registration for JEE Advanced 2017 will begin online on May 11, 2017
  • You need to sign in with your JEE Main Roll Number and password to proceed for the JEE Advanced 2017 registration process.
CategoryCut off Score
General Category100
Other Backward Class (OBC-NCL)70
Schedule Caste52
Schedule Tribe48

The All India Rank (AIR) for qualified candidates will be released on June 30, 2017 . The rank list for all categories will be based on normalization of marks obtained in JEE Main and marks in board exams. Based on these AIRs admissions will be granted in different NITs ,IIITs and other participating institutions.

Check Top NITs in JEE Main 2017


How to Check JEE Main 2017 Result

Candidates can check their result by following these simple steps. The JEE Mains Result would be announced in online mode only for the sucessfull candidates. JEE Main 

  • Enter your JEE Main 2017 Roll No. and Date of Birth
  • Click on ‘Submit’ button and your JEE Main 2017 Score card will appear
  • Download your JEE Main 2017 Score Card
JEE MAINJEE MAIN COUNSELLINGJoSAA COUNSELLINGCSAB NEUT COUNSELLINGJEE MAIN SEAT ALLOTMENT

LATEST B.TECH APPLICATION FORMS

Manipal University B.Tech Admission 2017: Apply Online Here
VIT University B.Tech Admission 2017: Apply Online Here
Jain University B.Tech Admission 2017: Apply Online Here
Chandigarh University B.Tech Admission 2017: Apply Online Here

JEE Main 2017 Score Card

The JEE Mains Result for Paper-I would be declared April 27, 2017. This score shall comprise the actual marks obtained in Paper-I of JEE Main 2017 along with the status of those who qualify for appearing in JEE Advanced provided and subject to other conditions of eligibility being met. Separately, on the basis of score in JEE Main 2017 and normalized score in Class 12th or equivalent qualifying exam (60% & 40% weightage respectively), separate rank lists will be prepared for admission to B.E/B. Tech. and B. Arch/ B. Planning (in institutions other than IITs). This will be declared by July 7, 2017.

JEE Main 2017 Rank Card indicating All India Rank, State Rank and Category Rank with total marks and marks in each paper will be available on JEE Main website after the declaration of result. Only the All India Rank (AIR) is used for admissions through Central Seat Allocation Board to NITs/IIITs/CFTIs/SFIs/Others, while other ranks are for information purposes.

No score card or rank card will be dispatched to the candidates and the candidates are advised to download their score card or rank cards from the JEE Main official website.


Expected JEE MAIN 2017 Cutoff for NITs and IIITs:

Based on the last year cutoff, you can guess that how many marks you required in JEE Main 2017 exam for admission to NITs and IIITs. In the below table we provide you the opening and closing rank of all NITs and IIITs, so that a candidate can easily understand that for getting admission what the strategy should be to crack the JEE Main exam.

Institute NameOpening RankClosing Rank
Dr. Bheem Rao Ambedkar NIT, Jalandhar430322540
Motilal Nehru NIT, Allahabad11058072
Maulana Azad NIT, Bhopal11528839
Malaviya NIT, Jaipur1625081
NIT Agartala408025737
NIT Calicut47017731
NIT Delhi70713423
NIT Durgapur737621206
NIT Goa257418383
NIT Hamirpur520012743
NIT Jamshedpur331813041
NIT Kurukshetra8748916
NIT Manipur1057738742
NIT Meghalaya1218534830
NIT Mizoram892934658
NIT Nagaland1246640511
NIT Patna686118903
NIT Puducherry537722276
NIT Raipur728017664
NIT Rourkela10056091
NIT Srinagar341129643
NIT Sikkim944631946
NIT Silchar279715557
NIT, Surathkal114468
NIT Tiruchirappalli105615416
NIT Uttarakhand553519466
NIT Warangal332100
AB Vajpayee IIIT and Management, Gwalior341920927
IIIT, Allahabad10914757
IIIT Amethi491518857
IIITDM Kancheepuram154217503
Birla Institute of Technology, Deogarh829039292
Birla Institute of Technology, Patna1677528684
Birla Institute of Technology, Ranchi481813285
Faculty of Engineering, Haridwar117036709
Institute of Technology, GG Vishwavidhalaya, Bilaspur1136636787
JK Institute of Applied Physics, Allahabad656227191

Important Points About JEE Main 2017 Rank List

JEE Main rank list will be declared on June 30, 2017 by JEE Apex Board. This list is prepared on the basis of JEE Main marks and marks in qualifying board exams. JEE Apex board releases rank list separately for all the categories, which include- Common Merit List (General), General PWD, OBC, OBC PWD, SC, SC PWD, ST, and ST PWD. Following are some important points regarding Rank List of JEE Main-

  • Normalization of marks is done on the basis of JEE Main marks and normalized score in 10+2 or equivalent board exam
  • Weightage of JEE Main score is 60% and qualifying exam marks is 40%
  • Candidates will be given one final chance to update their marks on JEE Main website before declaration of result
  • Based on this normalization, final score normalized score will be calculated and candidates will be placed in order of this score in JEE Main 2017 rank list

Admissions to NITs, IIITs and Government Funded Technical Institutes are done on the basis of these All India Ranks. Choice filling and Seat allocation process is conducted by JoSAA from 2015 onwards. Candidates are advised to download the JEE Main 2017 original score card and preserve it till the admission process is over.

JEE MAIN NITs SEAT INTAKEJEE MAIN IIITs SEAT INTAKE

Normalization Adopted in JEE Main for Admission to NITs/IIITs/CFTIs 

To normalize the marks of the qualifying examination the following formula is used. Note down the aggregate marks (A0) obtained by each student in JEE Main 2017. And now calculate the percentile (P) of each student on the basis of aggregate marks in his/her own board (B0) computed from the list of five subjects specified (each marked out of 100). Five subjects to be used for normalization:-

  1. Physics
  2. Mathematics
  3. Any one of the subjects Chemistry, Biology, Biotechnology and Technical Vocational subject
  4. One language
  5. Any subject other than the above four subjects.

In respect of 3, 4 and 5, the best mark in a given category will be chosen.

The percentile is to be computed among all students of the board whose subject combinations meet the eligibility criteria of JEE Main 2017. The variable B0 is only a base for calculating percentile (P), which is further used to get corresponding JEE (Main) marks.

Determine the JEE - Main aggregate marks corresponding to percentile (P) at the All- India level. Regard this as B1.

Also, determine the JEE - Main aggregate marks corresponding to percentile (P) among the set of aggregate scores obtained in the JEE-Main by the students of that board. Regard this as B2.

The normalized board score of the candidate is computed as:  Bfinal = 0.5 * (B1 + B2)

For the purpose of admission to CFTIs where it has been decided to use the JEE Main 2017 performance and the Normalized Board performance in the 60:40 ratio, the composite score for drawing the merit list will computed as: C = 0.6 * AO + 0.4 * Bfinal

READ MORE ABOUT Normalization of Marks in JEE Main 2017


JEE Main 2017 Counselling and Seat Allocation Process

The CSAB (Central Seat Allocation Board) is the major authority which holds admissions of candidates participating in JEE Main 2017 in the NIT’s, IIIT’s and other GFTI’s based on their JEE (Main) 2017 Rank. It has been constituted by National Institute of Technology, Patna and its Chairman is the Director of NIT, Patna. Candidates will be able to fill the choice Online for programmes/ Branches and Institutes at the time of Counselling registration. Candidates keep in touch with us by regular visit our website for latest information.

The candidates shall be allocated the seat based on their choice and the AIR (All India Rank) scored by them in Paper-I and Paper-II as given by them considered for admissions in Engineering and Architecture/Planning respectively. The candidates shall be provided with admissions in the first year of the Engineering and Architecture/Planning institutions participating in the CSAB through a centralized seat allocation procedure. The candidate must note that the process of seat allocation shall be strictly merit-based in the order of preference of choices filled by the candidate and availability of the seats.

READ MORE ABOUT JEE Main Counselling


JEE Main 2017 Document verification

The documents will be verified at the time of JEE Main 2017 Counselling/Seat allotment process i.e. admission process. If the rank is revised due to rechecking/revaluation/re-totaling of board marks, the revised board mark sheet will also be verified at the time of seat allotment process/ counselling process. The main purpose of the document verification is to verify the different records of like- qualifying examination, age, identification, category and disability (if any), and state of eligibility of the candidates. And SC/ ST/ OBC/ and PwD candidates must produce the original certificate issued by the competent authority at the time of Seat Allocation Process as well as at the time of admission.

Note: If candidates fail to produce any documents, such candidates will not be considered for counselling as well as admission.


JEE Main 2017 Answer Key

Candidates who are applying for JEE Main 2017 can check their JEE Main answer key and OMR on the official website from April 18, 2017-April 22, 2017.Note that OMR sheet will be available for Computer Based mode only.In case you find any discrepancies in the OMR sheet you can challenge it, For this you have to fill an online applicatilon form and have to pay Rs 1000/- per answer. 

Candidates can check the JEE Main Answer Key by logging in with the help of their Roll Number and Date of Birth to view the official answer key and scanned image of their OMR Response sheet.


Tie Breaking in JEE Main 2017

In case of a tie, i.e. when two or more candidates obtain equal marks (by giving 60% weightage to performance in JEE Main 2017 and 40% weightage to normalize marks in Class 12th or equivalent qualifying examination), inter-se merit of such candidates shall be decided in the following order:

Rank List for Admission to B.E/B.Tech (in Institutions other than IITs)

  • Candidate obtaining higher marks in mathematics in JEE Main 2017 exam will be given higher rank
  • Candidate obtaining higher marks in Physics in JEE Main 2017 exam will be given higher rank
  • Candidate obtaining higher marks in Chemistry in JEE Main 2017 exam will be given higher rank
  • Candidate obtaining higher total normalised qualifying exam marks will be given higher rank
  • Resolution by finding the ratio of positive marks and negative marks. Candidate having higher absolute value of the ratio will be given better rank.

Note: If the resolution is not possible after this criterion, candidates will be given the same rank.

JEE Main Rank List for admission to B. Arch/B.Planning (in Institutions other than IITs)

  • Candidate obtaining higher marks in Aptitude Test in Paper-2 (B. Arch./B. Planning.) in JEE Main 2017 exam will be given higher rank
  • Candidate obtaining higher marks in Drawing Test in Paper-2 (B. Arch./B. Planning.) in JEE Main 2017 exam will be given higher rank
  • Candidate obtaining higher marks in Chemistry in JEE Main 2017 exam will be given higher rank
  • Candidate obtaining higher total normalised qualifying exam marks will be given higher rank
  • Resolution by finding the ratio of positive marks and negative marks. Candidate having higher absolute value of the ratio will be given better rank.

Note: If the resolution is not possible after this criterion, candidates will be given the same rank.


JEE Main 2017 Frequently Asked Questions (FAQs)

Question1. What is Difference between Percentage of Marks and Percentile?

Ans. The percentile and percentage of marks are totally different entities; you should not be confused with both these entities. Normally the percentage is number out of 100 and A Percentile score is the value below which a certain percent of observations fall.  There is a formula exist to calculate the percentile which is given below. Percentile of a candidates in a board or in JEE Main 2017 exam will reflect many Candidates have scored below that candidate in his/her Board or JEE Main Examination. To understand about percentile in detail, read the below example:

Formula: The Percentile of a Candidate will be calculated as 100 X Number of candidates in the ‘group’ with aggregate marks less than the candidate/ Total number of the candidates in the ’group’.

For Example: Suppose in a particular Board: Number of Registered Candidates =13918 and Number of Appeared Candidates = 13711 then

a. A Candidate who has scored 50% marks in the Board and 2218 candidates have scored below him; his Percentile score will be calculated as follows

Percentile score for 50% marks in the Board = 2218*100/13711 = 16.18

b. A Candidate who has scored 60% marks in the Board and 6865 candidates have scored below him; his Percentile score will be calculated as follows

Percentile score for 60% marks in the Board = 6865*100/13711 = 50.07

c. A Candidate who has scored 90% marks in the Board and 13615 candidates have scored below him; his Percentile score will be calculated as follows

Percentile score for 90% marks in the Board = 13615*100/13711 = 99.30

Through this example you are cleared that percentage of marks obtained by a candidate (50%, 60% or 90%) is different from the percentile score (16.18, 50.07 or 99.30).

Question2. How to Calculate the Percentile?

Ans. The below formula is used for calculating the percentile, and to better understand also read the below example.

Formula: The Percentile of a Candidate will be calculated as 100 X Number of candidates in the ‘group’ with aggregate marks less than the candidate/ Total number of the candidates in the ’group’.

Question3. What Normalization is Adopted of Class 12th Qualifying Examination Marks in JEE Main 2017 for Admission to NITs/ IIITs?

Ans. Note down the aggregate marks (A0) obtained by each student in JEE- Main. Compute the percentile (P) of each student on the basis of aggregate marks in his/her own board (B0) computed from the list of five subjects specified (each marked out of 100). The percentile is to be computed among all students of the board whose subject combinations meet the eligibility criteria of JEE-Main.

Five Subjects to be Used for Normalization:-

  1. Physics
  2. Mathematics
  3. Any one of the subjects Chemistry, Biology, Biotechnology and Technical Vocational subject
  4. One language
  5. Any subject other than the above four subjects

In respect of 3, 4 and 5, the best mark in a given category will be chosen.

The variable B0 is only a base for calculating percentile (P), which is further used to get corresponding JEE (Main) marks. Now Determine the JEE - Main aggregate marks corresponding to percentile (P) at the All- India level. Regard this as B1.

Also, determine the JEE - Main aggregate marks corresponding to percentile (P) among the set of aggregate scores obtained in the JEE-Main by the students of that board. Regard this as B2. The normalized board score of the candidate was computed as:

Bfinal = 0.5 * (B1 + B2)

For the purpose of admission to CFTIs where it has been decided to use the JEE Main performance and the Normalized Board performance in the 60:40 ratio, the composite score for drawing the merit list was computed as:

C = 0.6 * AO + 0.4 * Bfinal


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