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JEE Main 2020 Result – NTA has announced JEE Main September result 2020. Along with it, JEE Main 2020 AIR (all India rank) will also be released. NTA releases JEE Main result at jeemain.nta.nic.in. Candidates can check it by entering application number and date of birth. On the result, NTA scores in B.Tech / B.Arch / B.Plan in physics, chemistry, maths, etc along with the total are mentioned. Other than this, AIR and category rank is also disclosed in it. NTA also announced category wise cut off to term 2.5 lakh toppers eligible for JEE Advanced exam. Based on JEE Main 2020 rank, candidates are called for JoSAA counselling and seat allotment in NITs, IIITs, CFTIs. Know how to check JEE Main 2020 result, date, details in it, and more details.
JEE Main 2020 Result
For JEE Main 2020 September exam, 953473 candidates registered. Whereas, for Jan exam 1066450 candidates applied. JEE Main rank list for all the candidates will be released now, for all the test takers. The important dates for JEE Main 2020 result and more are as follows:
|JEE Main 2020||Result Date|
|JEE Main 2020 Sep exam date||01 to 06 Sep 2020|
|JEE Main Result Date 2020||11 Sep 2020|
|Registration for JEE Advanced 2020 starts on||12 Sep 2020|
|JoSAA 2020 starts from||6 Oct 2020|
Check Result Here: Candidates can check JEE Main 2020 result as per the schedule from, jeemain.nta.nic.in.
How to Check JEE Main 2020 Result?
NTA only releases JEE Main 2020 result online, on jeemain.nta.nic.in. The notice for the release of the result is uploaded on the official website of NTA, which is nta.ac.in. Candidates can check the result from the JEE Main login, by following the steps below.
Step 1 – To check the result, first of all, candidates need to go to the official website.
- Candidates can directly access login by clicking on the link above.
Step 2 – In the official website, candidates need to look for “View JEE Main 2020 Result and Rank” button.
Step 3 – On clicking this link, candidate get 2 options for:
- Through application number and password.
- Through application number and date of birth.
Step 4 – Next, the candidates need to click on any of the 2 options.
- If the candidates select “through application number and password” then they need they enter application number, date of birth, and security pin.
- If candidates select “through application number and date of birth” then they need to enter application number, date of birth. and security pin.
Step 5 – After entering all the details, candidates need to click on the red “Submit” button.
Step 6 – On clicking it, the dashboard opens and from there candidates can check JEE Main 2020 result, rank, NTA score, etc.
Note: Candidates should download their JEE Main 2020 scorecard. It is needed during JoSAA counselling and final admission.
How is JEE Main 2020 Result and AIR Prepared?
The result for JEE Main 2020 is prepared by applying the method of normalisation since the exam is conducted on different dates and time. To make sure that no one is either no aspirants at disadvantage, NTA will rank the candidates based on the percentile score obtain din the entrance exam.
What is a percentile score?
The percentile score is obtained based on the relative performance of all the aspirants appearing in the exam. The topper in each session will get the same Percentile of 100 which is desirable. After which the highest and lowest scores are also converted to appropriate Percentiles. It is calculated by using the following formula:
100 x Number of candidates appeared in the ‘session’ with raw score equal to or less than the candidate / Total number of the candidates appeared in the ‘session’
Declaration of final JEE Mains result and releasing ranks
NTA conducts JEE Main entrance exam in multiple shifts on multiple days. The calculation of score to a single percentile is done by NTA through the normalisation process.
The formula to calculate JEE Main 2020 scores is:
JEE Main 2020 Normalization Process
In the process of normalization, candidates are identified for their actual merit / rank by applying the following methods:
Step 1: Distribution of examinees
Candidates are distributed randomly into four (04) sessions to ensure that there is equal distribution of the students. These four sessions are as provided below:
- Session-1: Day-1 Shift-1
- Session-2: Day-1 Shift-2
- Session-3: Day-2 Shift-1
- Session-4: Day-2 Shift-2
Step 2: Preparation of JEE Mains result
The result for JEE Main 2020 is prepared in the form of raw scores, percentile scores and the total score. The percentile score for each session is prepared using the formula as given below:
Step 3: Compiling JEE Main NTA score and preparation of overall merit / rank list
After this, the percentile score for all the session will b merged together to get NTA score. It is then further used for the preparation of overall merit / rank list.
Tie Breaker Criteria for JEE Main 2020
In case of two or more candidates obtain equal marks in the entrance exam, their marks are resolved by applying the following methods –
Tie Breaker Criteria for Paper 1 (B.Tech)
- Candidates who secure more percentile score in Mathematics will be placed higher.
- If there is still a tie, aspirants who secure more percentile score in Physics will be ranked higher.
- If the tie is still not resolved, those who secures more percentile score in Chemistry will be ranked higher.
- Lastly, if the tie still persists, the age of the candidate will be taken into consideration. Older candidate will be ranked higher over younger candidate.
Tie Breaker Criteria for Paper 2 (B.Arch)
- Candidates who secures higher marks in the mathematics will be given a higher rank.
- If there is still a tie, then the aspirant with a higher score in the aptitude test will be allotted a higher rank.
- If there is still a tie, then aspirants with higher scores in the drawing test will be given higher rank.
- In case, the above criteria does not resolve the tie, aspirants having lesser number of negative responses in the paper will be ranked higher.
- Lastly, if the tie still persists, the older candidate will be given preference over younger candidate.
Tie Breaker Criteria for Paper 3 (B.Plan)
- Those who obtain higher marks in the mathematics will be given a higher rank.
- If there is still a tie, then the aspirant with higher scores in the aptitude test will be given a higher rank.
- If the tie still persists, then the candidate will higher marks in planning based questions will be given the higher rank.
- Even if the above criteria do not break the tie, aspirants with less number of negative responses in the paper will be ranked higher.
- Even if the tie still remains, then the age of the candidate will be considered. Older candidates will be given preference over younger candidate.
Forgot JEE Main 2020 Login Details? Get here
In case if you do not remember JEE Main 2020 login details, then you can retrieve it by simply following the instructions as provided below.
How to Get Application Number for JEE Main Login?
To get the application number for JEE Main candidate login, follow these steps:
Step 1: Click here in case if you forgot JEE Main application number.
Step 2: On clicking the direct link, a login page appears in which you need to provide your name, father’s name, mother’s name, and date of birth.
Step 3: After this, click on the “get application number” button to submit the details and then your application number will appear on the screen.
Details Given in JEE Main 2020 Result
The JEE Main 2020 September result contains much more details than the 1st attempt result. Most crucial information mentioned in the it are as follows:
NTA score for Paper-I (B.E / B.Tech) and / or Paper II (B.Arch / B.Planning):
Those who appeared for UG engineering exam.
- NTA score in Jan physics, chemistry, maths.
- NTA score in Apr / Sep physics, chemistry, maths.
- Final NTA score (based on best of the two in total).
Those who appeared for UG architecture / planning exam.
- NTA score in Jan maths, aptitude test, and drawing test.
- NTA score in Apr / Sep maths, aptitude test, and drawing test.
- Final NTA score (based on best of the two in total).
JEE Main 2020 All India Rank based on Final NTA Score: It is mentioned category wise, for following categories:
Category Wise Cut Off Score for JEE (Advanced): This is the minimum qualifying marks that JEE Main aspirants need to be eligible for JEE Advanced exam. The cut off is also released category wise.
Other Details: Apart from this, following details are also mentioned on it:
- Application number
- Cabdidate’s name
- Mother’s name
- Father’s name
- State code of eligibility
- Roll number (for both exams, if applicable)
- PwD status
- Date of birth
- Bar code
- Few instructions
This is how the JEE Main 2020 scorecard looks like.
JEE Main 2020 Reservation
The reservation of seats in JEE Main 2020 for Central Government run Institutes is as follows:
What after JEE Main 2020 Result is Declared?
Once NTA declares JEE Mains result, a cut off score for JEE Advanced 2020 category-wise is released. Students can apply for admission in various private colleges for admission in B.Tech programmes. After this, JEE Advanced is conducted by the respective conducting body. Those who qualify JEE Advanced will be then eligible to participate in JoSAA 2020 which conducts counselling joint seat allocation process for admissions to 100 institutes including 23 IITs, 31 NITs, 25 IIITs and 28 Other GFTIs.
JEE Main 2020 Cut Off for JEE Advanced
In the scorecard of JEE Main 2020, qualifying cut off for JEE Advanced is also given. Until the actual cut off is announced, candidates can check the last year, cut off from below:
JEE Main Opening and Closing Rank
In the table below, we have provided opening and closing rank for the top NITS as per NIRF ranking 2019. It has been prepared based on previous year information and will be updated as soon as JoSAA counselling starts.
|Name of the NIT||Opening Rank||Closing rank|
|Motilal Nehru National Institute of Technology||05||32077|
JEE Main 2020 Statistics
Candidates can check the data for JEE Main 2020 for paper 1 and paper 2 from the table below. The data provided below as per the unofficial sources and will be updated with the press release by NTA.
|Particulars||JEE Main 2020 Jan||JEE Main 2020 Apr|
|Total number of Registered candidates||9,21,261 – B.E/B.Tech|
1.38 lakh – Paper 2
|Will be notified later|
|Total number of candidates appeared in exam||B.E/B.Tech – 8,69,010||Will be notified later|
|Total number of female candidates||2,64,026||Will be notified later|
|Total number of male candidates||6,04,981||Will be notified later|
|Total number of Transgender candidates||3||Will be notified later|
About JEE Main 2020
Joint Entrance Examination – Main earlier known as All India Engineering Entrance Examination (AIEEE) is a national level engineering entrance test. It is conducted by National Testing Agency (NTA) for admission to various undergraduate engineering and architecture courses in institutes accepting the JEE-Main score, which mainly consists of National Institutes of Technology (NITs), Indian Institutes of Information Technology (IIITs) and Government Funded Technical Institutes (GFTIs). The examination consists of three papers – Paper 1 for B.E./BTech courses, Paper 2 for B.Arch and Paper 3 for BPlan courses.
JEE Main 2020 Jan Session Result
- Click here to download press release in English for JEE Main B.Arch and B.Plan result .
- Click here to download press release in Hindi for JEE Main B.Arch and B.Plan result.
- Click here to download the notice in English for JEE Main B.Tech / B.E result.
- Click here to download the notice in Hindi for JEE Main B.Tech / B.E result.
- JEE Main 2020 Toppers
JEE Main 2020 Toppers for Jan Session
A total of 869010 candidates appeared for BE / BTech exam out of which a total of 09 students obtained perfect 100 percentile. Check the name of the 100 percentiles in JEE Main Jan 2020 exam below.
|Name of the 100 percentiler||State of eligibility|
|Landa Jitendra||Andhra Pradesh|
|Thandavarthi Vishnu Sri Sai Sankar||Andhra Pradesh|
|Nishant Agarwal||Delhi (NCT)|
|Rongala Arun Siddardha||Telangana|
|Chagari Koushal Kumar Reddy||Telangana|
JEE Main Jan 2020 State wise list of 100 percentilers
|State code of eligibility||Candidates name||NTA score|
|Andaman and Nicobar islands||Akshat Singh||97.3286395|
|Andhra Pradesh||Landa Jitendra||100.0000000|
|Andhra Pradesh||Thadavarthi Vishnu Sri Sai Sankar||100.0000000|
|Arunachal Pradesh||Apurba Nath||96.1831211|
|Chandigarh||Kunwar Preet Singh||99.9972380|
|Dadra and Nagar Haveli||Sharad Vishwakarma||99.2655484|
|Daman and Diu||Jain Anmol Rajesh Kumar||98.6482824|
|Delhi (NCT)||Nishant Agarwal||100.0000000|
|Himachal Pradesh||Sarthak Diwan||99.9683001|
|Jammu and Kashmir||Aryan Gupta||99.7338263|
|Madhya Pradesh||Akarsh Jain||99.9959069|
|Maharashtra||Vedang Dhirendra Asgaonkar||99.9972435|
|Meghalaya||P Alias Chakma||96.3974118|
|Odisha||Sourabh Soumyakanta Das||99.9924070|
|Outside India||Nilay Mankala||99.5839421|
|Tamil Nadu||Gaurav R Kochar||99.9959069|
|Telangana||Rongala Arun Siddardha||100.0000000|
|Telangana||Chagari Koushal Kumar reddy||100.0000000|
|Uttar Pradesh||l Gokulnath||99.9993050|
|West Bengal||Sreemanti Dey||99.9923468|
JEE Main 2020 Result Frequently Asked Questions
Ans : There is a normalization process. The formula used to normalize the score is available in the information brochure of JEE Main.
Ans : No. The counselling process is independent of NTA. Thus there is no change in JoSAA.
Ans : Class 12th boards Result only certifies the candidate’s eligibility for the exam. However, for taking admission to NITs, IIITs and CFTIs are based on All India Rank as explained above along with at least 75% marks in the 12th class exam or be in the top 20 percentile in the 12th class examination conducted by the respective Boards.
Ans : The candidates have to register for the JoSSA counselling at the official JoSSA website jossa.nic.in in order to be able to fill their choice of institutes in order of preference to be eligible for participating in admission counselling.
Ans : No, the JEE Main Results declared by NTA after the publishing of the Answer Key will be final and binding.
Ans : The candidates can retrieve their Application Number and Password by pressing the i can’t access my account button and select the option of forgot password or application number button. After filling in details like registered email, date of birth and father’s name, the candidate will be able to retrieve the application number and reset their password.
Ans : The process of normalization helps in creating a uniform and just parameter of judgment for all the candidates who give the exam in a different session. To give an equal opportunity to all the candidates the process of normalization has been introduced in the declaration of the JEE Main 2020 Result.
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