JEE Main Exam Pattern 2023 of Paper 1 (B.E/B.Tech)
Students should be aware of the JEE Main 2023 test pattern in order to understand the examination method, type of questions, total number of sections, marking scheme, and so on. The JEE Main 2023 exam pattern is available in the table below.
Paper 1 (B.E/B.Tech) Exam Pattern
Factors  Details 
Mode of Exam  CBT (Computerbased examination) 
Exam Duration  3 hours/180 minutes 
Language of Examination  English, Hindi, Marathi, Assamese, Tamil Bengali,Kannada,Punjabi, Gujarati, Odia, Telugu, Malayalam and Urdu. 
Type of Questions   Multiple choice questions (MCQs)
 Questions with numerical responses

No. of Sections  Three are three sections:  Physics
 Chemistry
 Mathematics

No. of Questions  Mathematics: 25 (20+10) 10 questions with numerical answers Five of the ten questions are required. Physics: 25 (20+10) points for physics 10 questions, each with a number answer. Five of the ten questions are required. Chemistry: 25 (20+10) 10 questions with numerical answers Five of the ten questions are required. 75 questions in total (25 questions each) 
Total Marks  300 Marks (100 marks for each section) 
JEE Main Marking Scheme  MCQs: Each correct answer will be worth four points, while each incorrect response will be worth one point. Questions with numerical value answers: Candidates will receive four marks for each correct answer and one mark for each incorrect response. 
Tie Breaking Policy for JEE Main Paper I
If two or more candidates score the same marks in JEE Main Paper I exam, the order of merit will be determined as per the following tiebreaking rules. In order to make the candidates have a detailed understanding of tiebreaking policy, we have explained the same with some examples.
Tie Breaker 1: Candidates scoring higher marks in Mathematics will be given priority in the merit list as per tiebreaker policy  1.
Example:
Let us assume that Candidate A and Candidate B have scored 280 out of 300 in JEE Main Paper I. However, Candidate A scored 80 marks in Mathematics, while Candidate B scored 82 in Mathematics. Here, Candidate B will be given priority in the merit list as per the Tie Breaker Policy 1.
Tie Breaker 2: This policy comes into effect only if two or more candidates score the same marks in Mathematics. Since two or more candidates have similar marks in Mathematics, the marks scored by candidates in Physics will be taken into consideration to determine the merit.
Example:
Let us assume that Candidate A and Candidate B have scored 280 out of 300 in JEE Main Paper I. Both of them have also scored 80 marks in Mathematics. However, Candidate A scored 85 in Physics and Candidate B scored 82 in Physics. As per the tiebreaker policy 2, Candidate A will be given higher preference in the merit list than Candidate B.
Tie Breaker 3: This policy comes into effect only if two or more candidates score similar marks in Physics. In such cases, marks scored by candidates in Chemistry will be taken into consideration to determine the merit.
Example:
Let us assume that Candidate A and Candidate B have scored 280 out of 300 marks in JEE Main Paper I. On the other hand, both these candidates scored the same marks in the Mathematics and Physics section of the exam. However, Candidate A scored 80 marks in Chemistry and Candidate B scored 81 marks. As per tiebreaker policy 3, Candidate B will be given a higher preference in the merit list than Candidate A.
Tie Breaker 4: This policy comes into effect only if two or more candidates score the same marks in Mathematics, Physics and Chemistry sections of JEE Maine Paper I. In such cases, the candidate with less number of negative responses in the exam will be given higher preference in the order of merit.
Example:
Let us assume that Candidate A and Candidate B have scored 250 marks in JEE Main Paper I. On the other hand, both these candidates score the same marks in Mathematics, Physics and Chemistry. However, the number of wrong answers by Candidate A is 16, while the number of wrong answers by Candidate B is 18. In order to determine the merit, Candidate A will be given higher preference in the order of merit, as the number of wrong answers by Candidate A is less than Candidate B.
Tie Breaker 5: This policy comes into effect only if a tie still persists after applying all the above rules. In such cases, candidates older in age will be given higher preference in the merit list. Check the example below for a better understanding.
Example:
Let us assume that Candidate A and Candidate B scored 240 marks in JEE Main Paper I. On the other hand, these candidates scored same marks in Mathematics, Physics and Chemistry. At the same time, the number of incorrect answers by these candidates is also the same. However, Candidate A is 17 years and 6 months old and Candidate B is 17 years and 8 months old. As per tiebreaker policy 5, Candidate B will be given higher preference in the merit list.