The Indian Institute of Technology (IIT), Guwahati, will release the Joint Entrance Examination ( JEE) Advanced 2023, cut-off marks for various categories on June 18. It will be available for the candidates on the official website of JEE Advanced— jeeadv.ac.in. Candidates will be able to check the cutoff to know the minimum marks which the General, OBC, EWS, SC, and ST category students will require to get admitted to IIT. The minimum marks will be released category-wise on the day of the result along with the final answer key.
The information brochure that determines the ideal cut-off for the Jee Advanced 2023 exam mentions that the SC/ST category applicants have to score a minimum of 17.5 per cent of the overall score whereas the General category has to achieve 31.5 per cent to get included in the rank list.
According to the 2022 cut-off of this nationwide entrance examination for engineering, the minimum aggregate marks for the General category was 50 while the cut-off for SC/ST candidates was 28 marks. However, to get placed in the rank list of the exam, the general candidate had to score a minimum of 13.89 per cent of the aggregate marks whereas SC/ST applicants had to score 7.78 per cent for the same
Only the candidates that appear in both the papers—-Paper I and Paper II will be considered for the ranking. The total score in any subject will be the sum of the scores in both papers. The applicant’s scores in Physics, Chemistry, and Mathematics together make up the candidate’s overall score in the JEE (Advanced) 2023. The rank lists are prepared on the basis of the aggregate scores of the students in the exam.
Candidates are said to clear the exam if they achieve the required percentage of marks in each subject and overall. Only qualified students will be included in the rank list when published. The maximum mark for the exam is 360 with each subject scoring a maximum of 120. Paper 1 and Paper 2 consisted of three sections — Mathematics, Chemistry, and Physics.
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