New Delhi: The Indian Institute of Technology, Kanpur has released the eligibility criteria for JEE Advanced 2018.
The total number of candidates can be slightly greater than 2,24,000 in the presence of “tied” ranks/scores in any category.
Candidates should have been born on or after October 1, 1993. Five years relaxation is given to SC, ST and PwD candidates, ie, these candidates should have been born on or after October 1, 1988.
A candidate can attempt JEE (Advanced) a maximum of two times in consecutive years.
A candidate should have appeared for the Class XII (or equivalent) examination for the first time in either 2017 or 2018.
A candidate should NOT have been admitted in an IIT irrespective of whether or not he/she continued in the program OR accepted an IIT seat by reporting at a reporting centre in the past. Candidates whose admission at IITs was cancelled after joining any IIT are also NOT eligible to appear in JEE (Advanced) 2018.
Examination will be held on May 20, 2018. The entire JEE (Advanced) 2018 Examination will be conducted in fully computer based test mode. The exam consists of two papers, Paper 1 and Paper 2, each of three hours’ duration, and will be held in two sessions. Both the papers are compulsory.
The candidates must have secured at least 75 percent aggregate marks in the Class XII (or equivalent) Board examination. The aggregate marks for SC, ST and PwD candidates should be at least 65 percent.
He/she must be within the category-wise top 20 percentile of successful candidates in their respective Class XII (or equivalent) board examination.