NEW DELHI: Overturning a five-decade-old tradition India’s premier institutes, the Indian Institutes of Technology (IITs) have disclosed admission cut-offs for the first time in the Joint Entrance Examination (JEE) held on April 8.Over 5.6 lakh students appeared for the IIT-JEE on Sunday.The minimum qualifying mark for general category for IIT-JEE 2012 is 10%, while it is 9% for OBCs and 5% for SC, ST and physically-disabled (PD) candidates. General category students appearing for JEE need 14 out of 136 in chemistry, physics and mathematics. The aggregate cut-off for the general candidate is 35% or 143 out of 408 to qualify for an all India rank.Similarly, the minimum marks for OBC candidates are 13 out of 136, while it is seven out of 136 for SC/ST and PD.In addition to these minimum marks, a student will have to obtain aggregate cut-off to be eligible to be part of all India ranking. The aggregate cut-off for general category students is 35%, OBCs (31.5 %) and SC/ST/PD (17.5%).The decision is a result of the Supreme Court’s observations in a case filed by IIT-Kharagpur professor Rajeev Kumar. The court in its judgment last year had ruled that the selection process needed to be ``upgraded’’ and ``fine-tuned’’ year after year with periodic changes so that the selection process and examination remain relevant and meaningful.The disclosure has ensured that candidates can evaluate their performance soon after appearing for the exam. The JEE is accepted as one of the toughest engineering examinations across the world but the process has come under fire recently.RTIs filed by Kumar on IIT-JEE revealed how the country’s top technology institute has changes its examination system. In 2006 to 2011, there were varying cut-off marks for different subjects unlike the uniform pattern adopted this year. It varied from 37 in math to 55 in chemistry. In 2011, math cut-off was 34, while it was 20 for chemistry.