New Delhi: Good news for engineering aspirants! For the first time ever, Indian Institutes of Technology (IITs) have revealed the Joint Entrance Examination (JEE) admission cut-offs. Not releasing the cut-offs before results was a 50 year old tradition, but while making many improvements, the institution has released the same early, this year.
The IIT-JEE 2012 examination was held on April 8, 2012. Statistically, more than 5.6 lakh students appeared for the IIT-JEE this year. As per the issued cut-offs, minimum qualifying mark for general category for IIT-JEE 2012 is 10%, while it is 9% for OBCs and 5% for SC, ST as well as physically-disabled (PD) candidates.
However, general category students appearing for JEE need 14 out of 136 in Chemistry, Physics and Mathematics. Whereas, the aggregate cut-off for the general candidate is 35% or 143 out of 408 to qualify for an all India rank. Likewise, the minimum marks for OBC candidates are 13 out of 136, while it is 7 out of 136 for SC/ST and PD.
Moreover, a student will need 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%, 31.5% for OBCs and 17.5% for SC / ST / PD. As per sources, the decision is a result of the Supreme Court's observations in a case filed by IIT-Kharagpur professor Rajeev Kumar.
In its last year’s judgment, Supreme Court 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.
However, this early disclosure of cut-offs has ensured that candidates can evaluate their performance soon after appearing for the exam. The JEE is known-to-be as one of the toughest engineering examinations across the world but the process has come under fire recently.
