The Joint Entrance Examination (JEE) Advanced 2019 has been conducted and over 1.65 lakh candidates registered to appear for the same. The result for JEE Advanced, according to the official website, will be released on June 14, 2019. While the cut-off to qualify for admissions to the Indian Institute of Technology (IITs) will be released thereafter, based on previous years’ data, it seems this year the cut-off will go down.The number of qualifying students went from 36,500 in 2016 to 51,000 in 2017 and 31,988 in 2018, based on the official data. Moreover, the minimum aggregate marks to qualify for inclusion in a rank list for the open category was 90 marks in 2018. For SC, ST category the same was 45 marks.
Last year, the cut-off declined massively in comparison to 2017 when the same was 128 marks. This year too, the trend is expected to continue. “The cut-offs for admissions to IITs has been 30-35 per cent of the total marks. This year, I am expecting it to go between 27-30 per cent – which will be roughly around 105+ marks out of 372,” said FIITJEE Noida branch head Ramesh Batlish. He added, “This is because the number of applicants is comparatively lower and the number of seats at IITs has been increased due to the construction of newer IITs and also the introduction of supernumerary seats.”
According to the National Academic Director (Engineering), Aakash Education Services, Ajay Kumar Sharma the increase of 12 marks in total will not have much impact on the cut-off. “Since the exam was too lengthy and students could not complete it inspite of knowing the answers, this would bring the cut-off down by 1-2 per cent from last year,” he said while adding, “As compared to previous years’ exams, the physics section was tougher and mathematics was lengthy.” He said the cut-off would not go beyond 80-85 marks.