The common rank list cut-off for IIT Main is 89.75 percentile

The wait is over for more than 11.4 lakh students who attempted the JEE Main this year. The National Testing Agency has released the results for the April edition of the examination, and declared a common rank list cut-off of 89.75 percentile for eligibility for the JEE Advanced 2019 examination.

Overall, 24 students scored a 100 percentile, with four students each from Telangana and Rajasthan achieving the perfect score, the highest among the states.

Students can check their scores as well as the final answer keys at jeemain.nic.in.

The JEE Main is used as an entrance examination for admission to the undergraduate engineering courses at National Institutes of Technology, Indian Institutes for Information Technology and other institutions run by some state governments. The top 2.24 lakh students will be eligible to appear for JEE Advanced 2019 which is the admission test for the Indian Institutes of Technology.

This was the first time that JEE was a completely digital examination, and also the first time it was held twice, once in January and once in April. Students were free to decide whether they wished to attempt the examination twice. Of the 11.47 lakh students, 6.08 lakh appeared in both. Of those, a little over half actually did better in their first attempt.

For economically weaker sections, the cut-off is an NTA score of 78.21, for Other Backward Classes, it was 74.31, for Scheduled Castes, it was 54.01, and for Scheduled Tribes, it was 44.33.