2,24,000 students to appear for JEE Advanced 2019

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JEE advanced 2019 will be conducted on May 19, 2019. 2,24,000 students are appearing for the examination. Candidates are selected to appear for JEE advanced after they pass JEE mains. The whole exam is conducted on computer. This year, JEE mains is being conducted twice. The cut off to be eligible for JEE Advanced 2019 will depend on the announcement of the result of second JEE mains 2019.

The last date to apply for JEE mains April 2019 is March 7, 2019. The exam will be conducted between April 7, 2019 and April 20, 2019. After the declaration of JEE mains April 2019 result after which the applications will be released for JEE advanced exam. JEE main January 2019 exam result has already been declared by NTA.

A student can score marks in the JEE Main exam way above than the cut off set for JEE Advanced eligibility and yet not be allowed to appear in the JEE Advanced exam. Why?

Because, there are some additional criteria which a student should fulfill in order to be eligible for the JEE Advanced exam.

Last year the eligibility criteria of JEE advanced exam was that, a candidate from general category should have scored 75% marks in his/her class 12th boards examination. Whereas, a candidate from SC, ST and PwD category should have scored 65% marks. While calculating, marks scored in Physics, Chemistry, Mathematics, any Language, and any other subject other than the ones mentioned before will be considered.

Also, the students must be in top 20 percentile in their class 12th board examination.

IIT Roorkee, which is conducting the exam this year, has made no announcements as to any change in the eligibility criteria. However, students are still advised to wait for the official brochure which shall be released in due time by the institute for detail information on the eligibility criteria for JEE Advanced exam.

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