While JEE Main 2020 would begin from January 6, about 12 lakh students appear for the examination in the hope of qualifying for the JEE Advanced and getting a chance at the prestigious IITs every year.
According to experts, cracking an entrance examination is 80 percent hard work, 5 percent luck and 15 percent strategy. With over a week left for the exam, below is a quick look at how much you need to score to qualify for the JEE Advanced 2020 to help you plan your approach.
The expected cut off and score required to qualify for the IIT JEE or JEE Advanced as follows.
According to last year’s statistics, about 2.25 lakh students would be shortlisted for the JEE Advanced 2020 based on their JEE Main 2020 rank/ results. The students would be from unreserved and reserved category. Students may please note that the students would get both percentile score and raw score. The cut off for JEE Advanced would be based on the percentile score. An expected raw score is also provided in the table.
Category Percentile Score Cut off (as per 2019) Tentative Scores for respective percentile score (out of 360)
- Common Rank List: 89.7548849 79+
- General EWS 78.2174869 74+
- OBC – NCL 74.3166557 45 – 50
- SC 54.0128155 28 – 40
- ST 44.3345172 24+
- PWD 0.11371730 -20 to 8
While students and parents please note that the expected cut off provided above is based on the cut off for 2019 and previous years, the exact cut off may vary.
Candidates should note that from 2019 JEE Main Ranks are calculated basis the percentile score. They should keep in mind that percentile scores of the students are normalized and calculated as per the formula available on the official website jeemain.nta.nic.in. This is because from 2019, JEE Main is conducted across multiple shifts in January and April.
The cut off for JEE Main would be released only after the JEE Man 2020 April examination along with the JEE Main 2020 ranks and final results.