JEE Main 2020 would begin from January 6. Every year 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. Any expert would tell you that cracking an entrance examination is 80 percent hard work, 5 percent luck and 15 percent strategy. With just 10 days left, here is a quick look at how much you need to score to qualify for the JEE Advanced 2020 to help you plan your approach. Check out the expected cut off and score required to qualify for the IIT JEE or JEE Advanced.
As per the last year estimates, 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.
||Percentile Score Cut off (as per 2019)
||Tentative Scores for respective percentile score (out of 360)
|Common Rank List
|OBC – NCL
||45 – 50
||28 – 40
||-20 to 8
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. However, depending on the number of students who appear and the general trends seen over the years, the expected cut off score remains the same with a variance range of about 5 marks either ways.
JEE Main Ranks and Scoring
From 2019, JEE Main Ranks are calculated basis the percentile score. The 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.
News Courtesy: Times Now News