JEE Main Paper 2
You can calculate your rank in JEE Main Paper 2 using the formula:
Rank = (100-percentile)*(No. of students appeared for the examination)/100
As the no. of students appeared for Paper 2 (Jan) are around 112679, your rank will be around 21500. However as the no. of students in April changes,
Hi Ongseng,
You got 59.9 percentile in JEE main 2 exam, your rank would be approximately in between 45000 to 50000. Also, Above is the range of your possible rank. The predicted rank may vary due to JEE Main tie breaker rules for the same score. Also, you can check
Hello Dipesh Mahato,
For your nta score you can predict your rank by using the formula:
(100-your total score)*112679/100.
and you can use careers360 college predictor to get to know your colleges:
https://engineering.careers360.com/jee-main-architecture-college-predictor
and you can also use the below link for further reference:
https://engineering.careers360.com/articles/jee-main-cutoff-for-b-arch-b-planning
Hope the answer is helpful.
Hey aspirant!
National Testing Agency (NTA) has declared the results of JEE Main 2020 Paper 2.
Now, to calculate your rank for B.Arch based on the percentile score that you have secured, you can use this formula:
(100 - your total percentile score) is the percentage of people ahead of
JEE Main paper 2 result has been declared by National Testing Agency. The candidates can now check and download their score card or result of JEE Main Paper 2 after logging into their account.
Check the detailed information: https://engineering.careers360.com/articles/jee-main-paper-2-result
Hello perimala,
Don't worry it may be due to the server problem you were not able to look the results. Now there are websites working properly. Kindly visit jee main page again and check your answers. Keep up your spirit and wish you luck.
It will between 20-23 of september
The Question containing Inaapropriate or Abusive Words
Question lacks the basic details making it difficult to answer
Topic Tagged to the Question are not relevant to Question
Question drives traffic to external sites for promotional or commercial purposes
The Question is not relevant to User
And never miss an important update
