Joint Entrance Exam Advanced
Heyy Habibul......
Sorry that's not possible. According to first criteria for JEE ADVANCE 2019 , candidates appearing for JEE Advanced 2019 must qualify JEE Main 2019 and secure a position among the top 2,24,000 candidates (including all categories).
To know more about criteria's visit below link :
https://engineering.careers360.com/articles/jee-advanced-eligibility-criteria
Heyy Uday.....
The rank required to be in IIT ROORKEE can't be predicted exactly but you can get idea from previous year cutoff, here's the link :
https://engineering.careers360.com/articles/jee-advanced-cutoff-for-iit-roorkee
And for the cutoff required to qualify for JEE ADVANCE as by TimeNow news is that a candidate needs to be in
hello rajendra kumar,
jee mains is enough to earn admission in DTU for b.tech courses,no need for jee advanced (jee advanced score is needed only to get admission in ug courses in iit's).for m.tech ,you'll be in need of GATE score.please check the given links:
all the best!
note:if
The cutoff of this year is not yet announced, Still you can go through the following article and give a check on the previous years cut-offs so that you have an idea how much do you neet to score to get into an IIT.
https://www.google.com/amp/s/engineering.careers360.com/articles/jee-advanced-cutoff
All the best.
Visit : https://engineering.careers360.com/result-predictor
you can check detailed information @ https://jeemainadmitcard2019.in/jee-main-exam-total-attempts/
JEE Main 2019 is conducted twice by NTA first time and the rank will be based on NTA percentile score so it would be difficult to provide you the exact information.
Based on this score your probable rank will be 165,204 - 185,204, You may not be eligible for JEE
It would be difficult to provide you the exact information as JEE Main 2019 is conducted twice by NTA first time and the rank will be based on NTA percentile score.
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