SRM University, Chennai Science and Humanities 2026
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Dear Aspirant,
With a percentile that good, you can expect to get admit in a fairly good NIT and a fairly productive branch. You should visit official websites of following NITs and look forward to get into one of these.
1. NIT Rourkela : http://www.nitrkl.ac.in/
2. VNIT Nagpur : http://vnit.ac.in/
Dear aspirant,
Your rank will be nearly (100-95.6)*145386/100=6396 based on previous year trends.And chances are there with NITs
You have chances with
NIT raipur
NIT Hamirpur
For last year NIT cutoffs go to
Https://www.google.com/amp/s/engineering.careers360.com/articles/jee-main-cutoff-for-b-arch-b-planning/amp
Good luck
Dear aspirant,
Your rank will be nearly (100-95.6)*145386/100=6396 based on previous year trends.And chances are there with NITs
You have chances with
NIT raipur
NIT Hamirpur
For last year NIT cutoffs go to
https://www.google.com/amp/s/engineering.careers360.com/articles/jee-main-cutoff-for-b-arch-b-planning
Good luck
Hi,
As 1,45,386 students appeared this year for the examination according to that by using the Rank Formula (100-Total percentile)*145386/100 your rank will be around 16719
This is your approximate rank.
According to your predicted rank you may get NIT Harimpur (Home quota) according to previous year cut offs .
As 1,45,386 students appeared this year for the examination according to that by using the Rank Formula (100-Total percentile)*145386/100 your rank will be around 16719
This is your approximate rank.
According to your predicted rank you may get NIT Harimpur (Home quota) according to previous year cut offs .
Hello Gautom,
207 must be your marks in JEE mains paper 2.
In order to get your rank you have to tell your percentile.
So please comment your percentile.
For more details you can get refrence from career 360 website :
https://engineering.careers360.com/articles/jee-main-cutoff-for-b-arch-b-planning
thanks
Hello,
Your rank would be around 46k by using the formula
145000*(100-percentile obtained)/100.
According to last year's cutoff you may not get a seat in any nit or central govt colleges.
Still you have a chance ,work hard for April paper
Good luck!
Your rank would be around 46k by using the formula
145000*(100-percentile obtained)/100.
According to last year's cutoff you may not get a seat in any nit or central govt colleges.
Still you have a chance ,work hard for April paper
Good luck!
Dear aspirant,
Your rank will be nearly (100-68.84)*145386/100=45302 based on previous year trends.And chances are less to be qualified to appear for jee advanced exam as experts predicted cutoff to be nearly 85 percentile for general candidate.
Good luck
Hi Vinslin,
This year a total of 1,45,386 aspirant appeared for the JEE B.Arch Examination. Therefore, basing upon the total number of aspirants the formula for calculating your rank would be ([100-your percentile] x 145386)/100. Therefore, approximately your rank will be around 1163. Calculating marks using percentile is not
Hi Ishika Soni,
Congratualtions on your succesfull results. This year a total of 1,45,386 aspirant appeared for the JEE B.Arch Examination. Therefore, basing upon the total number of aspirants the formula for calculating your rank would be ([100-your percentile] x 145386)/100. Therefore, approximately your rank will be around 382. Calculating
Hello Ishika,
Congratulations on scoring 99.73 in JEE mains mains paper 2 .
By using this formula :-
(100- your total percentile score) * 1,45,386/ 100.
(Note the rank calculated by this formula is approximate and should not be taken as actual rank, the actual rank by NTA may differ.)
Congratulations on scoring 99.73 in JEE mains mains paper 2 .
By using this formula :-
(100- your total percentile score) * 1,45,386/ 100.
(Note the rank calculated by this formula is approximate and should not be taken as actual rank, the actual rank by NTA may differ.)
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