I have got 88 percentile in paper 2 jee 2019. What's the best I can get??

The exact prediction would be difficult as JEE Main is being conducted twice by NTA for the first time. As a total No. of 1,45,386 candidates had appeared for January Session, Using the Formula (100-your score)* 1,45,386 / 100- Your Expected rank will be 17,446. With this rank the chances are less to get a seat in NITs, You can also prefer the previous year cutoff to take an Idea about the admission possibilities.

https://engineering.careers360.com/articles/jee-main-cutoff-for-b-arch-b-planning

NOTE: This is only a prediction. NTA will merge the scores obtained in first and second attempt by candidates of JEE Main Paper 2 2019 for the compilation and preparation of overall merit/rank list. The rank will be released on 30th April, 2019.

If you wish to take the admission in IITs ( IIT Kharagpur & IIT Roorkee) for Architecture programme, You need to appear for JEE Main Paper I after qualifying it you need to appear for JEE Advance AAT.

Good Luck!

Hello Aspirant,

Considering your percentile I have calculated you probable rank around 17,447 (considering total number of stidents appeared for this exam is 1,45,386). Remember it may change after second phase of Jee mains which will be conducted in april. Most probably you will qualify for AAT (architecture aptitude test(for admission in IITs)). So, if you do well in AAT you will have the options of IITs that Kanpur and Roorkee. Even if you not then there are 9 NITs and some GFTIs. NITs offer very less seats for architecture courses, so the only option you will be left with is to go for some government funded colleges like: school of planning and architecture Delhi, Bhopal, etc.

Best of Luck !!!