50 Views

My percentile is 65 what would be mark obtained?


Chiranth Gowda Y T 1st Feb, 2019
Answer (1)
manisha.gupta 1st Feb, 2019

The calculation of the marks would be difficult as per your score. You can check the answer key which has been announced to take an Idea about the marks on the basis of your percentile score.

https://engineering.careers360.com/articles/jee-main-paper-2-answer-key

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 50,885.


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.


Good Luck!


Related Questions

Amity University-Noida B.Tech...
Apply
Among top 100 Universities Globally in the Times Higher Education (THE) Interdisciplinary Science Rankings 2026
BML Munjal University | B.Tec...
Apply
A Hero Group Initiative | Up to 100% Scholarships | Highest CTC 32.99 LPA | Average CTC 8.45 LPA | Accepts JEE Score | Last Date: 31st Jan'26
Amity University-Noida BBA Ad...
Apply
Among top 100 Universities Globally in the Times Higher Education (THE) Interdisciplinary Science Rankings 2026
Amity University-Noida MBA Ad...
Apply
Ranked among top 10 B-Schools in India by multiple publications | Top Recruiters-Google, MicKinsey, Amazon, BCG & many more.
Amity University-Noida Law Ad...
Apply
Among top 100 Universities Globally in the Times Higher Education (THE) Interdisciplinary Science Rankings 2026
Amity University-Noida M.Tech...
Apply
Among top 100 Universities Globally in the Times Higher Education (THE) Interdisciplinary Science Rankings 2026
View All Application Forms

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books