Engineer banne ke liye kitna percentage number chahiye agar diploma karna hai to  diploma me kitna kharch hai

Hello, if you are talking about admission through JEE main score than for general category you need minimum 75% in 12 boards and for ST, SC, OBC AND PWD, you need minimum 65% in 12 boards for admission into various NITs, IIITs and GFTIs. But for other state funded colleges, the minimum percentage may vary accordingly.
Please visit the individual websites of interest universities to get more information about the minimum percentage.

Thank you

Hello Mohit
There are various Universities and colleges that offer B.Tech and each of them has its own guidelines and eligibility criteria.
To be on a safer side,you need to have more than 60%  in your 10+2 and more than 60% in PCM.
For admission in colleges through JEE Mains, you need to have minimum 75% aggregate in your 10+2.
You will be granted relaxation if you belong to any reserved categories.
Hope it makes sense to you

Hello Mohit,

For taking admission in any engineering college you are required to have secured minimum 45% in 10+2 or equivalent exam with mathematics and physics as main subjects along with Bio-Technology/chemistry/biology as the optional subject. Minimum marks required by SC/ST/OBC candidates is 40%.

Hope it helps
All the best!

Hi aspirant

Nationality  The candidates should be of Indian nationality.
Age Limit  There is no age limit for the candidates
Qualifying Examination  The candidates should have completed their 12th standard or equivalent examination from a recognized board/university with Physics, Chemistry and Mathematics. Those who will be appearing for the qualifying exam in 2019 will also be eligible
Marks Required  General category candidates should have secured minimum 50% aggregate marks in 12th standard examination (45% for SC/ ST/ OBC candidates).



For more information follow below link


https://www.google.co.in/url?q=https://engineering.careers360.com/articles/mht-cet-eligibility-criteria&sa=U&ved=2ahUKEwiy1ZP3zKniAhVZinAKHYEkCOYQFjADegQIChAG&usg=AOvVaw0MFV01HoYn1-UT32T_kTQB