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Hello aspirant
Minimum rank to qualify for jee main 2020 is difficult to estimate. But we can assume based on previous year's cut off.
For general category expected Jee main cut off 2020 is above 89.7548849 percentile. That means minimum 90 percentile. Now for 90 percentile you need to score
Hello,
To get an admission in IIT for M.Tech you need to have a minimum of 60% CGPA in B.Tech and you should get a very good score or All India Rank or a very good percentile in GATE exam that is conducted national wide for different fields of engineering.
Candidate should complete 10+2 in English Medium conducted by CBSE or equivalent examination from a recognised board with Physics, Chemistry and Mathematics in Science Stream with an overall aggregate of minimum 60%. Must appear in JEE Mains 2020. It is based on JEE Mains score but even then you should
Dear,
In case of B.Tech, you must have passed 10+2 (English Medium) examination from a recognized board in Science stream with an overall 60% aggregate and valid JEE mains/scorecard.Additionally, you should have secured not less than 50% marks in each of the subjects: (1) Math (2) Physics (3) English (4)
It's not possible for us to get your I'd but you will have details about KCET form in your mail I'd. While the time of registration, for confirmation purpose and after success confirmation you will receive a mail at the I'd you have provided. You can check your KCET application
If A and B are non empty sets, then A*B(I mean A into B) is called cartesian product of the two sets A and B. Any subset of this cartesian product is known as a relation.
A function is a relation which satisfies some conditions as follows
A function is
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