It's a Question of your Career!

Hey dear,

Hope you are doing well.

YES,  NIOS is valid as per the declaration released by the NTA which is responsible for the conduction of NEET exam.

Hope it helps.

Thank you!!!

Yes stipends are usually given in every medical colleges to the PGs and the interns may it be Government or Private. But yes, recently there's been a problem in the Telangana state specially. As more than 1200 students across the 12-15 private medical colleges did not receive the monthly stipends

Hello Aspirant.

National institute of technology(s) (NITs) are among of the most esteemed universities of our nation. There must be several other institutions that are way ahead in terms of education than NIT but college of engineering Ernakulam can't be considered more effective than NIT(s).

Best of luck !

Hello Sunil,

India happens to be one of the largest producers of engineers in the world. India, along with other south Asian countries, has a very high regard for engineers (only lower than doctors I guess) society. Aside the IITs ,IIITs and NITs, there are some Engineering colleges in India

Hello Aryan,

Yes, you can do BA B.ed integrated courses from DU ( Delhi University). The list  of some of the prestigious colleges are as follows:

1. Department of Education (CIE)

2. Lady Irwin College

3. Shyama Prasad Mukherjee College for Women

4. Maharishi Valmiki College of Education

5. Jesus

Hello  Aspirant,

Candidates can download their admit card by logging in to their accounts using their username and password.

Once downloaded, candidates need to take a printout of the CUCET 2020 admit card and carry it to their allotted test centres. Chandigarh University CUCET admit card will contain all relevant

Hello Aspirant.

The problem is that JNU does not provide the course of BTech/bsc in Biotechnology. It does provide postgraduate course of MSc in biotechnology.

To be eligible for MSc in biotechnology in JNU there is a common entrance exam that the candidate give to get selected for this course.

Let the equation of required line having intercepts a and b with the axes is x/a+y/b=1.

Now, according to given information, P is the mid-point of AB
Therefore, P = (a/2,b/2) => (-3,4). Then (a,b) will be (-6,8).

So, the equation is 4x-3y+24=0. (By substituting in the above equation)