Skip to main content

Updates ...

Visit the website for latest posts, courses, admission & more.

For guest/sponsored article(s), please check this link.

Examination Reforms

Click here!

Thinking in this direction in the past has given birth to AIEEE, which in due course has become just one more burden to the students. Ideally, there must be just one exam. The students who top that exam may opt for IIT, followed by NIT and thereafter state colleges/other colleges. This looks very difficult to implement but needs to be done keeping in view the long-term efficiency of the system. Here is how to go about it:

  • The boards like CBSE, ICSE, State etc. play the role of trainers and not examiners.
  • A central body is appointed as examiner.
  • This body takes the tests and prepares merit list and a minimum cut-off in each subject and the aggregate.
  • This test should be in line with JEE (may be a little easier) having a practical component as well as languages and other papers.
  • Students opt for engineering institutes and branches based on their marks in Physics and Mathematics. An institute can ask for more information from the candidate in addition to the marks obtained by him/her in this exam. However, the institute does not take any other written exam. Students opt for medical institues based on their marks in Biology, Chemistry and Physics.
  • Students are allotted institutes and branches depending upon their merit and availability.
  • The students who do not attain minimum marks in less than or equal to 2 subjects are allowed to take supplementary. If they fail in more than 2 subjects, they are declared fail and they will have to repeat the class.

Popular posts from this blog

A man starts walking from the point P (-3, 4) ....

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR} $ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1 $ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$