Mathematically, Tanuj's argument is spot on. Presently, I will secure a benefit of 2.5 marks for marking the same options for 40 questions, ie, 6.25% or 0.06. Earlier it was 10 marks, ie, proportion of .25 (25%).
However, it will work more accurately with greater number of questions, that is, the chances to secure positive score will increase as the number of questions which have to be guess-answered increases.
Example, I have 10 questions to guess answer, then simple rule of percentage will say, I ought to secure .6 marks (or to be more accurate somewhere around .25) However, the error margin is high enough, to also make the score zero or negative. Consider that out of the 10, only 2 were correct, then I score zero; and if 4 were correct, then I score 2.5.
However, if I have 60 questions which I have to guess answer, then the 6.25% will give more accurate score.
Why? Because, as the aggregate increases, the answer begins to shift towards the expected mean, that is 6.25%
Conclusion: In all, Tanuj's point will work if the candidate is in danger of leaving/failing to answer a large chunk of the paper. It would however, yield unpredictable result if the number of such questions is less, that is, somewhere around 10-20.
All in all, it toughens the answering strategy for kids vis-a-vis their earlier counterparts.
Hope it helps :)
H.