The CSIR-UGC NET paper in Mathematical Sciences is a single three-hour paper of 120 questions worth 200 marks. You do not attempt all of them, and understanding that is worth marks on its own.
The marks distribution, three parts at a glance
| Part | On the paper | You may attempt | Marks each | Negative |
|---|---|---|---|---|
| A — General Aptitude | 20 | 15 | 2 | −0.5 |
| B — Subject, single correct | 40 | 25 | 3 | −0.75 |
| C — Subject, multiple correct | 60 | 20 | 4.75 | none |
Multiply it out: 15×2 + 25×3 + 20×4.75 = 30 + 75 + 95 = 200 marks.
So the marks distribution is 30 / 75 / 95 across Parts A, B and C — not the 20 / 40 / 60 of the question counts. Part C allows fewer attempts than Part B and carries more marks than it, and that asymmetry is the shape of the whole paper: a plan weighted by how many questions each part has is weighted wrongly.
That is the distribution across the parts. For the distribution across the syllabus — which topics those Part B and C questions have actually come from — the topic-wise weightage is counted question by question from complete previous-year papers, with the coverage of each paper shown.
The rule most candidates learn too late
If you answer more than the permitted number in a part, the extra answers are not simply ignored in your favour. The evaluation takes the first N answered questions in paper order and discards the rest — regardless of which ones you were most confident about.
So if you answer all 20 Part A questions, only the first 15 count. If question 18 was your best answer and question 3 was a guess, that is bad luck: question 3 counts, question 18 does not.
The practical consequence is that deciding what to leave blank is part of the exam. A paper where you answer exactly the permitted number, chosen deliberately, beats one where you answer everything and let the ordering choose for you.
That is a skill, and it is the one no test series practises — they hand you exactly the number you are allowed to attempt, which removes the decision. The selection drill puts a real 40-question Part B in front of you and asks only attempt or skip, with no solving, then shows you which of your attempts this rule would have discarded.
Part C is not multiple-choice as you know it
In Part C, one or more of the four options may be correct, and you score only if you select exactly the correct set — every correct option, and no incorrect one. There is no partial credit and no negative marking.
That combination changes the strategy completely:
- Because there is no negative marking, a question you can fully reason about is always worth attempting.
- Because credit is all-or-nothing, a half-solved question is worth exactly zero. Three questions solved completely beat six solved halfway.
- Ruling an option out is as valuable as ruling one in — you need certainty about all four.
At 4.75 marks each, Part C carries 95 of the 200 marks from only 20 attempts. It is where the exam is won.
Negative marking, in perspective
Part A costs 0.5 for a wrong answer and Part B costs 0.75. Neither is severe enough to justify leaving a question blank when you have narrowed it to two options: a coin flip between two survivors has positive expected value in both parts. Blind guessing across four options does not.
What to do with this
To put your own numbers through the scheme above, the CSIR NET score calculator applies the per-part limits and the negative marking exactly as described here.
Sit a paper with the real limits enforced and the real ordering in place, and the selection skill becomes obvious. Our full-length mock tests put 120 questions in front of you and cap attempts at 15/25/20, exactly like the real paper — most practice sets quietly hand you only the number you are allowed to attempt, which trains the wrong habit.
Confirm the pattern against the current official notification at csirnet.nta.nic.in before your exam; the structure above has been stable for several cycles but the notification is the authority.