Paper II at a Glance: Pattern, Qualifying Rule and Scoring Arithmetic
Paper II, widely called the CSAT (Civil Services Aptitude Test), is the second of the two objective papers in the Civil Services (Preliminary) Examination. Its marks stay out of the merit ranking, yet a score below 66 of its 200 marks keeps a candidate out whatever the Paper I result. This chapter sets out the pattern, the qualifying rule and the scoring arithmetic that follows, so that later decisions about attempts, guessing and pacing rest on verified numbers.
KarmSakha preparation topics covered
- Stages of the Civil Services (Preliminary) Examination and where Paper II sits
- Paper II structure: 80 multiple-choice items, 200 marks, two hours, four options
- The 33 per cent qualifying minimum: 66 of 200 marks
- Negative marking: one-third of an item's marks for each wrong answer
- The scoring formula and the scores that clear or miss the line
- Break-even analysis of blind guessing and the effect of eliminating options
- Attempts, accuracy and the minimum number of correct answers
- Pacing 80 items in 120 minutes, with checkpoints and a skip-and-return rule
- Answer-sheet discipline and pre-submission checks
- The eight official Paper II heads and the chapters that teach them
Where Paper II Sits in the Civil Services Examination
The Civil Services Examination has three stages. The Preliminary Examination is objective and has two compulsory papers of 200 marks each: Paper I, General Studies, and Paper II, the CSAT. The Main Examination is written and descriptive, and a personality test (interview) comes last. The official syllabus states the Preliminary structure in one line: the examination comprises two compulsory papers of 200 marks each. Paper I marks decide who moves on to the Main stage; Paper II is a qualifying test with a fixed minimum.
The CSAT became a qualifying paper in 2015. The UPSC notice of that year made Paper II a qualifying paper with a 33 per cent minimum and did not count its marks in the merit ranking. Current summaries of the CSE 2026 pattern carry the same rule. The practical meaning is an asymmetry: the CSAT cannot raise a candidate's rank, but it can remove a candidate whatever the Paper I result. Every calculation in this chapter follows from that asymmetry, and the score is checked against one line and then set aside.
This edition describes the CSE 2026 cycle. Clearias reports that the notification was released on 4 February 2026, and that the Preliminary Examination was scheduled for Sunday, 24 May 2026. The UPSC calendar for 2027, reported on 20 May 2026, places the Preliminary Examination on 23 May 2027, plans the notification for 13 January 2027 and the Main Examination for 20 August 2027. The calendar says the dates may change, so the notice for the year a candidate sits is the final authority.
Paper II at a Glance: Items, Marks, Time and Language
Paper II has 80 multiple-choice items and 200 marks, so each correct answer carries 2.5 marks (200 divided by 80). The paper lasts two hours: the syllabus sets Paper II at 200 marks with a duration of two hours, and the syllabus describes the questions as multiple choice and objective. Each item offers four options with one correct answer, which this book labels A to D. Answers go on an OMR sheet, and each answer is scored against the question number it is shaded against. The table gathers the pattern in one place.
The Paper II syllabus text sets one language rule, and it is narrow. It says that questions on English Language Comprehension skills at Class X level, the last item of Paper II, “will be tested through passages from English language only without providing Hindi translation thereof in the question paper.” This wording is taken from published reproductions of the syllabus, so the current notice should be checked for its exact text. The text says nothing about the language of the other items, so the notice for the year a candidate sits must be read for those. Candidates who read more comfortably in Hindi should practise English-only passages early rather than assume a translation will appear.
Three numbers drive the decisions in this book: 2.5 marks for each correct item, a 66-mark qualifying line, and 1.5 minutes for each item. The first and third are plain arithmetic on the pattern above, and the second is the qualifying figure of 33 per cent of 200, which the 2015 notice and current summaries both state. The table's Basis column shows where each figure comes from, so a candidate can see which numbers depend on the notice and which are simple arithmetic.
| Feature | Value | Basis |
|---|---|---|
| Items | 80 multiple-choice, objective items | Syllabus Note 2; CSE notice |
| Marks | 200 in total; 2.5 for each correct item | 200 divided by 80 |
| Duration | Two hours (120 minutes) | Syllabus text, Paper II |
| Options | Four per item, one correct | CSE 2026 notification (check the notice for your cycle) |
| Wrong answer | Minus one-third of the item's marks: 5/6, about 0.83 | CSE 2026 notification (check the notice for your cycle) |
| Unanswered item | No deduction; scores zero | CSE 2026 notification (check the notice for your cycle) |
| Qualifying minimum | 33 per cent: 66 of 200 marks | 2015 notice; CSE 2026 summaries |
| Merit ranking | Paper II marks not counted | 2015 notice |
| Language | English Language Comprehension passages in English only | Syllabus, last item |
The 33 Per Cent Qualifying Rule: What 66 Marks Demands
Thirty-three per cent of 200 marks is 66 marks, and that is the line this book uses. The 2015 notice required at least 33 per cent in the paper to qualify for the Mains, so the rule is a floor on the paper's own total, not a rank. The floor is absolute: a candidate who scores 65 is out even with a strong Paper I result, and a candidate who clears 66 has the Paper I result counted. The Paper II score is therefore a gate, checked against the line and then set aside.
Translated into items, 66 marks is 26.4 items' worth of correct answers at 2.5 marks each. Twenty-seven correct answers with no wrong ones give 67.5 and clear the line; twenty-six correct with no wrong ones give exactly 65, one mark short. Each wrong answer takes away 5/6 of a mark, so every mistake raises the number of correct answers needed to stay above 66. The line applies to the total of marks, and each wrong answer lowers the total. That is why the line is best read as a running total rather than as a count of correct answers.
Every score on this paper is a multiple of 5/6, because S = (5/6)(3c − w). Since 66 is not a multiple of 5/6 (66 ÷ 5/6 = 79.2), no reachable score equals 66 exactly. The nearest reachable scores are 65.83, which fails, and 66.67, which qualifies, and each is a whole number of steps of 5/6 away from the line. The exact verdict therefore does not depend on whether the line is read as 66 or as 66.67: every score up to 65.83 fails and every score from 66.67 upwards qualifies. Scores of 66.67 come from 3c − w = 80, for example 28 correct with 4 wrong or 30 correct with 10 wrong.
The Scoring Formula and Break-Even Scenarios
Let c be the number of correct answers and w the number of wrong answers. Each correct answer adds 2.5 marks. Each wrong answer subtracts one-third of 2.5, which is 5/6 of a mark, or about 0.83. An unanswered item contributes nothing, so with u = 80 − c − w blank items the score is S = 2.5c − (5/6)w. Keep fractions exact while working and round only the final figure. The formula has no penalty for blank items, so a blank is never worse than a wrong answer.
The table applies the formula to five combinations. Two of them score 66.67: 28 correct with 4 wrong, and 30 correct with 10 wrong. Both clear 66, though not by much. One more wrong answer on top of the 28 correct lowers the score by 5/6 to 65.83, and the candidate fails. A pass and a fail on this paper can be separated by a single wrong answer. That is why the next two sections matter: they show when a guess helps and how many answers must be right for a given number of attempts.
Written in attempts, where a = c + w, the score becomes S = (10/3)c − (5/6)a. Turning a wrong answer into a correct one gains 10/3 marks, about 3.33 rather than 2.5, because the 5/6 penalty disappears as well. This is the version to use when reviewing answers or planning how many items to attempt, and it underlies the attempt-based examples later in the chapter. Both forms give the same score for every pair of values, which makes either a reliable check on the other.
| Correct | Wrong | Blank | Score | Result |
|---|---|---|---|---|
| 27 | 0 | 53 | 67.50 | Qualifies |
| 26 | 0 | 54 | 65.00 | Fails |
| 28 | 4 | 48 | 66.67 | Qualifies |
| 28 | 5 | 47 | 65.83 | Fails |
| 30 | 10 | 40 | 66.67 | Qualifies |
Guessing: When Elimination Makes a Guess Worth Taking
Take an item with four options and no clue to the answer. The chance of a correct answer is 1/4 and of a wrong one 3/4, so the expected mark is (1/4)(2.5) + (3/4)(−5/6) = 0.625 − 0.625 = 0. A blind guess on a four-option item is worth nothing on average. The break-even chance follows from the same arithmetic: a guess with chance p has expected value 2.5p − (5/6)(1 − p), which is zero at p = 1/4. For any item worth 2.5 marks with the one-third deduction, a guess must beat a 1 in 4 chance to add value.
Elimination changes the odds. With three options left, a guess succeeds with chance 1/3 and has expected value (1/3)(2.5) + (2/3)(−5/6) = 5/18, about +0.28 marks. With two options left the chance is 1/2 and the expected value rises to 5/6, about +0.83 marks. The table sets the three cases side by side. Removing one option lifts the expected mark from zero to 5/18, and removing a second lifts it again to 5/6, so the benefit of elimination grows as the field shrinks. Guessing improves the expected total only once elimination leaves three or fewer options, and each gain is small.
An expected value is an average over many guesses, not a result on one item. A single three-option guess returns either +2.5 or −5/6, never +0.28. Over a whole paper the average matters, but a candidate close to the 66-mark line needs the actual outcomes to fall on the right side of it. Guessing is therefore a planning tool, and a blind guess on four options is neutral on average, which makes it a last resort rather than a method.
