A test consists of 25 MCQs. Each correct answer gives +4 marks and incorrect answer gives −1 mark. If a candidate scores 74 marks, then how many questions were left unattempted?
- (a)4
- (b)3
- (c)5
- (d)9
Correct — A, 4. Let c be the number correct, w the number wrong and u the number left blank, with c + w + u = 25. The score condition is 4c − w = 74, so w = 4c − 74. For w to be zero or more, c must be at least 19; and since c + w = 5c − 74 cannot exceed 25, c cannot exceed 19. So c = 19 exactly, w = 4(19) − 74 = 2, and the attempts total 21. That leaves 25 − 21 = 4 questions unattempted.
- (b)3 — Three blanks would mean 22 attempts. No split of 22 into correct and wrong scores 74 — the nearest are 18 correct and 4 wrong for 68, or 19 correct and 3 wrong for 73.
- (c)5 — Five blanks leave 20 attempts. The best available score is 19 correct with one wrong, which comes to 75, and 18 correct with two wrong gives 70. Neither is 74.
- (d)9 — Nine blanks leave only 16 attempts, whose maximum possible score is 16 × 4 = 64 — below the stated 74 before any penalty is applied.
A marking-scheme problem is a pair of linear conditions in three unknowns, and the integer constraints do the rest. One condition fixes the total number of questions, the other fixes the score, and requiring both counts to be non-negative whole numbers usually pins the solution down uniquely.
The squeeze argument is faster than trial and error: w = 4c − 74 forces c ≥ 19 from below, and c + w ≤ 25 forces c ≤ 19 from above, so only one value survives. As printed the stem reads 'incorrect answer gives −1 mark' without repeating the word 'each', which is how the booklet prints it; the intended reading is a penalty of one mark for every wrong response, and the arithmetic only works on that reading. Working backwards from the options is the other reliable route, since a blank count fixes the number of attempts and you can test whether 74 is reachable.
- Set up c + w + u = total and (marks per correct)c − (penalty)w = score.
- Non-negativity of w and the cap on attempts often squeeze c to a single value.
- Here c = 19, w = 2, u = 4, giving 76 − 2 = 74 marks from 21 attempts.
- CAPF's own paper carries a negative marking of one-third of the marks allotted, so a wrong response costs 0.67 of the two marks a question is worth.
Two inequalities meeting at a single integer is what makes the answer unique.
- Assuming every question was attempted, which makes the system unsolvable in whole numbers.
- Reporting the number of correct answers, or of wrong ones, when the blanks were asked for.
- Dropping the sign on the penalty and solving 4c + w = 74.
Asked with a marking scheme, a total number of questions and a final score, with the correct, wrong or unattempted count as the target.
In an exam, a candidate attempts 20 questions and scores 72 marks. If 5 marks are awarded for each correct answer and 2 marks are deducted for each wrong answer, then how many questions were answered correctly by him?
- (a) 18
- (b) 17
- (c) 16
- (d) 15
Answer(c) 16
The same marking-scheme setup from CAPF's own 2020 paper, made easier by telling you how many questions were attempted. Here that figure is withheld, which is what turns a one-line equation into a bounding argument.
- practice — not a real PYQ
In a test of 30 questions, 3 marks are given for a correct answer and 1 mark is deducted for a wrong one. A candidate attempts all 30 and scores 62. The number of correct answers is
- (a)21
- (b)22
- (c)23
- (d)24
Answer(c) 23 — 3c − (30 − c) = 62 gives 4c = 92, so c = 23 and the remaining 7 are wrong.
- practice — not a real PYQ
A quiz awards 5 marks per correct answer and deducts 2 per wrong one. A student answers 20 questions and scores 58. The number of wrong answers is
- (a)4
- (b)6
- (c)8
- (d)10
Answer(b) 6 — 5c − 2(20 − c) = 58 gives 7c = 98, so c = 14 and 6 are wrong.