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?

Correct — C, 16. The candidate attempts 20 questions, so if c of them are right then 20 − c are wrong; nothing is left unattempted. Score is 5c − 2(20 − c) = 72. Expanding gives 5c − 40 + 2c = 72, so 7c = 112 and c = 16. Check it: 16 right earns 80, the remaining 4 wrong cost 8, and 80 − 8 = 72. There is a faster route for the hall. If every one of the 20 were right the score would be 100, and each question moved from right to wrong costs 5 marks gained plus 2 marks deducted, that is 7 marks. The shortfall is 100 − 72 = 28, and 28 ÷ 7 = 4 wrong answers, leaving 16 correct.

• (a)18 — Eighteen right and two wrong gives 90 − 4 = 86, not 72. The shortfall from 100 would have to be 14, which is two wrong answers, not the four the score requires.
• (b)17 — Seventeen right and three wrong gives 85 − 6 = 79. Note that no whole number of wrong answers can produce a score of 79 either way round — the score falls in steps of 7 from 100, so only 100, 93, 86, 79, 72 and so on are reachable.
• (d)15 — Fifteen right and five wrong gives 75 − 10 = 65, seven short. This is the answer you get by taking one step too many down the ladder of sevens.

Any scoring problem with a reward, a penalty and a fixed number of attempts reduces to one linear equation. Write the count of correct answers as the unknown, express the wrong answers as the remainder, and set the total equal to the score. The shortcut worth carrying is the step size: with +a for a correct answer and −b for a wrong one, every question that changes from right to wrong costs a + b marks, so the gap between the perfect score and the actual score is always a multiple of a + b.

Here a + b is 5 + 2 = 7, and the gap is 28, which is exactly four sevens. That single division answers the question without writing an equation at all, and it also tells you at once whether a stated score is even possible. This is the same arithmetic that governs the CAPF paper the candidate is sitting: two marks for a correct answer and one-third of two marks deducted for a wrong one, so each wrong answer costs two-thirds of a mark against the two it would have earned.

• With c correct out of 20 attempted, 5c − 2(20 − c) = 72 gives 7c = 112 and c = 16.
• The step size between adjacent outcomes is the reward plus the penalty, here 5 + 2 = 7.
• The gap from the perfect score is 100 − 72 = 28, and 28 ÷ 7 = 4 wrong answers.
• Verification: 16 × 5 = 80 earned, 4 × 2 = 8 deducted, net 72.
• Only scores of the form 100 − 7k are reachable when all twenty questions are attempted.

• Assuming some questions were left unattempted when the stem says twenty were attempted.
• Forgetting that a wrong answer costs the reward as well as the penalty when you reason from the perfect score.
• Solving 5c − 2c = 72 by mistake, which drops the constant and gives a non-integer.

Asked as a single linear equation dressed as an exam score; the reward-plus-penalty step size is what turns it into a ten-second item.

• practice — not a real PYQ

  A candidate attempts all 30 questions of a test and scores 60. Each correct answer earns 4 marks and each wrong answer loses 1 mark. How many did he get right?

  • (a)16
  • (b)18
  • (c)20
  • (d)22

  Answer(b) 18 — the perfect score is 120, each wrong costs 4 + 1 = 5, and (120 − 60) ÷ 5 = 12 wrong, leaving 18 right.

• practice — not a real PYQ

  In a test of 20 questions with 5 marks for a correct answer and 2 marks deducted for a wrong one, which of the following scores is NOT possible if all 20 are attempted?

  • (a)93
  • (b)86
  • (c)80
  • (d)72

  Answer(c) 80 — reachable scores fall in steps of 7 from 100, so 100, 93, 86, 79, 72 and so on; 80 is not among them.