EPFO APFC 2023 General Studies · Question 40 of 120

Correct — C, (c) 10. Do it in two steps, and then note the shortcut that makes the whole question a five-second problem. Step 1 — how many ticks are correct. The candidate cycles A, B, C, D, A, B, C, D and so on from Question 1, so the pattern repeats every four questions and C falls on the third question of every cycle: Questions 3, 7, 11, and so on, each four apart. The last such question not exceeding 50 is 47, so the number of C ticks is (47 − 3)/4 + 1 = 12. The same count comes out of the division: 50 = 4 × 12 + 2, so the candidate completes twelve full cycles, covering Questions 1 to 48, and then ticks A on Question 49 and B on Question 50. Twelve complete cycles mean twelve C ticks. Since C is the correct answer to every question, he has 12 correct answers and, because he ticks every question, 50 − 12 = 38 wrong ones. Step 2 — the score. Each correct answer carries 4 marks and each wrong answer carries −1 mark, so the total is 12 × 4 − 38 × 1 = 48 − 38 = 10. Now the shortcut. Suppose the candidate gets n answers right out of 50 attempted; his score is 4n − (50 − n) = 5n − 50, which is a multiple of 5 for every whole number n. So under this marking scheme, with all fifty questions attempted, the total score can only ever be a multiple of 5. Of the four values printed, only 10 is a multiple of 5 — 6, 8 and 12 are not — so option (c) is the answer without counting a single C. This is worth internalising as a technique rather than as a trick: when a marking scheme fixes the arithmetic, the set of achievable scores is often a lattice, and checking divisibility eliminates options far faster than computing the answer. Note two printing details reproduced from the booklet: the stem says the paper 'has got four options', and the minus sign in '−1 mark' is printed at en-dash length.

• (a)6 — A score of 6 cannot arise at all under the marking described. With all fifty questions attempted, the score is 4n − (50 − n) = 5n − 50 for n correct answers, and that expression is a multiple of 5 whatever whole number n takes: the achievable totals run −50, −45, −40 and so on up to 200, and 6 appears nowhere among them. The option is presumably meant for a candidate who has miscounted the C positions and then done the subtraction without checking, but it can be discarded before any counting begins. Testing an option against the arithmetic of the marking scheme, rather than against your own calculation, is the cheapest check available on any scoring problem.
• (b)8 — Like option (a), 8 is not a possible score. Every total under this scheme is of the form 5n − 50 and therefore ends in 0 or 5, so both 6 and 8 are ruled out on inspection. There is a second reason to be suspicious of small even numbers here: the marks per correct answer and the penalty per wrong answer are both whole numbers with a difference of 5 between them, and it is that difference, not the values themselves, that fixes the spacing between achievable scores. Working out the spacing of a scheme before working out the answer is a habit that pays across every quantitative paper with negative marking.
• (d)12 — 12 is the number of questions the candidate answers correctly, not his score, and it is the trap the whole item is built around: the counting is the hard part, and having done it a candidate is tempted to write the number down and move on. But the question asks for the total score, which still requires the marks to be applied — 12 correct at 4 marks each, less 38 wrong at 1 mark each, giving 10. Note also that 12 is not a multiple of 5 and so could not have been a score in any case. Answering the question actually asked, rather than the quantity you have just computed, is the discipline this option tests.

Two ideas combine in this question. The first is periodicity: a repeating cycle of length k places a given item at positions r, r + k, r + 2k and so on, and the number of such positions up to N is found either by the arithmetic-progression formula, (last − first)/k + 1, or by dividing N by k and reading the quotient and remainder. Here the cycle has length 4, C occupies the third slot, and 50 questions give 12 complete cycles with a remainder of 2, so C is ticked 12 times. The second idea is the arithmetic of a marking scheme. If a correct answer earns p marks and a wrong answer costs q, and all N questions are attempted, then n correct answers give a score of pn − q(N − n) = (p + q)n − qN. Two consequences follow immediately: the achievable scores form an arithmetic progression with common difference p + q, and every achievable score is congruent to −qN modulo (p + q). With p = 4 and q = 1 the spacing is 5, so the scores are the multiples of 5 from −50 to 200. This structural view is what turns a counting problem into a divisibility check, and it applies to the marking of the APFC paper itself, where each question carries 2·5 marks and a wrong answer forfeits one-third of that.

This is the lightest item in the quantitative block and one of the most valuable, because it can be answered correctly by two independent routes and because both routes are fast. It is also, quietly, a question about examination strategy — the candidate in the stem is doing exactly what a desperate examinee sometimes does, ticking a fixed pattern in the hope that chance will pay, and the arithmetic shows what that strategy is worth. Twelve right and thirty-eight wrong on a four-option paper with a one-mark penalty yields 10 marks out of a possible 200, which is a fair illustration of why blind pattern-marking is not a strategy at all. On the APFC paper the same reasoning applies with different numbers: 2·5 marks for a correct answer and a deduction of one-third of that for a wrong one, which means random guessing among four options has an expected value of zero, while guessing after eliminating one option has a positive expected value. Knowing where that break-even point lies is worth more on the day than any single fact in this block.

• In a repeating cycle of length 4 starting at Question 1, the third element falls on Questions 3, 7, 11 and so on; the count up to 50 is (47 − 3)/4 + 1 = 12, which is also the quotient in 50 = 4 × 12 + 2.
• With C correct on every question, the candidate has 12 correct answers and 38 wrong ones, giving 12 × 4 − 38 × 1 = 10 marks.
• If a correct answer earns p marks and a wrong answer costs q, and all N questions are attempted, the score is (p + q)n − qN for n correct answers, so achievable scores form an arithmetic progression with common difference p + q.
• With p = 4, q = 1 and N = 50 the spacing is 5 and the achievable scores are the multiples of 5 from −50 to 200, so of the printed options only 10 is attainable and the question can be answered by divisibility alone.
• The APFC paper's own marking is 2·5 marks for a correct answer with one-third of that deducted for a wrong one, which makes the expected value of a blind guess among four options zero and the expected value of a guess after eliminating one option positive.

• Reporting the number of correct answers instead of the score; the counting is the effortful part, which is precisely why 12 is offered as an option
• Miscounting the cycle positions by starting from Question 0 or by taking the last C position as 49 rather than 47; the division 50 = 4 × 12 + 2 settles it in one line
• Forgetting that every question is attempted, so that the wrong answers number 38 and each costs a mark — a candidate who penalises only the questions he thinks of as attempted will overstate the score
• Ignoring the structure of the marking scheme; with 4 for a correct answer and −1 for a wrong one, every score is a multiple of 5, which alone eliminates three of the four options

The quantitative block of this paper always carries one or two items that are more about careful reading than about mathematics, and this cyclic-marking question is the type specimen: the arithmetic is trivial, the counting is easy to fumble, and the option set contains the intermediate quantity as a decoy. Related versions change the cycle length, change which option is correct, leave some questions unattempted, or ask how many marks a given pattern would lose rather than what it scores. In every version the same two moves work — count the positions by division rather than by listing, and check the achievable-score lattice before computing anything.

No directly related past PYQ was found.

• practice — not a real PYQ

  A test has 60 questions, each with four options A, B, C and D. A candidate ticks in the pattern A, B, C, D, A, B, C, D and so on starting from Question No. 1, till the end. If option (B) is the correct answer to every question, and each correct answer carries 3 marks while each wrong answer carries −1 mark, what is his total score?

  • (a)0
  • (b)15
  • (c)30
  • (d)45

  Answer(a) 0 — B falls on the second question of each four-question cycle, that is on Questions 2, 6, 10 and so on up to 58, which is (58 − 2)/4 + 1 = 15 questions; equivalently 60 = 4 × 15 gives fifteen complete cycles. So the candidate has 15 correct answers and 45 wrong ones, and his score is 15 × 3 − 45 × 1 = 45 − 45 = 0.

• practice — not a real PYQ

  In a test of 50 questions where each correct answer carries 4 marks and each wrong answer carries −1 mark, a candidate attempts every question. Which one of the following cannot be his total score?

  • (a)0
  • (b)10
  • (c)12
  • (d)15

  Answer(c) 12 — with n correct answers out of 50 attempted, the score is 4n − (50 − n) = 5n − 50, which is a multiple of 5 for every whole number n, so the achievable totals are the multiples of 5 from −50 to 200. A score of 0 arises at n = 10, of 10 at n = 12 and of 15 at n = 13, but no value of n gives 12.