Konkur Percentage: Weighted Composite, Negative Marking and Guess Strategy
Parisa is estimating her Konkur percentage the night before the exam. The rules she is working with: each correct answer earns 3 points, each wrong answer subtracts 1 point, blanks score 0, and a section percentage is (3C − W) ÷ (3N) × 100, where N is the number of items in the section.
English: N = 30, four options per item. On her last mock she answered 24 items: 19 correct, 5 wrong, 6 blank.
Second foreign language: N = 20, three options per item. She answered 17: 12 correct, 5 wrong, 3 blank.
The composite in her target programme weights English twice and the second language once: composite = (2 × E + 1 × S) ÷ 3. Last year's cut-off in her region was 55.00.
Her plan for the last two minutes of each section is to fill every blank at random, reasoning that 'a blank earns nothing, so guessing can only help'. She also believes that leaving an item blank is always safer than guessing.
1. Compute E, S and the composite to two decimal places, and state whether she clears 55.00 - and how much margin she has in items.
Read the worked answer
English: (3 × 19 − 5) ÷ (3 × 30) = (57 − 5) ÷ 90 = 52 ÷ 90 = 0.5778 → 57.78.
Second language: (3 × 12 − 5) ÷ (3 × 20) = (36 − 5) ÷ 60 = 31 ÷ 60 = 0.5167 → 51.67.
Composite: (2 × 57.78 + 51.67) ÷ 3 = (115.56 + 51.67) ÷ 3 = 167.22 ÷ 3 = 55.74.
She clears the cut-off by 0.74 points - less than one item.
Margin check matters more than the headline number. One extra wrong answer in English changes E to 51 ÷ 90 = 56.67, and the composite to (113.33 + 51.67) ÷ 3 = 55.00 exactly. So a single careless error puts her precisely on last year's line, where the result depends on tie-breaking rules she does not know. Conversely one extra correct answer in the second language raises S to 34 ÷ 60 = 56.67 and the composite to (115.56 + 56.67) ÷ 3 = 57.41.
Conclusion: she is above the line on average, but with essentially zero buffer. The practical implication is that variance control in English matters as much as raw ability, because English is double-weighted.
2. Evaluate her 'a blank earns nothing, so guessing can only help' policy using expected value, given four options in English and three in the second language.
Read the worked answer
Expected value of a blind guess: EV = p(correct) × 3 + p(wrong) × (−1); a blank is worth 0.
Four options: (1/4)(3) + (3/4)(−1) = 0.75 − 0.75 = 0. Guessing is exactly neutral, not harmful - so 'a blank earns nothing, so it is safer' is false here, though guessing does not help either. If she can eliminate even one option: (1/3)(3) + (2/3)(−1) = 1 − 0.67 = +0.33, so any partial elimination converts a neutral guess into a profitable one.
Three options: (1/3)(3) + (2/3)(−1) = 1 − 0.67 = +0.33 per item. With three blanks left, expected gain is about +1 point, i.e. a third of an item - small but strictly positive, and free.
For comparison, five options: (1/5)(3) + (4/5)(−1) = 0.6 − 0.8 = −0.20: there, blank is correct.
The general rule: guessing pays when (options − 1) < 3, is neutral at exactly four options, and loses above four. Her policy is therefore right in the second-language section and harmless in English, but her justification ('can only help') is wrong: the sign of the expected value depends on the number of options, not on the fact that blanks score zero. One caveat: EV maximises the average score, not the probability of clearing a specific cut-off; sitting exactly at 55.00 with positive-EV gambles available, gambling is still the better play.
3. She wants 60.00 in English. Find the minimum number of (a) her six blanks that would have had to be correct, and (b) her five wrong answers that would have had to be blanks instead.
Read the worked answer
Target: 60% of 90 = 54 points. She currently has 52 ÷ 90 = 57.78.
(a) Blanks converted to correct. Each conversion adds the full 3 points: 52 + 3k ≥ 54, so k = 1 gives 55 ÷ 90 = 61.11, which clears 60. One correct answer from a blank is enough.
(b) Wrong answers that would have been blanks. Each conversion adds only 1 point (removing the −1 penalty): 52 + k ≥ 54, so k = 2 gives 54 ÷ 90 = 60.00 exactly. Two saved wrong answers are needed - and the result sits on the boundary, so three would be required for any real comfort.
Why the asymmetry: a correct answer is worth 3, a saved wrong answer is worth 1, a blank is worth 0. The exchange rate between the three outcomes is 3 : 1 : 0, which should drive her last-minute time allocation. With two minutes left, turning a blank into a checked answer is worth three times as much as re-checking one doubtful answer she has already marked - provided she is genuinely better than chance on that content, which is exactly the condition the expected-value calculation in the previous task formalises.
Note also the reverse reading: one wrong answer destroys three points' worth of nothing, so a random fill-in is not 'free' in a section with five options - but in English, at four options, it is.