Scholarship shortlist with a component floor
A college has two full scholarships. Five candidates are scored on three components-academic record, interview and community project-in that order. The selection rules are: weightage 40%, 30% and 30%; a candidate must score at least 60 in every component to be eligible; and if two eligible candidates tie on the weighted total, the higher community-project score wins. The scores are: A - 78, 64, 82; B - 88, 58, 76; C - 62, 72, 68; D - 70, 80, 60; E - 66, 68, 90. A member of the selection committee argues that B should still be considered because B has the highest academic score. Another member wants to replace the weights with a simple average of the three components. The chair asks for a defensible shortlist under the stated rules and for a check on whether the alternative method would change the outcome.
1. Compute weighted totals, apply eligibility, and name the two scholarship winners.
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Eligibility first: B has 58 in the interview, below the 60 floor, so B is excluded regardless of the 88 in academics. (If eligibility were ignored, B's weighted total would be 35.2 + 17.4 + 22.8 = 75.4, which would rank first - the floor is decisive.)\n\nWeighted totals = 0.4 × academic + 0.3 × interview + 0.3 × project:\nA = 31.2 + 19.2 + 24.6 = 75.0\nC = 24.8 + 21.6 + 20.4 = 66.8\nD = 28.0 + 24.0 + 18.0 = 70.0\nE = 26.4 + 20.4 + 27.0 = 73.8\n\nRanking among eligible candidates: A 75.0, E 73.8, D 70.0, C 66.8. Winners: A and E. The academic-score argument cannot override a mandatory eligibility rule.
2. If B's interview score is re-marked to 61 and all else stays the same, does the shortlist change? Show the new total.
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B's new total = 0.4 × 88 + 0.3 × 61 + 0.3 × 76 = 35.2 + 18.3 + 22.8 = 76.3. B is now eligible and ranks first. E (73.8) drops out, while A (75.0) remains second. New winners: B and A.\n\nThis is a threshold effect: a 3-mark change in one component both clears the eligibility floor and moves B from excluded to first. The order of operations matters - eligibility is checked before weighted ranking, not after.
3. Would a simple average of the three components change the set of winners? Compare and explain.
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Simple averages: A = (78 + 64 + 82) / 3 = 224 / 3 = 74.67; C = (62 + 72 + 68) / 3 = 202 / 3 = 67.33; D = (70 + 80 + 60) / 3 = 210 / 3 = 70.00; E = (66 + 68 + 90) / 3 = 224 / 3 = 74.67. B remains ineligible.\n\nA and E tie at 74.67, so the tie-break (higher community project) gives E first (90 vs 82) and A second. The set {A, E} is unchanged, but the ranking changes and the tie-break now does real work. Under the weighted method A led E by 1.2 points. The chair should note that the outcome is stable in membership but not in rank.