Scholarship Ranking with Eligibility and Weights
A scholarship committee ranks four candidates using weighted scores: academic record 60%, interview 25%, extracurricular 15%. Each component is scored 0–100. Candidate A has academic 82, interview 70, extracurricular 90. Candidate B has academic 88, interview 60, extracurricular 75. Candidate C has academic 79, interview 85, extracurricular 80. Candidate D has academic 84, interview 80, extracurricular 60. The committee also has an eligibility rule: a candidate must score at least 80 on academic record to be considered. The top eligible candidate receives the only scholarship. Scores are combined as a weighted average; there is no rounding until the final total. The committee is considering changing the weights to 50% academic, 30% interview, and 20% extracurricular for future years, but this year's decision must use the current weights.
1. Calculate the weighted scores for all four candidates under the current weights and identify the winner after applying the eligibility rule.
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A: 0.60(82) + 0.25(70) + 0.15(90) = 49.2 + 17.5 + 13.5 = 80.2. B: 0.60(88) + 0.25(60) + 0.15(75) = 52.8 + 15.0 + 11.25 = 79.05. C: 0.60(79) + 0.25(85) + 0.15(80) = 47.4 + 21.25 + 12.0 = 80.65. D: 0.60(84) + 0.25(80) + 0.15(60) = 50.4 + 20.0 + 9.0 = 79.4. Eligibility requires academic ≥ 80, so C (79) is excluded. Among A, B, and D, A has the highest total (80.2 vs 79.4 vs 79.05). A wins.
2. What interview score would B need to tie A for the scholarship, assuming B's other scores and A's total stay unchanged?
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A's total is 80.2. B's current total is 79.05, a gap of 1.15 points. Interview counts for 25% of the total, so each additional interview point adds 0.25 to B's total. B needs 1.15 ÷ 0.25 = 4.6 more interview points. B's interview score must be 60 + 4.6 = 64.6 to tie A, all else unchanged. If the committee breaks ties in A's favor, B needs slightly more than 64.6 to win outright.
3. If the committee used the proposed 50/30/20 weights this year, would the winner change? Show totals.
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Proposed totals: A = 0.50(82) + 0.30(70) + 0.20(90) = 41 + 21 + 18 = 80.0. B = 0.50(88) + 0.30(60) + 0.20(75) = 44 + 18 + 15 = 77.0. C = 0.50(79) + 0.30(85) + 0.20(80) = 39.5 + 25.5 + 16 = 81.0, but C is still ineligible. D = 0.50(84) + 0.30(80) + 0.20(60) = 42 + 24 + 12 = 78.0. Among eligible candidates A leads with 80.0, so the winner does not change. The weights change margins, not the eligibility outcome.