Weighted Admission Index and a Misleading Headline
Three applicants compete for the last seat in a college's MCA programme. The admission index is 0.50 × CET percentile + 0.30 × graduation percentage + 0.20 × Class 12 percentage, and is rounded to two decimals only at the end. Applicant A: CET 92.40, graduation 68.50%, Class 12 74.00%. Applicant B: CET 88.90, graduation 81.20%, Class 12 66.00%. Applicant C: CET 95.10, graduation 60.00%, Class 12 71.50%. The declared general-category cutoff is an index of 81.50. Ties are broken in favour of the higher CET percentile. The college also places any candidate whose graduation percentage is below 65 on a 'provisional' list regardless of index. A newspaper report claims that C is certain to get the seat because C has the highest CET percentile, and a parents' group argues that A is safe because A clears the cutoff.
1. Compute all three indices, apply the cutoff, the provisional rule and the tie-break, and decide who gets the seat.
Read the worked answer
A: 0.50 × 92.40 = 46.20; 0.30 × 68.50 = 20.55; 0.20 × 74.00 = 14.80 → 81.55 B: 0.50 × 88.90 = 44.45; 0.30 × 81.20 = 24.36; 0.20 × 66.00 = 13.20 → 82.01 C: 0.50 × 95.10 = 47.55; 0.30 × 60.00 = 18.00; 0.20 × 71.50 = 14.30 → 79.85
- C is below the 81.50 cutoff, and C's graduation percentage (60) is below 65, so C is provisional on two independent grounds. The newspaper's reasoning treats one component as if it were the whole index.
- A clears the cutoff by only 0.05, which is inside plausible rounding noise; A is eligible but not 'safe' in any robust sense.
- B has the highest index and clears the cutoff comfortably, so B gets the seat.
The lesson: a weighted index cannot be ranked by looking at its heaviest component alone.
2. Holding the other components fixed, what CET percentile would A need to at least tie B, and who would win that tie?
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A's non-CET contribution is 20.55 + 14.80 = 35.35. B's index is 82.01. Set 0.50x + 35.35 = 82.01 → 0.50x = 46.66 → x = 93.32.
So A needs a CET percentile of at least 93.32. At exactly 93.32 the indices tie, and the tie-break goes to the higher CET percentile - A's 93.32 against B's 88.90 - so A would win the tie. Any value below 93.32 leaves A behind.
This is a useful sanity check on the earlier result: A's deficit to B (82.01 − 81.55 = 0.46 index points) is worth 0.46 / 0.50 = 0.92 CET percentile points, which is why a modest CET improvement would have reversed the outcome.
3. The college proposes rounding each component percentage to the nearest integer before applying the weights. Does the ranking change, and what general principle does this illustrate?
Read the worked answer
Recomputing with pre-rounded components:
- A: graduation 68.50 → 69 → 20.70; Class 12 74 → 74 → 14.80; index = 46.20 + 20.70 + 14.80 = 81.70
- B: 81.20 → 81 → 24.30; 66 → 66 → 13.20; index = 44.45 + 24.30 + 13.20 = 81.95
- C: 60 → 60 → 18.00; 71.50 → 72 → 14.40; index = 47.55 + 18.00 + 14.40 = 79.95
Ranking is unchanged (B > A > C), but the margins move: A's cushion over the cutoff rises from 0.05 to 0.20, and C gains 0.10 for no real improvement.
Principle: rounding an intermediate quantity injects an error of up to half a unit, which the weight then magnifies or shrinks (here up to 0.15 index points per component). Near a cutoff, that is enough to flip an outcome. Round once, at the end. The same principle explains why a 0.05 margin should never be announced as 'safe'.