IAS 2026 — CSAT Maths

The correct option is (a) - 5.1.

Solution

The question asks for the cutoff score such that exactly three candidates are selected. The final score for each candidate is calculated using the weighted average formula:
Final Score = (Score 1 * 0.2) + (Score 2 * 0.3) + (Score 3 * 0.5)

Let's calculate the final score for each candidate:
Candidate 1: $(5 \times 0.2) + (4 \times 0.3) + (6 \times 0.5) = 1.0 + 1.2 + 3.0$(5×0.2)+(4×0.3)+(6×0.5)=1.0+1.2+3.0 = 5.2
Candidate 2: $(4 \times 0.2) + (6 \times 0.3) + (5 \times 0.5) = 0.8 + 1.8 + 2.5$(4×0.2)+(6×0.3)+(5×0.5)=0.8+1.8+2.5 = 5.1
Candidate 3: $(3 \times 0.2) + (2 \times 0.3) + (8 \times 0.5) = 0.6 + 0.6 + 4.0$(3×0.2)+(2×0.3)+(8×0.5)=0.6+0.6+4.0 = 5.2
Candidate 4: $(9 \times 0.2) + (4 \times 0.3) + (3 \times 0.5) = 1.8 + 1.2 + 1.5$(9×0.2)+(4×0.3)+(3×0.5)=1.8+1.2+1.5 = 4.5
Candidate 5: $(8 \times 0.2) + (8 \times 0.3) + (2 \times 0.5) = 1.6 + 2.4 + 1.0$(8×0.2)+(8×0.3)+(2×0.5)=1.6+2.4+1.0 = 5.0

Now, let's arrange the scores in descending order:
1. Candidate 1: 5.2
2. Candidate 3: 5.2
3. Candidate 2: 5.1
4. Candidate 5: 5.0
5. Candidate 4: 4.5

To select exactly three candidates (Candidates 1, 3, and 2), the cutoff must be the score of the third-ranked candidate. If the cutoff is set at 5.1, candidates with scores ≥ 5.1 are selected. This includes Candidate 1 (5.2), Candidate 3 (5.2), and Candidate 2 (5.1), totaling exactly three candidates.

Why the other options are incorrect

• Option (b) - 5.2: If the cutoff were 5.2, only Candidate 1 and Candidate 3 would be selected (2 candidates), failing the requirement of selecting exactly three.
• Option (c) - 5.3: If the cutoff were 5.3, no candidates would be selected as the highest score achieved is 5.2.
• Option (d) - 5.4: Similar to 5.3, a cutoff of 5.4 is higher than all individual scores, resulting in zero selections.

Key Concept

Weighted Average and Ranking: The cutoff for selecting 'n' candidates is determined by the score of the n-th candidate when all scores are sorted in descending order.