CAT 2023 Slot 1 · DILR · Question 20
CAT 2023 Slot 1 previous year question on Caselet DI, with the answer and a step-by-step solution.
Set 4: Election for the Dean — this question belongs to a set. Read the full set with all 5 questions →
Faculty members in a management school can belong to one of four departments - Finance and Accounting (F&A), Marketing and Strategy (M&S), Operations and Quants (O&Q) and Behaviour and Human Resources (B&H). The numbers of faculty members in F&A, M&S, O&Q and B&H departments are 9, 7, 5 and 3 respectively.
Prof. Pakrasi, Prof. Qureshi, Prof. Ramaswamy and Prof. Samuel are four members of the school's faculty who were candidates for the post of the Dean of the school. Only one of the candidates was from O&Q.
Every faculty member, including the four candidates, voted for the post. In each department, all the faculty members who were not candidates voted for the same candidate. The rules for the election are listed below.
1. There cannot be more than two candidates from a single department.
2. A candidate cannot vote for himself/herself.
3. Faculty members cannot vote for a candidate from their own department.
After the election, it was observed that Prof. Pakrasi received 3 votes, Prof. Qureshi received 14 votes, Prof. Ramaswamy received 6 votes and Prof. Samuel received 1 vote. Prof. Pakrasi voted for Prof. Ramaswamy, Prof. Qureshi for Prof. Samuel, Prof. Ramaswamy for Prof. Qureshi and Prof. Samuel for Prof. Pakrasi.
Which of the following statements is/are true?
Statement A: Non-candidates from M&S voted for Prof. Qureshi.
Statement B: Non-candidates from F&A voted for Prof. Qureshi.
- ABoth statements A and B
- BOnly statement B
- COnly statement A
- DNeither statement A nor statement B
Answer & solution
Answer: Option B
Solving the set
Step 1 – Votes from candidates: Pakrasi → Ramaswamy, Qureshi → Samuel, Ramaswamy → Qureshi, Samuel → Pakrasi. So each candidate gets exactly one vote from the candidates; the rest come from department blocks (all non-candidates of a department vote together).
Department | Size | Non-candidates if 0 / 1 / 2 candidates |
F&A | 9 | 9 / 8 / 7 |
M&S | 7 | 7 / 6 / 5 |
O&Q | 5 | — / 4 / — (exactly one candidate) |
B&H | 3 | 3 / 2 / 1 |
Step 2 – Match the block sizes.
• Samuel: 1 vote, all from Qureshi. No block voted for him.
• Pakrasi: 3 − 1 = 2 block votes. Only a B&H block with one candidate has exactly 2 members. So B&H has one candidate and its block voted for Pakrasi.
• Ramaswamy: 6 − 1 = 5 block votes. A block of 5 is possible only for M&S with two candidates (O&Q has exactly one candidate, so its block is 4).
• Qureshi: 14 − 1 = 13 block votes = F&A (9, no candidate) + O&Q (4).
Candidate | Votes | From candidates | From department blocks |
Pakrasi | 3 | 1 (Samuel) | 2 = B&H block (B&H has one candidate) |
Qureshi | 14 | 1 (Ramaswamy) | 13 = F&A (9, no candidate) + O&Q (4) |
Ramaswamy | 6 | 1 (Pakrasi) | 5 = M&S block (M&S has two candidates) |
Samuel | 1 | 1 (Qureshi) | 0 |
Step 3 – Departments. Candidates: two in M&S, one in O&Q, one in B&H, none in F&A. Ramaswamy is not in M&S (M&S votes for him). Rule 3 means a candidate cannot vote for someone from the same department, so Samuel is not with Qureshi or Pakrasi. The M&S pair must therefore be Pakrasi and Qureshi. Ramaswamy and Samuel take O&Q and B&H in some order (Pakrasi is not in B&H, which voted for him; Qureshi is not in O&Q, which voted for him).
Candidate | Department | Votes for |
Pakrasi | M&S | Ramaswamy |
Qureshi | M&S | Samuel |
Ramaswamy | O&Q or B&H | Qureshi |
Samuel | B&H or O&Q (the other one) | Pakrasi |
This question
From Step 2, the M&S block (5) voted for Ramaswamy, so statement A is false. The F&A block (9) voted for Qureshi, so statement B is true.
Hence, the answer is option 2.
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