CAT 2023 Slot 1 — DILR Question 6

Mixed PracticeEasy

Passage / Data

Answer the following questions based on the information given below:

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 two candidates can belong to the same department?

Answer & solution

Prof. Pakrasi and Prof. Qureshi
B
Prof. Qureshi and Prof. Ramaswamy
C
Prof. Pakrasi and Prof. Samuel
D
Prof. Ramaswamy and Prof. Samuel

Solution

Easy

One set-up solves all five questions. Reconstruct the vote flow: each department's non-candidates vote as a single block for one candidate, no one votes within their own department, and the four candidates' cross-votes are given. Pin down every block's size and destination first.

Departments and sizes: F&A $=9$ , M&S $=7$ , O&Q $=5$ , B&H $=3$ ; total $=24$ faculty, so total votes $=24=3+14+6+1$ ✓. Candidates' cross-votes: P $\to$ R, Q $\to$ S, R $\to$ Q, S $\to$ P. Removing these, the department blocks must add to each candidate's tally: Pakrasi $3-1=2$ , Qureshi $14-1=13$ , Ramaswamy $6-1=5$ , Samuel $1-1=0$ . The 20 non-candidates split into exactly these block sums.

Size each block. O&Q has exactly one candidate, so its block $=5-1=4$ . A department with $k$ candidates contributes a block of (size $-\,k$ ). The only split of $\{F\&A,\,M\&S,\,O\&Q,\,B\&H\}$ blocks into $\{2,13,5,0\}$ is:

\begin{aligned} &\text{B\&H}=2\ (\text{1 cand.})\ \to\ \text{Pakrasi}\\ &\text{M\&S}=5\ (\text{2 cand.})\ \to\ \text{Ramaswamy}\\ &\text{F\&A}=9\ (\text{0 cand.})\ \to\ \text{Qureshi}\\ &\text{O\&Q}=4\ (\text{1 cand.})\ \to\ \text{Qureshi} \end{aligned}

Check: Q $=9+4+1=14$ , R $=5+1=6$ , P $=2+1=3$ , S $=0+1=1$ ✓. So candidates number $0,2,1,1$ across F&A, M&S, O&Q, B&H.

Locate the M&S pair. M&S holds 2 candidates and its block votes Ramaswamy, so by Rule 3 Ramaswamy $\notin$ M&S — the pair is two of $\{P,Q,S\}$ . Two candidates in one department cannot vote for each other. Testing the three pairs: $\{P,S\}$ fails (S $\to$ P, same dept) and $\{Q,S\}$ fails (Q $\to$ S, same dept). Only $\{P,Q\}$ survives.

So Pakrasi and Qureshi share M&S. That is exactly the pair this question asks for.

Prof. Pakrasi and Prof. Qureshi.