### Brain Teasers

# The Chairman

Logic
Logic puzzles require you to think. You will have to be logical in your reasoning.

Five men among A, B, C, D and E sitting in this order at a round table had to decide as to who among them would be the chairman by voting. None of the five members voted for himself or his neighbors. The first ballot was a tie off. The second time, C voted for E while the others stuck to their old choices, thereby resulting in E's victory. Who voted for B in the first ballot?

### Answer

D voted for B in the first ballot.A, B, C, D and E are sitting in this order at a round table.

The observations given are

1) None of the five members voted for himself or his neighbors.

2) The first ballot was a tie off.

3) The second time, C voted for E while the others stuck to their old choices, thereby resulting in E's victory.

From observation 1, we can deduce the following table for the first ballot:-

Person --> Voted for

A --> D or C

B --> E or D

C --> A or E

D --> A or B

E --> B or C

From observation 2, all of them must have got one vote each since the first ballot was a tie off. Now the above table can be split into one of the following two tables using observation 2.

Table 1:

Person --> Voted for

A --> D

B --> E

C --> A

D --> B

E --> C

OR

Table 2:

Person --> Voted for

A --> C

B --> D

C --> E

D --> A

E --> B

The two tables above would ensure a tie-off for the first ballot.

Now from observation 3, the second time, C voted for E while the others stuck to their old choices, thereby resulting in E's victory. This situation can emerge only if table 1 is the initial voting pattern and not table 2 (since from table 2, C anyway votes for E. If the others stuck to their choices, the second ballot would also result in a tie off). Keeping the initial pattern to be table 1, E won the second ballot because he received two votes from B and C.

Hence if table 1 was the initial voting pattern, then D voted for B in the first ballot.

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