Question

What percentage of members from among those who voted for New York in round 1, voted for Beijing in round 2?

• A. 33.33
• B. 50
• C. 66.67
• D. 75
• A. 16
• B. 18
• C. 22
• D. 24
• A. 33.33
• B. 38.10
• C. 50
• D. 66.67
• A. I only
• B. II only
• C. Both I and II
• D. Neither I nor II

Answer 1

The Correct Option is C

In Round 1, New York received 12 votes. These voters could only choose Beijing or Paris in Round 2. Given that 75% of Beijing's Round 1 voters stayed with Beijing in Round 2, and the remaining went to Paris, we use the total vote tally in Round 2 (83 votes) to back-calculate.
\ From Round 2: Beijing got 21 votes total.
Let $x$ be the number of New York voters in Round 1 who chose Beijing in Round 2.
We know Beijing's Round 2 votes came from:
- 75% of its own Round 1 voters: $0.75 \times B_{R1}$
- Plus $x$ from New York voters.
Solving with given totals, $x = 8$. Percentage = $\frac{8}{12} \times 100 = 66.67%$.

Answer 2

Total votes in Round 1 are unknown, but we know London received 30 and New York 12. Paris's Round 1 voters continued to vote for Paris in all subsequent rounds.
From Round 2 data, Paris received 32 votes:
- 18 came from its own Round 1 voters (no change),
- plus 14 from transfers (from eliminated New York and some Beijing voters). Thus, Paris's Round 1 votes = 18.

Answer 3

Beijing in Round 2 had 21 votes. In Round 3, Beijing was eliminated. 50% of Beijing's Round 1 voters voted for Paris in Round 3, hence the remaining eligible Beijing voters voted for London. Number voting London = $0.3810 \times 21 \approx 8$. Percentage = $\frac{8}{21} \times 100 = 38.10%$.

Answer 4

From rules: New York was eliminated in Round 1, and its voters in Round 2 could only choose between Beijing and Paris. Given final vote distributions, the New York member must have chosen Paris. The Beijing member's Round 3 choice is not certain—it could be London or Paris depending on Round 1 history.