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%

Answer 1

The correct option is D 75%

L = London, Paris = P, New York= NY, Beijing = B In round III, one of the two cities, either London or Paris will get 38 votes and the other 37.
Further:
1. The persons representing London, Paris, Beijing and New York cannot vote as long as their own cities are in contention. In round I, New York gets eliminated and hence the representative from NY becomes eligible for voting in the II round hence increasing the total votes by 1. This means the total votes in the first round must be 83 - 1 = 82.

2. After round II, the representative from Beijing votes in the III round. This should have increased the number of total votes by 1 and the total votes must have become 83 + 1 =84.

3. We are given that the total votes in round III are 75 only. We conclude that 84 - 75 = 9 people in who vote n round I and II have become ineligible for voting in round III.

4. 9 people who have voted in round II and II become ineligible for voting in round III. The reason of their ineligibility is that till round and II. they have already voted for two different cities which are not available for contention in round III. All of these 9 voters ore those who voted for NY in round I and then voted for Beijing in round II.

5. Beijing's vote in round II is 21. This includes 9 votes from people who voted for NY in the first round. So 21 - 9 = 12 people voted for Beijing in both round I and II

6. We are given that 75% of the people who voted for Beijing in round I. voted again for Beijing in round II as well. So, 16 people must have voted for Beijing in round I.

7. In round I we have:
82 -L + P + B +NY or 82 -30 + P + 16 + 12
Giving P = 24

8. In round II, we have:
83= L+ 32 + 21, giving L = 30

NY had 12 votes in round 1. 9 of these votes went to B(see point 2, again). The rest 3 went to P.

9. 16 votes for B in round I. 12 of them still vote for B. The rest 4 voted for either L or P. As L has the same number of votes in both the rounds I and II. This means in round II, these 4 votes must have gone to Paris only.

10. The representative from NY did not vote in round I but has voted in round II. As L has the same people voting for it (30 votes in both the rounds I and II) and we know the exact break up of B in II. This NY representative vote must go to Paris only. Further. in order to avoid ineligibility, this NY rep must vote for Paris only in round III also.

11. Paris (in round II) break up is:
32 = 24 (from round I, who voted for Paris ) +4 (out of the 16, who voted for Beijing in round I) + 3(out of 12, who voted for NY in round I) + 1 (NY -Rep)

12. Beijing gets eliminated in round II. So, the rep of Beijing can vote in round III.

13. 12 People (out of 21) who voted for Beijing in round II are still eligible for vote in round III.

14. 50% of people who voted for Beijing in I ( i.e. 8 People) voted for Paris in round III. These 8 People include 4 of those who voted for Paris in round II also. Therefore 4(out of 12 who voted for Beijing in round II and are still eligible for vote in round III ) people have voted for Paris in round III.

15. This implies that the rest 8(out of 12 who voted for Beijing in round II and are still eligible for vote in round III) can vote for London only. This makes London's vote = 30+ 8 or 38 in round III which implies that Paris got 37 votes.

16. The Beijing Rep who is eligible to vote in round III must have voted for Paris only.

$\begin{array}{cccccc}Round& TotalVotes& London\left(L\right)& Paris\left(P\right)& Beijing\left(B\right)& NewYork\left(NY\right)\\ I& 82& 30& 24& 16& 12\\ II& 83& 30& 32=(24+4+3+1ofNY-rep)& 21(12+9)& \times \\ III& 75& 38=(30+8)& 37=(32+4+1ofB-rep)& \times & \times \end{array}$
Required percentage $=\frac{9}{12}\times 100=75\%$