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100 families planned wedding on 21st December 2020 at 9 pm, an auspicious day and time in New Delhi and booked a marriage hall online in Chattarpur Farms. Due to a technical glitch the hall was booked simultaneously for 64 families who received a confirmation with a deposit of ₹ 1 Lakh each. The managementon identifying the glitch was troubled and contacted the families for a refund but the families were not ready to accommodate. The managementand the families then mutually agreed to have a toss in order to decide which family will be finally allotted the venue. Certain directions regarding holdingthe toss are as follows:
1. The families were allotted a number from 1 to 64 in alphabetical ascending order of their names.
2. It was further decided, Family 1 will have a toss with Family 64 andis considered as Toss 1. Toss 2 will be between Family 2 and Family 63 and Toss 3 between Family 3 and Family 62 and so on.
3. The winner of each toss moves to the next round and the winner of Toss 1 will now have a toss with winner of Toss 32, winner of Toss 2 with Toss 31 and so on.
4. The remaining rounds of toss are in ascending order i.e. winner of Toss 1 with Toss 2, winner of Toss 3 with Toss 4 and so till there is one winner.
5. It was also decided that if the final winner withdrawsand hands over the venue to the other finalist, the winner will be given ₹ 1.5 Lakh over and above the refund amount.
If family 51 won the toss, then with which family can they not have a toss in the finals?
1. 27
2. 24
3. 29
4. 37
The below table shows the matching of families in each round as per the given conditions.
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First, we will trace all the matches of 51 in various rounds.
Round 1 against 14, round 2 against 19, round 3 against 13, round 4 against 15, round 5 against 9, and round 6 against 1.
Thus, 51 takes the place of 9 in the final round.
The family that will also take the place of 9 in the finals cannot have a toss with 51.
1. 27
Round 1 against 38, round 2 against 6, round 3 against 5, round 4 against 7, round 5 against 1, and round 6 against 9. Thus, it will take the place of 1 in the final round. Hence, it can play against 51.
Similarly, 29 and 37 can also play against 51 in the final round. Thus, all three can be ruled out.
2. 24.
Round 1 against 41, round 2 against 9, round 3 against 10, round 4 against 11, round 5 against 13, round 6 against 1.
Thus, 24 will also take the place of 9 in the final round.
Hence, 24 can never toss against 51.
Hence, the answer is option B.
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