Games and Tournaments Questions for CAT
Games and Tournaments is one of the most important topics in the LRDI Section of the CAT Exam. You can check out these Games and Tournaments Passages in CAT Previous year papers. In this Section, you can watch these Games and Tournaments basics to improve your RC Skills in CAT. This article will look into some important Games and Tournaments Questions for CAT. These are good sources for practice; If you want to practice these questions, you can download this CAT Games and Tournaments Questions PDF below, which is completely Free.
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Instructions
10 players – P1, P2, … , P10 – competed in an international javelin throw event. The number (after P) of a player reflects his rank at the beginning of the event, with rank 1 going to the topmost player. There were two phases in the event with the first phase consisting of rounds 1, 2, and 3, and the second phase consisting of rounds 4, 5, and 6. A throw is measured in terms of the distance it covers (in meters, up to one decimal point accuracy), only if the throw is a ‘valid’ one. For an invalid throw, the distance is taken as zero. A player’s score at the end of a round is the maximum distance of all his throws up to that round. Players are re-ranked after every round based on their current scores. In case of a tie in scores, the player with a prevailing higher rank retains the higher rank. This ranking determines the order in which the players go for their throws in the next round.
In each of the rounds in the first phase, the players throw in increasing order of their latest rank, i.e. the player ranked 1 at that point throws first, followed by the player ranked 2 at that point and so on. The top six players at the end of the first phase qualify for the second phase. In each of the rounds in the second phase, the players throw in decreasing order of their latest rank i.e. the player ranked 6 at that point throws first, followed by the player ranked 5 at that point and so on. The players ranked 1, 2, and 3 at the end of the sixth round receive gold, silver, and bronze medals respectively.
All the valid throws of the event were of distinct distances (as per stated measurement accuracy). The tables below show distances (in meters) covered by all valid throws in the first and the third round in the event.
Distances covered by all the valid throws in the first round
Distances covered by all the valid throws in the third round
The following facts are also known.
i. Among the throws in the second round, only the last two were valid. Both the throws enabled these players to qualify for the second phase, with one of them qualifying with the least score. None of these players won any medal.
ii. If a player throws first in a round AND he was also the last (among the players in the current round) to throw in the previous round, then the player is said to get a double. Two players got a double.
iii. In each round of the second phase, exactly one player improved his score. Each of these improvements was by the same amount.
iv. The gold and bronze medalists improved their scores in the fifth and the sixth rounds respectively. One medal winner improved his score in the fourth round.
v. The difference between the final scores of the gold medalist and the silver medalist, as well as the difference between the final scores of the silver medalist and the bronze medalist was 1.0 m.
Question 1: By how much did the gold medalist improve his score (in m) in the second phase?
a) 1.0
b) 2.0
c) 2.4
d) 1.2
1) Answer (C)
Solution:
Let us arrange the players in the order in which they throw in each round.
Round 1: Here the players throw in order of their initial seeds so the order is as follows:
So, their rank at the start of Round 2 is in order of their throws in the first round. Also, we need to consider the same order for people having invalid throws.
Round 2: P2, P4, P8, P10 had the same relative rankings since they all have invalid throws, that is, 0 metres. Rest are arranged as per their throw distances.
Now, in round 2, only P8 and P10 had valid throws. Hence, their order will change at the start of Round 3, however, the remaining order stays the same. That is, P8 and P10 will move up in the table and occupy some higher places, whereas some of the others may move down consequently.
Round 3: In Round 3, we can see that P1 improved his score from 82.9 to 88.6. The other 2 participants did not improve their scores. Also after Round 3, P8 and P10 qualify, where one of P8 or P10 is at the sixth position. So at the end of Round 3, we can say that P6, P3, P2, and P4 are at the bottom 4 positions(ranks). One of P8 or P10 is at the sixth position. P1 > P7 > P5 > P9.
So at the end of round 3, the ranks are as follows:
The other person between P8/P10 can go anywhere between Rank 1 and Rank 5.
Now let us consider the two players who got a double. Doubles happen in the transition between rounds.
1 -> 2 – Not possible
2 -> 3 – Possible if P10 reaches Rank 1 after round 2.
3 -> 4 – P8/P10 who is the last among qualifying will be the first to throw. So, here it definitely happens.
4 – > 5 AND 5 – > 6 not possible.
So, after Round 2, definitely, P10 reaches the top of the ranking. P8 is at the bottom. Hence, after Round 3, P10 either retains rank 1 or P1 surpasses him and P10 becomes Rank 2.
So, two combinations are possible at the end of Round 3:
Now, we know that in each of the rounds in phase 2, only one player improves his score. Also, P8 and P10 cannot win medals. Hence, in case 1, three of P1, P7, P5 and P9 will improve their scores by x and reach the top 3 positions. However, the top 3 positions’ distances are in AP.
P1 – 88.6 + x
P7 – 87.2 + x
P5 – 86.4 + x
P9 – 84.1 + x
The differences do not satisfy the condition. Hence, case 1 is invalidated.
Case 2: Here, P1 definitely wins a medal, and P10 does not. So, two of P7, P5 and P9 jumps above P10. Now, if we have three different people increasing their scores or distances in each of the three rounds, again we would not get a difference of 1 among the Gold, Silver and Bronze medallists. Hence, one of them increases his score twice and the other increases his score twice and none of them is P1.
Let us take the cases where P1 is individually the G, S and B medallists.
Case 1: P1 is a G medallist.
P1 – 88.6
The silver medallist is 87.6 and the bronze medallist is 86.6 metres. However, P10 has thrown for a distance that is greater than 87.2 metres. Hence, in this case, he would be the B medallist. Hence, this is not the right case.
Case 2: P1 is the S medallist.
P1 – 88.6
G – 89.6
B – 87.6
Now, if we see the differences
89.6 – 87.2 = 2.4
87.6 – 86.4 = 1.2
This satisfies the condition that P7 has increased his score twice to become the gold winner and P5 has increased it once to become the bronze winner.
Hence, P1 – Silver
P7 – Gold
P5 – Bronze
P7 improved by a total of 2.4.
Question 2: Which of the following can be the final score (in m) of P8?
a) 81.9
b) 0
c) 82.7
d) 85.1
2) Answer (C)
Solution:
Let us arrange the players in the order in which they throw in each round.
Round 1: Here the players throw in order of their initial seeds so the order is as follows:
So, their rank at the start of Round 2 is in order of their throws in the first round. Also, we need to consider the same order for people having invalid throws.
Round 2: P2, P4, P8, P10 had the same relative rankings since they all have invalid throws, that is, 0 metres. Rest are arranged as per their throw distances.
Now, in round 2, only P8 and P10 had valid throws. Hence, their order will change at the start of Round 3, however, the remaining order stays the same. That is, P8 and P10 will move up in the table and occupy some higher places, whereas some of the others may move down consequently.
Round 3: In Round 3, we can see that P1 improved his score from 82.9 to 88.6. The other 2 participants did not improve their scores. Also after Round 3, P8 and P10 qualify, where one of P8 or P10 is at the sixth position. So at the end of Round 3, we can say that P6, P3, P2, and P4 are at the bottom 4 positions(ranks). One of P8 or P10 is at the sixth position. P1 > P7 > P5 > P9.
So at the end of round 3, the ranks are as follows:
The other person between P8/P10 can go anywhere between Rank 1 and Rank 5.
Now let us consider the two players who got a double. Doubles happen in the transition between rounds.
1 -> 2 – Not possible
2 -> 3 – Possible if P10 reaches Rank 1 after round 2.
3 -> 4 – P8/P10 who is the last among qualifying will be the first to throw. So, here it definitely happens.
4 – > 5 AND 5 – > 6 not possible.
So, after Round 2, definitely, P10 reaches the top of the ranking. P8 is at the bottom. Hence, after Round 3, P10 either retains rank 1 or P1 surpasses him and P10 becomes Rank 2.
So, two combinations are possible at the end of Round 3:
Now, we know that in each of the rounds in phase 2, only one player improves his score. Also, P8 and P10 cannot win medals. Hence, in case 1, three of P1, P7, P5 and P9 will improve their scores by x and reach the top 3 positions. However, the top 3 positions’ distances are in AP.
P1 – 88.6 + x
P7 – 87.2 + x
P5 – 86.4 + x
P9 – 84.1 + x
The differences do not satisfy the condition. Hence, case 1 is invalidated.
Case 2: Here, P1 definitely wins a medal, and P10 does not. So, two of P7, P5 and P9 jumps above P10. Now, if we have three different people increasing their scores or distances in each of the three rounds, again we would not get a difference of 1 among the Gold, Silver and Bronze medallists. Hence, one of them increases his score twice and the other increases his score twice and none of them is P1.
Let us take the cases where P1 is individually the G, S and B medallists.
Case 1: P1 is a G medallist.
P1 – 88.6
The silver medallist is 87.6 and the bronze medallist is 86.6 metres. However, P10 has thrown for a distance that is greater than 87.2 metres. Hence, in this case, he would be the B medallist. Hence, this is not the right case.
Case 2: P1 is the S medallist.
P1 – 88.6
G – 89.6
B – 87.6
Now, if we see the differences
89.6 – 87.2 = 2.4
87.6 – 86.4 = 1.2
This satisfies the condition that P7 has increased his score twice to become the gold winner and P5 has increased it once to become the bronze winner.
Hence, P1 – Silver
P7 – Gold
P5 – Bronze
P8 comes between P9 and P6.
Hence, 82. 5 < P8 < 84.1
P8 = 82.7
Question 3: What was the final score (in m) of the silver-medalist?
a) 89.6
b) 88.4
c) 88.6
d) 87.2
3) Answer (C)
Solution:
Let us arrange the players in the order in which they throw in each round.
Round 1: Here the players throw in order of their initial seeds so the order is as follows:
So, their rank at the start of Round 2 is in order of their throws in the first round. Also, we need to consider the same order for people having invalid throws.
Round 2: P2, P4, P8, P10 had the same relative rankings since they all have invalid throws, that is, 0 metres. Rest are arranged as per their throw distances.
Now, in round 2, only P8 and P10 had valid throws. Hence, their order will change at the start of Round 3, however, the remaining order stays the same. That is, P8 and P10 will move up in the table and occupy some higher places, whereas some of the others may move down consequently.
Round 3: In Round 3, we can see that P1 improved his score from 82.9 to 88.6. The other 2 participants did not improve their scores. Also after Round 3, P8 and P10 qualify, where one of P8 or P10 is at the sixth position. So at the end of Round 3, we can say that P6, P3, P2, and P4 are at the bottom 4 positions(ranks). One of P8 or P10 is at the sixth position. P1 > P7 > P5 > P9.
So at the end of round 3, the ranks are as follows:
The other person between P8/P10 can go anywhere between Rank 1 and Rank 5.
Now let us consider the two players who got a double. Doubles happen in the transition between rounds.
1 -> 2 – Not possible
2 -> 3 – Possible if P10 reaches Rank 1 after round 2.
3 -> 4 – P8/P10 who is the last among qualifying will be the first to throw. So, here it definitely happens.
4 – > 5 AND 5 – > 6 not possible.
So, after Round 2, definitely, P10 reaches the top of the ranking. P8 is at the bottom. Hence, after Round 3, P10 either retains rank 1 or P1 surpasses him and P10 becomes Rank 2.
So, two combinations are possible at the end of Round 3:
Now, we know that in each of the rounds in phase 2, only one player improves his score. Also, P8 and P10 cannot win medals. Hence, in case 1, three of P1, P7, P5 and P9 will improve their scores by x and reach the top 3 positions. However, the top 3 positions’ distances are in AP.
P1 – 88.6 + x
P7 – 87.2 + x
P5 – 86.4 + x
P9 – 84.1 + x
The differences do not satisfy the condition. Hence, case 1 is invalidated.
Case 2: Here, P1 definitely wins a medal, and P10 does not. So, two of P7, P5 and P9 jumps above P10. Now, if we have three different people increasing their scores or distances in each of the three rounds, again we would not get a difference of 1 among the Gold, Silver and Bronze medallists. Hence, one of them increases his score twice and the other increases his score twice and none of them is P1.
Let us take the cases where P1 is individually the G, S and B medallists.
Case 1: P1 is a G medallist.
P1 – 88.6
The silver medallist is 87.6 and the bronze medallist is 86.6 metres. However, P10 has thrown for a distance that is greater than 87.2 metres. Hence, in this case, he would be the B medallist. Hence, this is not the right case.
Case 2: P1 is the S medallist.
P1 – 88.6
G – 89.6
B – 87.6
Now, if we see the differences
89.6 – 87.2 = 2.4
87.6 – 86.4 = 1.2
This satisfies the condition that P7 has increased his score twice to become the gold winner and P5 has increased it once to become the bronze winner.
Hence, P1 – Silver
P7 – Gold
P5 – Bronze
P1 – 88.6 m.
Question 4: Who threw the last javelin in the event?
a) P7
b) P1
c) P9
d) P10
4) Answer (A)
Solution:
Let us arrange the players in the order in which they throw in each round.
Round 1: Here the players throw in order of their initial seeds so the order is as follows:
So, their rank at the start of Round 2 is in order of their throws in the first round. Also, we need to consider the same order for people having invalid throws.
Round 2: P2, P4, P8, P10 had the same relative rankings since they all have invalid throws, that is, 0 metres. Rest are arranged as per their throw distances.
Now, in round 2, only P8 and P10 had valid throws. Hence, their order will change at the start of Round 3, however, the remaining order stays the same. That is, P8 and P10 will move up in the table and occupy some higher places, whereas some of the others may move down consequently.
Round 3: In Round 3, we can see that P1 improved his score from 82.9 to 88.6. The other 2 participants did not improve their scores. Also after Round 3, P8 and P10 qualify, where one of P8 or P10 is at the sixth position. So at the end of Round 3, we can say that P6, P3, P2, and P4 are at the bottom 4 positions(ranks). One of P8 or P10 is at the sixth position. P1 > P7 > P5 > P9.
So at the end of round 3, the ranks are as follows:
The other person between P8/P10 can go anywhere between Rank 1 and Rank 5.
Now let us consider the two players who got a double. Doubles happen in the transition between rounds.
1 -> 2 – Not possible
2 -> 3 – Possible if P10 reaches Rank 1 after round 2.
3 -> 4 – P8/P10 who is the last among qualifying will be the first to throw. So, here it definitely happens.
4 – > 5 AND 5 – > 6 not possible.
So, after Round 2, definitely, P10 reaches the top of the ranking. P8 is at the bottom. Hence, after Round 3, P10 either retains rank 1 or P1 surpasses him and P10 becomes Rank 2.
So, two combinations are possible at the end of Round 3:
Now, we know that in each of the rounds in phase 2, only one player improves his score. Also, P8 and P10 cannot win medals. Hence, in case 1, three of P1, P7, P5 and P9 will improve their scores by x and reach the top 3 positions. However, the top 3 positions’ distances are in AP.
P1 – 88.6 + x
P7 – 87.2 + x
P5 – 86.4 + x
P9 – 84.1 + x
The differences do not satisfy the condition. Hence, case 1 is invalidated.
Case 2: Here, P1 definitely wins a medal, and P10 does not. So, two of P7, P5 and P9 jumps above P10. Now, if we have three different people increasing their scores or distances in each of the three rounds, again we would not get a difference of 1 among the Gold, Silver and Bronze medallists. Hence, one of them increases his score twice and the other increases his score twice and none of them is P1.
Let us take the cases where P1 is individually the G, S and B medallists.
Case 1: P1 is a G medallist.
P1 – 88.6
The silver medallist is 87.6 and the bronze medallist is 86.6 metres. However, P10 has thrown for a distance that is greater than 87.2 metres. Hence, in this case, he would be the B medallist. Hence, this is not the right case.
Case 2: P1 is the S medallist.
P1 – 88.6
G – 89.6
B – 87.6
Now, if we see the differences
89.6 – 87.2 = 2.4
87.6 – 86.4 = 1.2
This satisfies the condition that P7 has increased his score twice to become the gold winner and P5 has increased it once to become the bronze winner.
Hence, P1 – Silver
P7 – Gold
P5 – Bronze
P7 the gold winner had already received rank 1 at the end of Round 5. Hence, he was the last one to throw in the tournament.
Question 5: Who won the silver medal?
a) P5
b) P7
c) P9
d) P1
5) Answer (D)
Solution:
Let us arrange the players in the order in which they throw in each round.
Round 1: Here the players throw in order of their initial seeds so the order is as follows:
So, their rank at the start of Round 2 is in order of their throws in the first round. Also, we need to consider the same order for people having invalid throws.
Round 2: P2, P4, P8, P10 had the same relative rankings since they all have invalid throws, that is, 0 metres. Rest are arranged as per their throw distances.
Now, in round 2, only P8 and P10 had valid throws. Hence, their order will change at the start of Round 3, however, the remaining order stays the same. That is, P8 and P10 will move up in the table and occupy some higher places, whereas some of the others may move down consequently.
Round 3: In Round 3, we can see that P1 improved his score from 82.9 to 88.6. The other 2 participants did not improve their scores. Also after Round 3, P8 and P10 qualify, where one of P8 or P10 is at the sixth position. So at the end of Round 3, we can say that P6, P3, P2, and P4 are at the bottom 4 positions(ranks). One of P8 or P10 is at the sixth position. P1 > P7 > P5 > P9.
So at the end of round 3, the ranks are as follows:
The other person between P8/P10 can go anywhere between Rank 1 and Rank 5.
Now let us consider the two players who got a double. Doubles happen in the transition between rounds.
1 -> 2 – Not possible
2 -> 3 – Possible if P10 reaches Rank 1 after round 2.
3 -> 4 – P8/P10 who is the last among qualifying will be the first to throw. So, here it definitely happens.
4 – > 5 AND 5 – > 6 not possible.
So, after Round 2, definitely, P10 reaches the top of the ranking. P8 is at the bottom. Hence, after Round 3, P10 either retains rank 1 or P1 surpasses him and P10 becomes Rank 2.
So, two combinations are possible at the end of Round 3:
Now, we know that in each of the rounds in phase 2, only one player improves his score. Also, P8 and P10 cannot win medals. Hence, in case 1, three of P1, P7, P5 and P9 will improve their scores by x and reach the top 3 positions. However, the top 3 positions’ distances are in AP.
P1 – 88.6 + x
P7 – 87.2 + x
P5 – 86.4 + x
P9 – 84.1 + x
The differences do not satisfy the condition. Hence, case 1 is invalidated.
Case 2: Here, P1 definitely wins a medal, and P10 does not. So, two of P7, P5 and P9 jumps above P10. Now, if we have three different people increasing their scores or distances in each of the three rounds, again we would not get a difference of 1 among the Gold, Silver and Bronze medallists. Hence, one of them increases his score twice and the other increases his score twice and none of them is P1.
Let us take the cases where P1 is individually the G, S and B medallists.
Case 1: P1 is a G medallist.
P1 – 88.6
The silver medallist is 87.6 and the bronze medallist is 86.6 metres. However, P10 has thrown for a distance that is greater than 87.2 metres. Hence, in this case, he would be the B medallist. Hence, this is not the right case.
Case 2: P1 is the S medallist.
P1 – 88.6
G – 89.6
B – 87.6
Now, if we see the differences
89.6 – 87.2 = 2.4
87.6 – 86.4 = 1.2
This satisfies the condition that P7 has increased his score twice to become the gold winner and P5 has increased it once to become the bronze winner.
Hence, P1 – Silver
P7 – Gold
P5 – Bronze
Hence, P1 won the silver medal.
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Question 6: Which two players got the double?
a) P1, P8
b) P2, P4
c) P8, P10
d) P1, P10
6) Answer (C)
Solution:
Let us arrange the players in the order in which they throw in each round.
Round 1: Here the players throw in order of their initial seeds so the order is as follows:
So, their rank at the start of Round 2 is in order of their throws in the first round. Also, we need to consider the same order for people having invalid throws.
Round 2: P2, P4, P8, P10 had the same relative rankings since they all have invalid throws, that is, 0 metres. Rest are arranged as per their throw distances.
Now, in round 2, only P8 and P10 had valid throws. Hence, their order will change at the start of Round 3, however, the remaining order stays the same. That is, P8 and P10 will move up in the table and occupy some higher places, whereas some of the others may move down consequently.
Round 3: In Round 3, we can see that P1 improved his score from 82.9 to 88.6. The other 2 participants did not improve their scores. Also after Round 3, P8 and P10 qualify, where one of P8 or P10 is at the sixth position. So at the end of Round 3, we can say that P6, P3, P2, and P4 are at the bottom 4 positions(ranks). One of P8 or P10 is at the sixth position. P1 > P7 > P5 > P9.
So at the end of round 3, the ranks are as follows:
The other person between P8/P10 can go anywhere between Rank 1 and Rank 5.
Now let us consider the two players who got a double. Doubles happen in the transition between rounds.
1 -> 2 – Not possible
2 -> 3 – Possible if P10 reaches Rank 1 after round 2.
3 -> 4 – P8/P10 who is the last among qualifying will be the first to throw. So, here it definitely happens.
4 – > 5 AND 5 – > 6 not possible.
So, after Round 2, definitely, P10 reaches the top of the ranking. P8 is at the bottom. Hence, after Round 3, P10 either retains rank 1 or P1 surpasses him and P10 becomes Rank 2.
So, two combinations are possible at the end of Round 3:
Now, we know that in each of the rounds in phase 2, only one player improves his score. Also, P8 and P10 cannot win medals. Hence, in case 1, three of P1, P7, P5 and P9 will improve their scores by x and reach the top 3 positions. However, the top 3 positions’ distances are in AP.
P1 – 88.6 + x
P7 – 87.2 + x
P5 – 86.4 + x
P9 – 84.1 + x
The differences do not satisfy the condition. Hence, case 1 is invalidated.
Case 2: Here, P1 definitely wins a medal, and P10 does not. So, two of P7, P5 and P9 jumps above P10. Now, if we have three different people increasing their scores or distances in each of the three rounds, again we would not get a difference of 1 among the Gold, Silver and Bronze medallists. Hence, one of them increases his score twice and the other increases his score twice and none of them is P1.
Let us take the cases where P1 is individually the G, S and B medallists.
Case 1: P1 is a G medallist.
P1 – 88.6
The silver medallist is 87.6 and the bronze medallist is 86.6 metres. However, P10 has thrown for a distance that is greater than 87.2 metres. Hence, in this case, he would be the B medallist. Hence, this is not the right case.
Case 2: P1 is the S medallist.
P1 – 88.6
G – 89.6
B – 87.6
Now, if we see the differences
89.6 – 87.2 = 2.4
87.6 – 86.4 = 1.2
This satisfies the condition that P7 has increased his score twice to become the gold winner and P5 has increased it once to become the bronze winner.
Hence, P1 – Silver
P7 – Gold
P5 – Bronze
Hence, P8 and P10 got the doubles.
Instructions
The game of Chango is a game where two people play against each other; one of them wins and the other loses, i.e., there are no drawn Chango games. 12 players participated in a Chango championship. They were divided into four groups: Group A consisted of Aruna, Azul, and Arif; Group B consisted of Brinda, Brij, and Biju; Group C consisted of Chitra, Chetan, and Chhavi; and Group D consisted of Dipen, Donna, and Deb.
Players within each group had a distinct rank going into the championship. The players have NOT been listed necessarily according to their ranks. In the group stage of the game, the second and third ranked players play against each other, and the winner of that game plays against the first ranked player of the group. The winner of this second game is considered as the winner of the group and enters a semi-final.
The winners from Groups A and B play against each other in one semi-final, while the winners from Groups C and D play against each other in the other semi-final. The winners of the two semi-finals play against each other in the final to decide the winner of the championship.
It is known that:
1. Chitra did not win the championship.
2. Aruna did not play against Arif. Brij did not play against Brinda.
3. Aruna, Biju, Chitra, and Dipen played three games each, Azul and Chetan played two games each, and the remaining players played one game each.
Question 7: Who among the following did NOT play against Chitra in the championship?
a) Aruna
b) Chetan
c) Dipen
d) Biju
7) Answer (D)
Solution:
Group A :
Since Aruna played 3 games if she belongs to rank 2 or rank 3 in her group she must have reached the semifinals and lost in the semifinals. But for this case, she must play against rank 1 and rank 3 in her group. But she did not play against Arif from her group.
Hence Aruna was ranked 1 in her group, Among Azul and Arif one of them was ranked 2 and the other was ranked 3. Azul defeated Arif in the first round and in the second round lost to Aruna. Aruna played her first round with Azul and won the round and played against the winner from group B and defeated them and moved to the finals.
Group B :
Brij did not play against Brinda. Biju played three games, Brij and Brinda played one game each.
Since Brij and Brinda played only one game each one of them was ranked 1 and the other was ranked 2 and 3. Brij did not play against Brinda. We are aware that Aruna reached finals and hence the person from Group B did not reach the finals. Biju played three games and hence must have played with Brij/Brinda in the first round and won the round. Plays with Brij/ Brinda and wins the second round. Plays with Aruna and loses the third round.
Group – C :
Chitra played 2 matches and Chetan played 2 matches. For Chitra to play 2 matches if she is rank 2 or rank 3 in her group. She must at least reach the semifinals. But in this case, Chetan will be defeated in his first round. So Chitra must be ranked 1 in her group and Chetan must be ranked 2 or rank 3 in his group. He defeats Chhavi in his first round and loses to Chitra in his second round. Chitra plays Chetan in her first round, wins over the winner of group D in her second round, and loses in the final against Aruna as per condition 1.
Group -D :
The person from Group D did not reach the finals because Chitra reached the finals. In order for Dipen to play 3 matches before his finals. Dipen must be ranked 2 or rank 3 in his group and plays Deb and Donna in the first two rounds in any order and wins over both of them. Dipen loses to Chitra in his third round.
Aruna plays Chitra in the finals and wins the final round.
Brij was the player from group B who played Chitra. Aruna played in finals, Chetan in round 2, and Dipen in semi finals
Question 8: Who won the championship?
a) Chitra
b) Aruna
c) Brij
d) Cannot be determined
8) Answer (B)
Solution:
Group A :
Since Aruna played 3 games if she belongs to rank 2 or rank 3 in her group she must have reached the semifinals and lost in the semifinals. But for this case, she must play against rank 1 and rank 3 in her group. But she did not play against Arif from her group.
Hence Aruna was ranked 1 in her group, Among Azul and Arif one of them was ranked 2 and the other was ranked 3. Azul defeated Arif in the first round and in the second round lost to Aruna. Aruna played her first round with Azul and won the round and played against the winner from group B and defeated them and moved to the finals.
Group B :
Brij did not play against Brinda. Biju played three games, Brij and Brinda played one game each.
Since Brij and Brinda played only one game each one of them was ranked 1 and the other was ranked 2 and 3. Brij did not play against Brinda. We are aware that Aruna reached finals and hence the person from Group B did not reach the finals. Biju played three games and hence must have played with Brij/Brinda in the first round and won the round. Plays with Brij/ Brinda and wins the second round. Plays with Aruna and loses the third round.
Group – C :
Chitra played 2 matches and Chetan played 2 matches. For Chitra to play 2 matches if she is rank 2 or rank 3 in her group. She must at least reach the semifinals. But in this case, Chetan will be defeated in his first round. So Chitra must be ranked 1 in her group and Chetan must be ranked 2 or rank 3 in his group. He defeats Chhavi in his first round and loses to Chitra in his second round. Chitra plays Chetan in her first round, wins over the winner of group D in her second round, and loses in the final against Aruna as per condition 1.
Group -D :
The person from Group D did not reach the finals because Chitra reached the finals. In order for Dipen to play 3 matches before his finals. Dipen must be ranked 2 or rank 3 in his group and plays Deb and Donna in the first two rounds in any order and wins over both of them. Dipen loses to Chitra in his third round.
Aruna plays Chitra in the finals and wins the final round.
Aruna is the winner.
Question 9: Which of the following pairs must have played against each other in the championship?
a) Deb, Donna
b) Azul, Biju
c) Donna, Chetan
d) Chitra, Dipen
9) Answer (D)
Solution:
Group A :
Since Aruna played 3 games if she belongs to rank 2 or rank 3 in her group she must have reached the semifinals and lost in the semifinals. But for this case, she must play against rank 1 and rank 3 in her group. But she did not play against Arif from her group.
Hence Aruna was ranked 1 in her group, Among Azul and Arif one of them was ranked 2 and the other was ranked 3. Azul defeated Arif in the first round and in the second round lost to Aruna. Aruna played her first round with Azul and won the round and played against the winner from group B and defeated them and moved to the finals.
Group B :
Brij did not play against Brinda. Biju played three games, Brij and Brinda played one game each.
Since Brij and Brinda played only one game each one of them was ranked 1 and the other was ranked 2 and 3. Brij did not play against Brinda. We are aware that Aruna reached finals and hence the person from Group B did not reach the finals. Biju played three games and hence must have played with Brij/Brinda in the first round and won the round. Plays with Brij/ Brinda and wins the second round. Plays with Aruna and loses the third round.
Group – C :
Chitra played 2 matches and Chetan played 2 matches. For Chitra to play 2 matches if she is rank 2 or rank 3 in her group. She must at least reach the semifinals. But in this case, Chetan will be defeated in his first round. So Chitra must be ranked 1 in her group and Chetan must be ranked 2 or rank 3 in his group. He defeats Chhavi in his first round and loses to Chitra in his second round. Chitra plays Chetan in her first round, wins over the winner of group D in her second round, and loses in the final against Aruna as per condition 1.
Group -D :
The person from Group D did not reach the finals because Chitra reached the finals. In order for Dipen to play 3 matches before his finals. Dipen must be ranked 2 or rank 3 in his group and plays Deb and Donna in the first two rounds in any order and wins over both of them. Dipen loses to Chitra in his third round.
Aruna plays Chitra in the finals and wins the final round.
Chitra and Dipen played in the semifinals
Question 10: Who among the following was DEFINITELY NOT ranked first in his/her group?
a) Dipen
b) Aruna
c) Brij
d) Chitra
10) Answer (A)
Solution:
Group A :
Since Aruna played 3 games if she belongs to rank 2 or rank 3 in her group she must have reached the semifinals and lost in the semifinals. But for this case, she must play against rank 1 and rank 3 in her group. But she did not play against Arif from her group.
Hence Aruna was ranked 1 in her group, Among Azul and Arif one of them was ranked 2 and the other was ranked 3. Azul defeated Arif in the first round and in the second round lost to Aruna. Aruna played her first round with Azul and won the round and played against the winner from group B and defeated them and moved to the finals.
Group B :
Brij did not play against Brinda. Biju played three games, Brij and Brinda played one game each.
Since Brij and Brinda played only one game each one of them was ranked 1 and the other was ranked 2 and 3. Brij did not play against Brinda. We are aware that Aruna reached finals and hence the person from Group B did not reach the finals. Biju played three games and hence must have played with Brij/Brinda in the first round and won the round. Plays with Brij/ Brinda and wins the second round. Plays with Aruna and loses the third round.
Group – C :
Chitra played 2 matches and Chetan played 2 matches. For Chitra to play 2 matches if she is rank 2 or rank 3 in her group. She must at least reach the semifinals. But in this case, Chetan will be defeated in his first round. So Chitra must be ranked 1 in her group and Chetan must be ranked 2 or rank 3 in his group. He defeats Chhavi in his first round and loses to Chitra in his second round. Chitra plays Chetan in her first round, wins over the winner of group D in her second round, and loses in the final against Aruna as per condition 1.
Group -D :
The person from Group D did not reach the finals because Chitra reached the finals. In order for Dipen to play 3 matches before his finals. Dipen must be ranked 2 or rank 3 in his group and plays Deb and Donna in the first two rounds in any order and wins over both of them. Dipen loses to Chitra in his third round.
Aruna plays Chitra in the finals and wins the final round.
Dipen was ranked 2 or 3 in his group
Instructions
The Hi-Lo game is a four-player game played in six rounds. In every round, each player chooses to bid Hi or Lo. The bids are made simultaneously. If all four bid Hi, then all four lose 1 point each. If three players bid Hi and one bids Lo, then the players bidding Hi gain 1 point each and the player bidding Lo loses 3 points. If two players bid Hi and two bid Lo, then the players bidding Hi gain 2 points each and the players bidding Lo lose 2 points each. If one player bids Hi and three bid Lo, then the player bidding Hi gains 3 points and the players bidding Lo lose 1 point each. If all four bid Lo, then all four gain 1 point each. Four players Arun, Bankim, Charu, and Dipak played the Hi-Lo game. The following facts are known about their game:
1. At the end of three rounds, Arun had scored 6 points, Dipak had scored 2 points, Bankim and Charu had scored -2 points each.
2. At the end of six rounds, Arun had scored 7 points, Bankim and Dipak had scored -1 point each, and Charu had scored -5 points.
3. Dipak’s score in the third round was less than his score in the first round but was more than his score in the second round.
4. In exactly two out of the six rounds, Arun was the only player who bid Hi.
Question 11: In which of the following rounds, was Arun DEFINITELY the only player to bid Hi?
a) Second
b) Third
c) Fourth
d) First
11) Answer (A)
Solution:
Let ‘H’ represents Hi and ‘L’ represents Lo.
Given if they bid
Case 1: HHHH then all players gets -1 points.
Case 2: HHHL => H gets +1 and L gets -3.
Case 3: HHLL => H gets +2 and L gets -2.
Case 4: HLLL => H gets +3 and L gets -1.
Case 5: LLLL => every player gets +1.
From the given information we can draw the following table:
**T1 is the cumulative of points till Round 3 and T2 is sum of points till round 6.
**Arun, Bankim, Charu, and Dipak are represented by A, B, C, D respectively.
From point 3, D1>D2>D3
D scored 2 points till round R3 and D1>D3>D2 the possible scenarios are :
Case D1: 3,2,-3
In this case the points of A in R1, R3, R2 will be -1,2/-2, 1 in any
possible combination the sum will not be 6. So, this case is invalid.
Case D2: 2,1,-1
In this case the points of A in R1, R3, R2 will be 2/-2, 1/-3, -1/3
so, if the points in R1, R3, R2 are 2,1,3 the case is valid and no other
cases are possible.
Case D3: 3,1,-2
In this case the points of A in R1, R3, R2 will be -1, 1/-3/1, 2/-2
in any possible combination the sum will not be 6. So, this case is
invalid.
.’. Points of A,D in (R1,R2,R3) are (2,3,1) and (2,-1,1) respectively.
Since A got +3 in R2, he is only the one to bid h in R2 and points of B and C in round 2 are -1,-1 i.e they bid L, L.
Since A and D got 2 points each in R1, C and B must have got -2, -2 i.e they bid L, L.
Since A and D got 1 point in R3, C and B must also have got 1 in R3 i.e they bid L, L.
With this data, the table now looks like:
No information is given about the individual scores in R4, R5, R6.
Given In exactly two out of the six rounds, Arun was the only player who bid Hi.
Let R.x, R.y, R.z represent R4, R5, R6 in any order.
Let A bid H in R.x=> B,C,D bid L.
The table now looks like:
For A, R.x+R.y+R.z=1 => R.y+R.z=-2
For B, R.x+R.y+R.z=1 => R.y+R.z=2
For C, R.x+R.y+R.z=-3 => R.y+R.z=-2
For D, R.x+R.y+R.z=-3.=> R.y+R.z=-2
(R.y, R.z) for A can be (-3,1) or (-1,-1)
Case A1:
If for A, (R.y, R.z)=(-3,1)
Since for both C,D: R.y+R.z=-2
We can’t get any combination such that the total points of B,C,D are obtained.
Case A2:
If for A, (R.y, R.z)=(-1,-1).the (R.y, R.z) of B,C,D can be (3,-1),
(-1,-1), (-1,-1) and they must have bid (H,H), (L,H), (L,H) respectively
while A must have bid (L, H)
Hence this case is valid.
The final table looks like:
R2 is correct answer.
Question 12: In how many rounds did Dipak gain exactly 1 point?
12) Answer: 1
Solution:
Let ‘H’ represents Hi and ‘L’ represents Lo.
Given if they bid
Case 1: HHHH then all players gets -1 points.
Case 2: HHHL => H gets +1 and L gets -3.
Case 3: HHLL => H gets +2 and L gets -2.
Case 4: HLLL => H gets +3 and L gets -1.
Case 5: LLLL => every player gets +1.
From the given information we can draw the following table:
**T1 is the cumulative of points till Round 3 and T2 is sum of points till round 6.
**Arun, Bankim, Charu, and Dipak are represented by A, B, C, D respectively.
From point 3, D1>D2>D3
D scored 2 points till round R3 and D1>D3>D2 the possible scenarios are :
Case D1: 3,2,-3
In this case the points of A in R1, R3, R2 will be -1,2/-2, 1 in any
possible combination the sum will not be 6. So, this case is invalid.
Case D2: 2,1,-1
In this case the points of A in R1, R3, R2 will be 2/-2, 1/-3, -1/3
so, if the points in R1, R3, R2 are 2,1,3 the case is valid and no other
cases are possible.
Case D3: 3,1,-2
In this case the points of A in R1, R3, R2 will be -1, 1/-3/1, 2/-2
in any possible combination the sum will not be 6. So, this case is
invalid.
.’. Points of A,D in (R1,R2,R3) are (2,3,1) and (2,-1,1) respectively.
Since A got +3 in R2, he is only the one to bid h in R2 and points of B and C in round 2 are -1,-1 i.e they bid L, L.
Since A and D got 2 points each in R1, C and B must have got -2, -2 i.e they bid L, L.
Since A and D got 1 point in R3, C and B must also have got 1 in R3 i.e they bid L, L.
With this data, the table now looks like:
No information is given about the individual scores in R4, R5, R6.
Given In exactly two out of the six rounds, Arun was the only player who bid Hi.
Let R.x, R.y, R.z represent R4, R5, R6 in any order.
Let A bid H in R.x=> B,C,D bid L.
The table now looks like:
For A, R.x+R.y+R.z=1 => R.y+R.z=-2
For B, R.x+R.y+R.z=1 => R.y+R.z=2
For C, R.x+R.y+R.z=-3 => R.y+R.z=-2
For D, R.x+R.y+R.z=-3.=> R.y+R.z=-2
(R.y, R.z) for A can be (-3,1) or (-1,-1)
Case A1:
If for A, (R.y, R.z)=(-3,1)
Since for both C,D: R.y+R.z=-2
We can’t get any combination such that the total points of B,C,D are obtained.
Case A2:
If for A, (R.y, R.z)=(-1,-1).the (R.y, R.z) of B,C,D can be (3,-1),
(-1,-1), (-1,-1) and they must have bid (H,H), (L,H), (L,H) respectively
while A must have bid (L, H)
Hence this case is valid.
The final table looks like:
Deepak got exactly one point in only R3.
Hence 1 is correct answer.
Question 13: In how many rounds did all four players make identical bids?
13) Answer: 2
Solution:
Let ‘H’ represents Hi and ‘L’ represents Lo.
Given if they bid
Case 1: HHHH then all players gets -1 points.
Case 2: HHHL => H gets +1 and L gets -3.
Case 3: HHLL => H gets +2 and L gets -2.
Case 4: HLLL => H gets +3 and L gets -1.
Case 5: LLLL => every player gets +1.
From the given information we can draw the following table:
**T1 is the cumulative of points till Round 3 and T2 is sum of points till round 6.
**Arun, Bankim, Charu, and Dipak are represented by A, B, C, D respectively.
From point 3, D1>D2>D3
D scored 2 points till round R3 and D1>D3>D2 the possible scenarios are :
Case D1: 3,2,-3
In this case the points of A in R1, R3, R2 will be -1,2/-2, 1 in any
possible combination the sum will not be 6. So, this case is invalid.
Case D2: 2,1,-1
In this case the points of A in R1, R3, R2 will be 2/-2, 1/-3, -1/3
so, if the points in R1, R3, R2 are 2,1,3 the case is valid and no other
cases are possible.
Case D3: 3,1,-2
In this case the points of A in R1, R3, R2 will be -1, 1/-3/1, 2/-2
in any possible combination the sum will not be 6. So, this case is
invalid.
.’. Points of A,D in (R1,R2,R3) are (2,3,1) and (2,-1,1) respectively.
Since A got +3 in R2, he is only the one to bid h in R2 and points of B and C in round 2 are -1,-1 i.e they bid L, L.
Since A and D got 2 points each in R1, C and B must have got -2, -2 i.e they bid L, L.
Since A and D got 1 point in R3, C and B must also have got 1 in R3 i.e they bid L, L.
With this data, the table now looks like:
No information is given about the individual scores in R4, R5, R6.
Given In exactly two out of the six rounds, Arun was the only player who bid Hi.
Let R.x, R.y, R.z represent R4, R5, R6 in any order.
Let A bid H in R.x=> B,C,D bid L.
The table now looks like:
For A, R.x+R.y+R.z=1 => R.y+R.z=-2
For B, R.x+R.y+R.z=1 => R.y+R.z=2
For C, R.x+R.y+R.z=-3 => R.y+R.z=-2
For D, R.x+R.y+R.z=-3.=> R.y+R.z=-2
(R.y, R.z) for A can be (-3,1) or (-1,-1)
Case A1:
If for A, (R.y, R.z)=(-3,1)
Since for both C,D: R.y+R.z=-2
We can’t get any combination such that the total points of B,C,D are obtained.
Case A2:
If for A, (R.y, R.z)=(-1,-1).the (R.y, R.z) of B,C,D can be (3,-1),
(-1,-1), (-1,-1) and they must have bid (H,H), (L,H), (L,H) respectively
while A must have bid (L, H)
Hence this case is valid.
The final table looks like:
All the players made identical bids in R3 and R.z
Question 14: In how many rounds did Bankim bid Lo?
14) Answer: 4
Solution:
Let ‘H’ represents Hi and ‘L’ represents Lo.
Given if they bid
Case 1: HHHH then all players gets -1 points.
Case 2: HHHL => H gets +1 and L gets -3.
Case 3: HHLL => H gets +2 and L gets -2.
Case 4: HLLL => H gets +3 and L gets -1.
Case 5: LLLL => every player gets +1.
From the given information we can draw the following table:
**T1 is the cumulative of points till Round 3 and T2 is sum of points till round 6.
**Arun, Bankim, Charu, and Dipak are represented by A, B, C, D respectively.
From point 3, D1>D2>D3
D scored 2 points till round R3 and D1>D3>D2 the possible scenarios are :
Case D1: 3,2,-3
In this case the points of A in R1, R3, R2 will be -1,2/-2, 1 in any possible combination the sum will not be 6. So, this case is invalid.
Case D2: 2,1,-1
In this case the points of A in R1, R3, R2 will be 2/-2, 1/-3, -1/3 so, if the points in R1, R3, R2 are 2,1,3 the case is valid and no other cases are possible.
Case D3: 3,1,-2
In this case the points of A in R1, R3, R2 will be -1, 1/-3/1, 2/-2 in any possible combination the sum will not be 6. So, this case is invalid.
.’. Points of A,D in (R1,R2,R3) are (2,3,1) and (2,-1,1) respectively.
Since A got +3 in R2, he is only the one to bid h in R2 and points of B and C in round 2 are -1,-1 i.e they bid L, L.
Since A and D got 2 points each in R1, C and B must have got -2, -2 i.e they bid L, L.
Since A and D got 1 point in R3, C and B must also have got 1 in R3 i.e they bid L, L.
With this data, the table now looks like:
No information is given about the individual scores in R4, R5, R6.
Given In exactly two out of the six rounds, Arun was the only player who bid Hi.
Let R.x, R.y, R.z represent R4, R5, R6 in any order.
Let A bid H in R.x=> B,C,D bid L.
The table now looks like:
For A, R.x+R.y+R.z=1 => R.y+R.z=-2
For B, R.x+R.y+R.z=1 => R.y+R.z=2
For C, R.x+R.y+R.z=-3 => R.y+R.z=-2
For D, R.x+R.y+R.z=-3.=> R.y+R.z=-2
(R.y, R.z) for A can be (-3,1) or (-1,-1)
Case A1:
If for A, (R.y, R.z)=(-3,1)
Since for both C,D: R.y+R.z=-2
We can’t get any combination such that the total points of B,C,D are obtained.
Case A2:
If for A, (R.y, R.z)=(-1,-1).the (R.y, R.z) of B,C,D can be (3,-1), (-1,-1), (-1,-1) and they must have bid (H,H), (L,H), (L,H) respectively while A must have bid (L, H)
Hence this case is valid.
The final table looks like:
Bikram bid Lo in R1,R2,R3,R.x. Hence 4 is correct answer.
Question 15: In how many rounds did Arun bid Hi?
15) Answer: 4
Solution:
Let ‘H’ represents Hi and ‘L’ represents Lo.
Given if they bid
Case 1: HHHH then all players gets -1 points.
Case 2: HHHL => H gets +1 and L gets -3.
Case 3: HHLL => H gets +2 and L gets -2.
Case 4: HLLL => H gets +3 and L gets -1.
Case 5: LLLL => every player gets +1.
From the given information we can draw the following table:
**T1 is the cumulative of points till Round 3 and T2 is sum of points till round 6.
**Arun, Bankim, Charu, and Dipak are represented by A, B, C, D respectively.
From point 3, D1>D2>D3
D scored 2 points till round R3 and D1>D3>D2 the possible scenarios are :
Case D1: 3,2,-3
In this case the points of A in R1, R3, R2 will be -1,2/-2, 1 in any
possible combination the sum will not be 6. So, this case is invalid.
Case D2: 2,1,-1
In this case the points of A in R1, R3, R2 will be 2/-2, 1/-3, -1/3
so, if the points in R1, R3, R2 are 2,1,3 the case is valid and no other
cases are possible.
Case D3: 3,1,-2
In this case the points of A in R1, R3, R2 will be -1, 1/-3/1, 2/-2
in any possible combination the sum will not be 6. So, this case is
invalid.
.’. Points of A,D in (R1,R2,R3) are (2,3,1) and (2,-1,1) respectively.
Since A got +3 in R2, he is only the one to bid h in R2 and points of B and C in round 2 are -1,-1 i.e they bid L, L.
Since A and D got 2 points each in R1, C and B must have got -2, -2 i.e they bid L, L.
Since A and D got 1 point in R3, C and B must also have got 1 in R3 i.e they bid L, L.
With this data, the table now looks like:
No information is given about the individual scores in R4, R5, R6.
Given In exactly two out of the six rounds, Arun was the only player who bid Hi.
Let R.x, R.y, R.z represent R4, R5, R6 in any order.
Let A bid H in R.x=> B,C,D bid L.
The table now looks like:
For A, R.x+R.y+R.z=1 => R.y+R.z=-2
For B, R.x+R.y+R.z=1 => R.y+R.z=2
For C, R.x+R.y+R.z=-3 => R.y+R.z=-2
For D, R.x+R.y+R.z=-3.=> R.y+R.z=-2
(R.y, R.z) for A can be (-3,1) or (-1,-1)
Case A1:
If for A, (R.y, R.z)=(-3,1)
Since for both C,D: R.y+R.z=-2
We can’t get any combination such that the total points of B,C,D are obtained.
Case A2:
If for A, (R.y, R.z)=(-1,-1).the (R.y, R.z) of B,C,D can be (3,-1),
(-1,-1), (-1,-1) and they must have bid (H,H), (L,H), (L,H) respectively
while A must have bid (L, H)
Hence this case is valid.
The final table looks like:
Arun bid high in R1,R2, R.x, R.z hence, 4 is correct answer.
Question 16: What were the bids by Arun, Bankim, Charu and Dipak, respectively in the first round?
a) Hi, Lo, Lo, Hi
b) Hi, Lo, Lo, Lo
c) Hi, Hi, Lo, Lo
d) Lo, Lo, Lo, Hi
16) Answer (A)
Solution:
Let ‘H’ represents Hi and ‘L’ represents Lo.
Given if they bid
Case 1: HHHH then all players gets -1 points.
Case 2: HHHL => H gets +1 and L gets -3.
Case 3: HHLL => H gets +2 and L gets -2.
Case 4: HLLL => H gets +3 and L gets -1.
Case 5: LLLL => every player gets +1.
From the given information we can draw the following table:
**T1 is the cumulative of points till Round 3 and T2 is sum of points till round 6.
**Arun, Bankim, Charu, and Dipak are represented by A, B, C, D respectively.
From point 3, D1>D2>D3
D scored 2 points till round R3 and D1>D3>D2 the possible scenarios are :
Case D1: 3,2,-3
In this case the points of A in R1, R3, R2 will be -1,2/-2, 1 in any
possible combination the sum will not be 6. So, this case is invalid.
Case D2: 2,1,-1
In this case the points of A in R1, R3, R2 will be 2/-2, 1/-3, -1/3
so, if the points in R1, R3, R2 are 2,1,3 the case is valid and no other
cases are possible.
Case D3: 3,1,-2
In this case the points of A in R1, R3, R2 will be -1, 1/-3/1, 2/-2
in any possible combination the sum will not be 6. So, this case is
invalid.
.’. Points of A,D in (R1,R2,R3) are (2,3,1) and (2,-1,1) respectively.
Since A got +3 in R2, he is only the one to bid h in R2 and points of B and C in round 2 are -1,-1 i.e they bid L, L.
Since A and D got 2 points each in R1, C and B must have got -2, -2 i.e they bid L, L.
Since A and D got 1 point in R3, C and B must also have got 1 in R3 i.e they bid L, L.
With this data, the table now looks like:
No information is given about the individual scores in R4, R5, R6.
Given In exactly two out of the six rounds, Arun was the only player who bid Hi.
Let R.x, R.y, R.z represent R4, R5, R6 in any order.
Let A bid H in R.x=> B,C,D bid L.
The table now looks like:
For A, R.x+R.y+R.z=1 => R.y+R.z=-2
For B, R.x+R.y+R.z=1 => R.y+R.z=2
For C, R.x+R.y+R.z=-3 => R.y+R.z=-2
For D, R.x+R.y+R.z=-3.=> R.y+R.z=-2
(R.y, R.z) for A can be (-3,1) or (-1,-1)
Case A1:
If for A, (R.y, R.z)=(-3,1)
Since for both C,D: R.y+R.z=-2
We can’t get any combination such that the total points of B,C,D are obtained.
Case A2:
If for A, (R.y, R.z)=(-1,-1).the (R.y, R.z) of B,C,D can be (3,-1), (-1,-1), (-1,-1) and they must have bid (H,H), (L,H), (L,H) respectively while A must have bid (L, H)
Hence this case is valid.
The final table looks like:
The bids by Arun, Bankim, Charu and Dipak, respectively in the first round are HLLH.
Hence Option A is correct.
