CAT Previous Year Questions: Games and Tournaments (July 18) - CAT MCQ

Let the total score of day 1, day 2, day 3, day 4, and day 5 are d1, d2, d3, d4, and d5, respectively. 

The table shows that d1+d2 = 30 ... eq (1), d2+d3 = 31 ... eq (2), d3+d4 = 32 .... eq(3), d4+d5 = 34 ... eq(4)

It is given that participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. All participants with a tied score are awarded the best available rank if there is a tie.

It is given that the total score on Day 3 is the same as the total score on Day 4. 

Therefore, d3 = d4 => d3 = d4 = 16, which implies d2 = 15, d5 = 18, and d1 = 15.

The day-wise score is given below:

It is known that Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil’s score on Day 4.

Hence, only Chatur scored 9 (one time) on Day 2, and no other person scored 9 on any of the given 5 days. Chatur scored 3 only one time, which was on Day 1. Therefore, the scores obtained by Chatur on Day 3, Day 4, and Day 5 are 6, 6, and 6, respectively. It is also known that Akhil's score on Day 4 is the same as the score obtained by Chatur on Day 1. Hence, Akhil's score on Day 4 is 3.

Hence, we get the following table:

From Table 2, we see that the rank of Bimal and Akhil is the same, which is 2. Hence, The score obtained by Akhil and Bimal is the same. Let the score be x. Therefore, 6+2x = 16 => x = 5

The rank of Chatur on Day 5 is 2, and the rank of Bimal is 1, which implies the score obtained by Bimal will be more than Chatur. Hence, Bimal can score either 7 or 8 on Day 5. Therefore, the score obtained by Akhil on Day 5 is either 5 or 4. 

It is given that Bimal’s scores are the same on Day 1 and Day 3. Hence, the score obtained by Bimal on Day 1 is 5, which implies The score obtained by Akhil is 7 on Day 1.

From Table 2, we can see that the rank of Bimal is 3 on Day 2, and the rank of Akhil is 2 on Day 2. Hence, the score of Bimal will be lower than Akhil on Day 2. 

Let the score of Akhil be a, and the score of Bimal be b. Then 9+a+b = 15, and a > b

=> a+b =6, and a> b

Hence, the value of a can be 4/5, and the value of b can be 2/1

Therefore, the final table is given below:

From the table, we can see that the score of Akhil is 7 on day 1.

The correct option is B

Let the total score of day 1, day 2, day 3, day 4, and day 5 are d1, d2, d3, d4, and d5, respectively.

The table shows that d1+d2 = 30 ... eq (1), d2+d3 = 31 ... eq (2), d3+d4 = 32 .... eq(3), d4+d5 = 34 ... eq(4)

It is given that participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. All participants with a tied score are awarded the best available rank if there is a tie.

It is given that the total score on Day 3 is the same as the total score on Day 4.

Therefore, d3 = d4 => d3 = d4 = 16, which implies d2 = 15, d5 = 18, and d1 = 15.

The day-wise score is given below:

It is known that Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil’s score on Day 4.

Hence, only Chatur scored 9 (one time) on Day 2, and no other person scored 9 on any of the given 5 days. Chatur scored 3 only one time, which was on Day 1. Therefore, the scores obtained by Chatur on Day 3, Day 4, and Day 5 are 6, 6, and 6, respectively. It is also known that Akhil's score on Day 4 is the same as the score obtained by Chatur on Day 1. Hence, Akhil's score on Day 4 is 3.

Hence, we get the following table:

From Table 2, we see that the rank of Bimal and Akhil is the same, which is 2. Hence, The score obtained by Akhil and Bimal is the same. Let the score be x. Therefore, 6+2x = 16 => x = 5

The rank of Chatur on Day 5 is 2, and the rank of Bimal is 1, which implies the score obtained by Bimal will be more than Chatur. Hence, Bimal can score either 7 or 8 on Day 5. Therefore, the score obtained by Akhil on Day 5 is either 5 or 4.

It is given that Bimal’s scores are the same on Day 1 and Day 3. Hence, the score obtained by Bimal on Day 1 is 5, which implies The score obtained by Akhil is 7 on Day 1.

From Table 2, we can see that the rank of Bimal is 3 on Day 2, and the rank of Akhil is 2 on Day 2. Hence, the score of Bimal will be lower than Akhil's on Day 2.

Let the score of Akhil be a, and the score of Bimal be b. Then 9+a+b = 15, and a > b

=> a+b =6, and a> b

Hence, the value of a can be 4/5, and the value of b can be 2/1

Therefore, the final table is given below:

From the table, we can see that the maximum score is obtained by Chatur. 

The correct option is A

Let the total score of day 1, day 2, day 3, day 4, and day 5 are d1, d2, d3, d4, and d5, respectively.

The table shows that d1+d2 = 30 ... eq (1), d2+d3 = 31 ... eq (2), d3+d4 = 32 .... eq(3), d4+d5 = 34 ... eq(4)

It is given that participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. All participants with a tied score are awarded the best available rank if there is a tie.

It is given that the total score on Day 3 is the same as the total score on Day 4.

Therefore, d3 = d4 => d3 = d4 = 16, which implies d2 = 15, d5 = 18, and d1 = 15.

The day-wise score is given below:

It is known that Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil’s score on Day 4.

Hence, only Chatur scored 9 (one time) on Day 2, and no other person scored 9 on any of the given 5 days. Chatur scored 3 only one time, which was on Day 1. Therefore, the scores obtained by Chatur on Day 3, Day 4, and Day 5 are 6, 6, and 6, respectively. It is also known that Akhil's score on Day 4 is the same as the score obtained by Chatur on Day 1. Hence, Akhil's score on Day 4 is 3.

Hence, we get the following table:

From Table 2, we see that the rank of Bimal and Akhil is the same, which is 2. Hence, The score obtained by Akhil and Bimal is the same. Let the score be x. Therefore, 6+2x = 16 => x = 5

The rank of Chatur on Day 5 is 2, and the rank of Bimal is 1, which implies the score obtained by Bimal will be more than Chatur. Hence, Bimal can score either 7 or 8 on Day 5. Therefore, the score obtained by Akhil on Day 5 is either 5 or 4.

It is given that Bimal’s scores are the same on Day 1 and Day 3. Hence, the score obtained by Bimal on Day 1 is 5, which implies The score obtained by Akhil is 7 on Day 1.

From Table 2, we can see that the rank of Bimal is 3 on Day 2, and the rank of Akhil is 2 on Day 2. Hence, the score of Bimal will be lower than Akhil on Day 2.

Let the score of Akhil be a, and the score of Bimal be b. Then 9+a+b = 15, and a > b

=> a+b =6, and a> b

Hence, the value of a can be 4/5, and the value of b can be 2/1

Therefore, the final table is given below:

From the table, we can see that the minimum score obtained by Bimal is 25.

Let the total score of day 1, day 2, day 3, day 4, and day 5 are d1, d2, d3, d4, and d5, respectively.

The table shows that d1+d2 = 30 ... eq (1), d2+d3 = 31 ... eq (2), d3+d4 = 32 .... eq(3), d4+d5 = 34 ... eq(4)

It is given that participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. All participants with a tied score are awarded the best available rank if there is a tie.

It is given that the total score on Day 3 is the same as the total score on Day 4.

Therefore, d3 = d4 => d3 = d4 = 16, which implies d2 = 15, d5 = 18, and d1 = 15.

The day-wise score is given below:

It is known that Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil’s score on Day 4.

Hence, only Chatur scored 9 (one time) on Day 2, and no other person scored 9 on any of the given 5 days. Chatur scored 3 only one time, which was on Day 1. Therefore, the scores obtained by Chatur on Day 3, Day 4, and Day 5 are 6, 6, and 6, respectively. It is also known that Akhil's score on Day 4 is the same as the score obtained by Chatur on Day 1. Hence, Akhil's score on Day 4 is 3.

Hence, we get the following table:

From Table 2, we see that the rank of Bimal and Akhil is the same, which is 2. Hence, The score obtained by Akhil and Bimal is the same. Let the score be x. Therefore, 6+2x = 16 => x = 5

The rank of Chatur on Day 5 is 2, and the rank of Bimal is 1, which implies the score obtained by Bimal will be more than Chatur. Hence, Bimal can score either 7 or 8 on Day 5. Therefore, the score obtained by Akhil on Day 5 is either 5 or 4.

It is given that Bimal’s scores are the same on Day 1 and Day 3. Hence, the score obtained by Bimal on Day 1 is 5, which implies The score obtained by Akhil is 7 on Day 1.

From Table 2, we can see that the rank of Bimal is 3 on Day 2, and the rank of Akhil is 2 on Day 2. Hence, the score of Bimal will be lower than Akhil on Day 2.

Let the score of Akhil be a, and the score of Bimal be b. Then 9+a+b = 15, and a > b

=> a+b =6, and a> b

Hence, the value of a can be 4/5, and the value of b can be 2/1

Therefore, the final table is given below:

In the question, it is given that the total score obtained by Bimal is a multiple of 3, which implies the total score obtained by Bimal is 27, which implies the total score obtained by Akhil is 23.

Akhil will score 23, when his scores on Days 1, 2, 3, 4, and 5 are 7, 4, 5, 3, 4, respectively.

Hence, the score obtained by him on Day 2 is 4.

The correct option is A

Let the total score of day 1, day 2, day 3, day 4, and day 5 are d1, d2, d3, d4, and d5, respectively.

The table shows that d1+d2 = 30 ... eq (1), d2+d3 = 31 ... eq (2), d3+d4 = 32 .... eq(3), d4+d5 = 34 ... eq(4)

It is given that participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. All participants with a tied score are awarded the best available rank if there is a tie.

It is given that the total score on Day 3 is the same as the total score on Day 4.

Therefore, d3 = d4 => d3 = d4 = 16, which implies d2 = 15, d5 = 18, and d1 = 15.

The day-wise score is given below:

It is known that Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil’s score on Day 4.

Hence, only Chatur scored 9 (one time) on Day 2, and no other person scored 9 on any of the given 5 days. Chatur scored 3 only one time, which was on Day 1. Therefore, the scores obtained by Chatur on Day 3, Day 4, and Day 5 are 6, 6, and 6, respectively. It is also known that Akhil's score on Day 4 is the same as the score obtained by Chatur on Day 1. Hence, Akhil's score on Day 4 is 3.

Hence, we get the following table:

From Table 2, we see that the rank of Bimal and Akhil is the same, which is 2. Hence, The score obtained by Akhil and Bimal is the same. Let the score be x. Therefore, 6+2x = 16 => x = 5

The rank of Chatur on Day 5 is 2, and the rank of Bimal is 1, which implies the score obtained by Bimal will be more than Chatur. Hence, Bimal can score either 7 or 8 on Day 5. Therefore, the score obtained by Akhil on Day 5 is either 5 or 4.

It is given that Bimal’s scores are the same on Day 1 and Day 3. Hence, the score obtained by Bimal on Day 1 is 5, which implies The score obtained by Akhil is 7 on Day 1.

From Table 2, we can see that the rank of Bimal is 3 on Day 2, and the rank of Akhil is 2 on Day 2. Hence, the score of Bimal will be lower than Akhil on Day 2.

Let the score of Akhil be a, and the score of Bimal be b. Then 9+a+b = 15, and a > b

=> a+b =6, and a> b

Hence, the value of a can be 4/5, and the value of b can be 2/1

Therefore, the final table is given below:

In the question, it is given that the score obtained by Akhil is 24, which implies the score obtained by Bimal is 26. 

The answer is 26

From the given information, we can see that the points scored by the players in a round has the following possibilities:
HHHH: (-1, -1, -1, -1)
HHHL: (1, 1, 1, -3)
HHLL: (2, 2, -2, -2)
HLLL: (3, -1, -1, -1)
LLLL: (1, 1, 1, 1)
Also, the total points scored by the four players in a round can only be -4 or 0 or 4.
From (1), the total points scored by the four players combined in the first three rounds is 6 + 2 - 2 - 2 = 4.
Hence, in the first three rounds, the total points scored by the four players must be either (- 4, 4, 4) OR (0, 0, 4), in any order.
Also, from (1), in the first three rounds, Arun scored 6 points. And from (2), Arun scored 7 points at the end of round 6. Hence, in the 4th , 5th and 6th rounds, he must have scored 1 point.
From (4), Arun scored 3 points in exactly 2 rounds.
These two rounds cannot both be among 4th , 5th and 6th rounds because he scored a net of only 1 point in these three rounds combined.
Hence, Arun must have scored 3 points in one round among 1st , 2nd and 3 rd rounds. If Arun scored 3 points in the first three rounds, then in that round, the total points scored by the four players combined must be 0 (in the case of HLLL).
Hence, the total points scored by the four players in the first three rounds must be (0, 0, 4). Among the first three rounds, in one round, the three players must have scored (3, -1, -1, - 1), with Arun scoring 3 points.
Since in another round, the four players scored a total of 4 points, they must have bid LLLL (as it is the only case in which they can score 4 points in total). They must have scored (1, 1, 1, 1).
Since Arun scored a total of 6 points in the first three rounds, and he scored 3 points and 1 point in two of these rounds, he must have score 2 points in the other round. This is possible only if the players bid HHLL and the scores of the four players must be (2, 2, - 2, - 2).
In the round that the players scored (3, - 1, - 1, - 1), Dipak must have scored - 1 points (since Arun scored 3 points).
In the round that the players scored (1, 1, 1, 1), Dipak must have scored 1 point.
In the round that the players scored (2, 2, - 2, - 2), Dipak must have scored 2 points (since the total points that Dipak scored in the first three rounds is 2).
From (3), Dipak must have scored 2 points in the first round, - 1 points in the second round and 1 point in the third round.
From this, we can fill the points for the first three rounds, as shown below

(Note that with this information, the first question of the set can be answered)
In the next three rounds, from (1) and (2), Arun must have scored 1 point, Bankim must have scored 1 point, Charu must have scored - 3 points and Dipak must have scored - 3 points.
The total points scored by the four players are - 4. This is possible if the total points scored by the four players in the three rounds are (0, 0, - 4) OR (4, - 4, - 4) in any order.
However, we know that Arun must have scored 3 points in one of these three rounds (from (4)). Hence, the total points scored by the players in this round must be 0.
Hence, the four players must have scored (0, 0, - 4) points in these three rounds. In one round the points scored by the players must be (3, - 1, - 1, - 1), with Arun scoring 3 points.
In the round in which the total points scored by the four players is 4, they must have scored ( - 1, - 1, - 1, - 1).
Since Bankim scored a total of 1 point in these three rounds, and he scored - 1 point and - 1 point in the two rounds mentioned above, he must have scored 3 points in the other third.
Hence, in the remaining round, the four players must have scored (3, - 1, - 1, - 1), with Bankim scoring 3 points.
However, with the given information, we cannot deduce the round number corresponding to the above rounds.
Hence, we get the following table:

The bids by Arun, Bankim, Charu, Dipak in the first round were Hi, Lo, Lo, Hi.

From the given information, we can see that the points scored by the players in a round has the following possibilities:
HHHH: (-1, -1, -1, -1)
HHHL: (1, 1, 1, -3)
HHLL: (2, 2, -2, -2)
HLLL: (3, -1, -1, -1)
LLLL: (1, 1, 1, 1)
Also, the total points scored by the four players in a round can only be -4 or 0 or 4.
From (1), the total points scored by the four players combined in the first three rounds is 6 + 2 - 2 - 2 = 4.
Hence, in the first three rounds, the total points scored by the four players must be either (- 4, 4, 4) OR (0, 0, 4), in any order.
Also, from (1), in the first three rounds, Arun scored 6 points. And from (2), Arun scored 7 points at the end of round 6. Hence, in the 4th , 5th and 6th rounds, he must have scored 1 point.
From (4), Arun scored 3 points in exactly 2 rounds.
These two rounds cannot both be among 4th , 5th and 6th rounds because he scored a net of only 1 point in these three rounds combined.
Hence, Arun must have scored 3 points in one round among 1st , 2nd and 3 rd rounds. If Arun scored 3 points in the first three rounds, then in that round, the total points scored by the four players combined must be 0 (in the case of HLLL).
Hence, the total points scored by the four players in the first three rounds must be (0, 0, 4). Among the first three rounds, in one round, the three players must have scored (3, -1, -1, - 1), with Arun scoring 3 points.
Since in another round, the four players scored a total of 4 points, they must have bid LLLL (as it is the only case in which they can score 4 points in total). They must have scored (1, 1, 1, 1).
Since Arun scored a total of 6 points in the first three rounds, and he scored 3 points and 1 point in two of these rounds, he must have score 2 points in the other round. This is possible only if the players bid HHLL and the scores of the four players must be (2, 2, - 2, - 2).
In the round that the players scored (3, - 1, - 1, - 1), Dipak must have scored - 1 points (since Arun scored 3 points).
In the round that the players scored (1, 1, 1, 1), Dipak must have scored 1 point.
In the round that the players scored (2, 2, - 2, - 2), Dipak must have scored 2 points (since the total points that Dipak scored in the first three rounds is 2).
From (3), Dipak must have scored 2 points in the first round, - 1 points in the second round and 1 point in the third round.
From this, we can fill the points for the first three rounds, as shown below

(Note that with this information, the first question of the set can be answered)
In the next three rounds, from (1) and (2), Arun must have scored 1 point, Bankim must have scored 1 point, Charu must have scored - 3 points and Dipak must have scored - 3 points.
The total points scored by the four players are - 4. This is possible if the total points scored by the four players in the three rounds are (0, 0, - 4) OR (4, - 4, - 4) in any order.
However, we know that Arun must have scored 3 points in one of these three rounds (from (4)). Hence, the total points scored by the players in this round must be 0.
Hence, the four players must have scored (0, 0, - 4) points in these three rounds. In one round the points scored by the players must be (3, - 1, - 1, - 1), with Arun scoring 3 points.
In the round in which the total points scored by the four players is 4, they must have scored ( - 1, - 1, - 1, - 1).
Since Bankim scored a total of 1 point in these three rounds, and he scored - 1 point and - 1 point in the two rounds mentioned above, he must have scored 3 points in the other third.
Hence, in the remaining round, the four players must have scored (3, - 1, - 1, - 1), with Bankim scoring 3 points.
However, with the given information, we cannot deduce the round number corresponding to the above rounds.
Hence, we get the following table:

Bankim bid Lo in 4 rounds

From the given information, we can see that the points scored by the players in a round has the following possibilities:
HHHH: (-1, -1, -1, -1)
HHHL: (1, 1, 1, -3)
HHLL: (2, 2, -2, -2)
HLLL: (3, -1, -1, -1)
LLLL: (1, 1, 1, 1)
Also, the total points scored by the four players in a round can only be -4 or 0 or 4.
From (1), the total points scored by the four players combined in the first three rounds is 6 + 2 - 2 - 2 = 4.
Hence, in the first three rounds, the total points scored by the four players must be either (- 4, 4, 4) OR (0, 0, 4), in any order.
Also, from (1), in the first three rounds, Arun scored 6 points. And from (2), Arun scored 7 points at the end of round 6. Hence, in the 4th , 5th and 6th rounds, he must have scored 1 point.
From (4), Arun scored 3 points in exactly 2 rounds.
These two rounds cannot both be among 4th , 5th and 6th rounds because he scored a net of only 1 point in these three rounds combined.
Hence, Arun must have scored 3 points in one round among 1st , 2nd and 3 rd rounds. If Arun scored 3 points in the first three rounds, then in that round, the total points scored by the four players combined must be 0 (in the case of HLLL).
Hence, the total points scored by the four players in the first three rounds must be (0, 0, 4). Among the first three rounds, in one round, the three players must have scored (3, -1, -1, - 1), with Arun scoring 3 points.
Since in another round, the four players scored a total of 4 points, they must have bid LLLL (as it is the only case in which they can score 4 points in total). They must have scored (1, 1, 1, 1).
Since Arun scored a total of 6 points in the first three rounds, and he scored 3 points and 1 point in two of these rounds, he must have score 2 points in the other round. This is possible only if the players bid HHLL and the scores of the four players must be (2, 2, - 2, - 2).
In the round that the players scored (3, - 1, - 1, - 1), Dipak must have scored - 1 points (since Arun scored 3 points).
In the round that the players scored (1, 1, 1, 1), Dipak must have scored 1 point.
In the round that the players scored (2, 2, - 2, - 2), Dipak must have scored 2 points (since the total points that Dipak scored in the first three rounds is 2).
From (3), Dipak must have scored 2 points in the first round, - 1 points in the second round and 1 point in the third round.
From this, we can fill the points for the first three rounds, as shown below

(Note that with this information, the first question of the set can be answered)
In the next three rounds, from (1) and (2), Arun must have scored 1 point, Bankim must have scored 1 point, Charu must have scored - 3 points and Dipak must have scored - 3 points.
The total points scored by the four players are - 4. This is possible if the total points scored by the four players in the three rounds are (0, 0, - 4) OR (4, - 4, - 4) in any order.
However, we know that Arun must have scored 3 points in one of these three rounds (from (4)). Hence, the total points scored by the players in this round must be 0.
Hence, the four players must have scored (0, 0, - 4) points in these three rounds. In one round the points scored by the players must be (3, - 1, - 1, - 1), with Arun scoring 3 points.
In the round in which the total points scored by the four players is 4, they must have scored ( - 1, - 1, - 1, - 1).
Since Bankim scored a total of 1 point in these three rounds, and he scored - 1 point and - 1 point in the two rounds mentioned above, he must have scored 3 points in the other third.
Hence, in the remaining round, the four players must have scored (3, - 1, - 1, - 1), with Bankim scoring 3 points.
However, with the given information, we cannot deduce the round number corresponding to the above rounds.
Hence, we get the following table:

Dipak gained exactly 1 point in 1 round

From the given information, we can see that the points scored by the players in a round has the following possibilities:
HHHH: (-1, -1, -1, -1)
HHHL: (1, 1, 1, -3)
HHLL: (2, 2, -2, -2)
HLLL: (3, -1, -1, -1)
LLLL: (1, 1, 1, 1)
Also, the total points scored by the four players in a round can only be -4 or 0 or 4.
From (1), the total points scored by the four players combined in the first three rounds is 6 + 2 - 2 - 2 = 4.
Hence, in the first three rounds, the total points scored by the four players must be either (- 4, 4, 4) OR (0, 0, 4), in any order.
Also, from (1), in the first three rounds, Arun scored 6 points. And from (2), Arun scored 7 points at the end of round 6. Hence, in the 4th , 5th and 6th rounds, he must have scored 1 point.
From (4), Arun scored 3 points in exactly 2 rounds.
These two rounds cannot both be among 4th , 5th and 6th rounds because he scored a net of only 1 point in these three rounds combined.
Hence, Arun must have scored 3 points in one round among 1st , 2nd and 3 rd rounds. If Arun scored 3 points in the first three rounds, then in that round, the total points scored by the four players combined must be 0 (in the case of HLLL).
Hence, the total points scored by the four players in the first three rounds must be (0, 0, 4). Among the first three rounds, in one round, the three players must have scored (3, -1, -1, - 1), with Arun scoring 3 points.
Since in another round, the four players scored a total of 4 points, they must have bid LLLL (as it is the only case in which they can score 4 points in total). They must have scored (1, 1, 1, 1).
Since Arun scored a total of 6 points in the first three rounds, and he scored 3 points and 1 point in two of these rounds, he must have score 2 points in the other round. This is possible only if the players bid HHLL and the scores of the four players must be (2, 2, - 2, - 2).
In the round that the players scored (3, - 1, - 1, - 1), Dipak must have scored - 1 points (since Arun scored 3 points).
In the round that the players scored (1, 1, 1, 1), Dipak must have scored 1 point.
In the round that the players scored (2, 2, - 2, - 2), Dipak must have scored 2 points (since the total points that Dipak scored in the first three rounds is 2).
From (3), Dipak must have scored 2 points in the first round, - 1 points in the second round and 1 point in the third round.
From this, we can fill the points for the first three rounds, as shown below

(Note that with this information, the first question of the set can be answered)
In the next three rounds, from (1) and (2), Arun must have scored 1 point, Bankim must have scored 1 point, Charu must have scored - 3 points and Dipak must have scored - 3 points.
The total points scored by the four players are - 4. This is possible if the total points scored by the four players in the three rounds are (0, 0, - 4) OR (4, - 4, - 4) in any order.
However, we know that Arun must have scored 3 points in one of these three rounds (from (4)). Hence, the total points scored by the players in this round must be 0.
Hence, the four players must have scored (0, 0, - 4) points in these three rounds. In one round the points scored by the players must be (3, - 1, - 1, - 1), with Arun scoring 3 points.
In the round in which the total points scored by the four players is 4, they must have scored ( - 1, - 1, - 1, - 1).
Since Bankim scored a total of 1 point in these three rounds, and he scored - 1 point and - 1 point in the two rounds mentioned above, he must have scored 3 points in the other third.
Hence, in the remaining round, the four players must have scored (3, - 1, - 1, - 1), with Bankim scoring 3 points.
However, with the given information, we cannot deduce the round number corresponding to the above rounds.
Hence, we get the following table:

Arun bid Hi in 4 rounds 

From the given information, we can see that the points scored by the players in a round has the following possibilities:
HHHH: (-1, -1, -1, -1)
HHHL: (1, 1, 1, -3)
HHLL: (2, 2, -2, -2)
HLLL: (3, -1, -1, -1)
LLLL: (1, 1, 1, 1)
Also, the total points scored by the four players in a round can only be -4 or 0 or 4.
From (1), the total points scored by the four players combined in the first three rounds is 6 + 2 - 2 - 2 = 4.
Hence, in the first three rounds, the total points scored by the four players must be either (- 4, 4, 4) OR (0, 0, 4), in any order.
Also, from (1), in the first three rounds, Arun scored 6 points. And from (2), Arun scored 7 points at the end of round 6. Hence, in the 4th , 5th and 6th rounds, he must have scored 1 point.
From (4), Arun scored 3 points in exactly 2 rounds.
These two rounds cannot both be among 4th , 5th and 6th rounds because he scored a net of only 1 point in these three rounds combined.
Hence, Arun must have scored 3 points in one round among 1st , 2nd and 3 rd rounds. If Arun scored 3 points in the first three rounds, then in that round, the total points scored by the four players combined must be 0 (in the case of HLLL).
Hence, the total points scored by the four players in the first three rounds must be (0, 0, 4). Among the first three rounds, in one round, the three players must have scored (3, -1, -1, - 1), with Arun scoring 3 points.
Since in another round, the four players scored a total of 4 points, they must have bid LLLL (as it is the only case in which they can score 4 points in total). They must have scored (1, 1, 1, 1).
Since Arun scored a total of 6 points in the first three rounds, and he scored 3 points and 1 point in two of these rounds, he must have score 2 points in the other round. This is possible only if the players bid HHLL and the scores of the four players must be (2, 2, - 2, - 2).
In the round that the players scored (3, - 1, - 1, - 1), Dipak must have scored - 1 points (since Arun scored 3 points).
In the round that the players scored (1, 1, 1, 1), Dipak must have scored 1 point.
In the round that the players scored (2, 2, - 2, - 2), Dipak must have scored 2 points (since the total points that Dipak scored in the first three rounds is 2).
From (3), Dipak must have scored 2 points in the first round, - 1 points in the second round and 1 point in the third round.
From this, we can fill the points for the first three rounds, as shown below

(Note that with this information, the first question of the set can be answered)
In the next three rounds, from (1) and (2), Arun must have scored 1 point, Bankim must have scored 1 point, Charu must have scored - 3 points and Dipak must have scored - 3 points.
The total points scored by the four players are - 4. This is possible if the total points scored by the four players in the three rounds are (0, 0, - 4) OR (4, - 4, - 4) in any order.
However, we know that Arun must have scored 3 points in one of these three rounds (from (4)). Hence, the total points scored by the players in this round must be 0.
Hence, the four players must have scored (0, 0, - 4) points in these three rounds. In one round the points scored by the players must be (3, - 1, - 1, - 1), with Arun scoring 3 points.
In the round in which the total points scored by the four players is 4, they must have scored ( - 1, - 1, - 1, - 1).
Since Bankim scored a total of 1 point in these three rounds, and he scored - 1 point and - 1 point in the two rounds mentioned above, he must have scored 3 points in the other third.
Hence, in the remaining round, the four players must have scored (3, - 1, - 1, - 1), with Bankim scoring 3 points.
However, with the given information, we cannot deduce the round number corresponding to the above rounds.
Hence, we get the following table:

In the second round, Arun was the only player to bid Hi.