Three participants – Akhil, Bimal and Chatur participate in a random draw competition for five days. Every day, each participant randomly picks up a ball numbered between 1 and 9. The number on the ball determines his score on that day. The total score of a participant is the sum of his scores attained in the five days. The total score of a day is the sum of participants' scores on that day. The 2-day average on a day, except on Day 1, is the average of the total scores of that day and of the previous day. For example, if the total scores of Day 1 and Day 2 are 25 and 20, then the 2-day average on Day 2 is calculated as 22.5. Table 1 gives the 2-day averages for Days 2 through 5.
Participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. If there is a tie, all participants with the tied score are awarded the best available rank. For example, if on a day Akhil, Bimal, and Chatur score 8, 7 and 7 respectively, then their ranks will be 1, 2 and 2 respectively on that day. These ranks are given in Table 2. The following information is also known. 1. 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. 2. The total score on Day 3 is the same as the total score on Day 4. 3. Bimal's scores are the same on Day 1 and Day 3. [2023]
Question for CAT Previous Year Questions: Games and Tournaments
Try yourself:What is Akhil's score on Day 1?
Explanation
Let the total score of day 1, day 2, day 3, day 4, and day 5 are d1, d2, d3, d4, and d5, respectively.
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.
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
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Question for CAT Previous Year Questions: Games and Tournaments
Try yourself:Who attains the maximum total score?
Explanation
Let the total score of day 1, day 2, day 3, day 4, and day 5 are d1, d2, d3, d4, and d5, respectively.
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.
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
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Question for CAT Previous Year Questions: Games and Tournaments
Try yourself:What is the minimum possible total score of Bimal?
Correct Answer : 25
Explanation
Let the total score of day 1, day 2, day 3, day 4, and day 5 are d1, d2, d3, d4, and d5, respectively.
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.
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.
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Question for CAT Previous Year Questions: Games and Tournaments
Try yourself:If the total score of Bimal is a multiple of 3, what is the score of Akhil on Day 2?
Explanation
Let the total score of day 1, day 2, day 3, day 4, and day 5 are d1, d2, d3, d4, and d5, respectively.
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.
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
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Question for CAT Previous Year Questions: Games and Tournaments
Try yourself:If Akhil attains a total score of 24, then what is the total score of Bimal?
Correct Answer : 26
Explanation
Let the total score of day 1, day 2, day 3, day 4, and day 5 are d1, d2, d3, d4, and d5, respectively.
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.
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
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Passage
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 for CAT Previous Year Questions: Games and Tournaments
Try yourself:What were the bids by Arun, Bankim, Charu and Dipak, respectively in the first round?
Explanation
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.
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Question for CAT Previous Year Questions: Games and Tournaments
Try yourself:In how many rounds did Bankim bid Lo?
Correct Answer : 4
Explanation
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
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Question for CAT Previous Year Questions: Games and Tournaments
Try yourself:In how many rounds did Dipak gain exactly 1 point?
Correct Answer : 1
Explanation
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
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Question for CAT Previous Year Questions: Games and Tournaments
Try yourself:In how many rounds did Arun bid Hi?
Correct Answer : 4
Explanation
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
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*Answer can only contain numeric values
Question for CAT Previous Year Questions: Games and Tournaments
Try yourself:In how many rounds did all four players make identical bids?
Correct Answer : 2
Explanation
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:
All four players made identical bids in 2 rounds
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Question for CAT Previous Year Questions: Games and Tournaments
Try yourself:In which of the following rounds, was Arun DEFINITELY the only player to bid Hi?
Explanation
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.
FAQs on Games and Tournaments CAT Previous Year Questions with Answer PDF
1. What are the different types of game theory in tournaments?
Ans. Game theory in tournaments can be classified into cooperative game theory, non-cooperative game theory, and evolutionary game theory.
2. How are tournaments structured in game theory?
Ans. Tournaments in game theory are typically structured as a series of rounds where participants compete against each other to advance to the next round based on their performance.
3. What is the significance of Nash equilibrium in tournament games?
Ans. Nash equilibrium is a key concept in tournament games as it represents a stable state where no player has an incentive to deviate from their strategy given the strategies of other players.
4. How do players strategize in tournament games to maximize their chances of winning?
Ans. Players in tournament games strategize by considering their opponents' likely moves, predicting potential outcomes, and adjusting their strategies accordingly to increase their chances of winning.
5. Can game theory be applied to real-life tournaments and competitions?
Ans. Yes, game theory can be applied to analyze and strategize in a wide range of real-life tournaments and competitions, including sports tournaments, business competitions, and political elections.
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