Comprehension
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.
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: 1
What were the bids by Arun, Bankim, Charu and Dipak, respectively in the first round?
What were the bids by Arun, Bankim, Charu and Dipak, respectively in the first round?
Updated On: Jan 15, 2026
- Hi, Lo, Lo, Lo
- Lo, Lo, Lo, Hi
- Hi, Hi, Lo, Lo
- Hi, Lo, Lo, Hi
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The Correct Option is D
Solution and Explanation
Lets follow the table :
| R1 | R2 | R3 | T1 | RX | RY | RZ | T2 | |
| A | 2 H | 3 H | 1 L | 6 | 3 H | 7 | ||
| B | -2 L | -1 L | 1 L | -2 | -1 L | -1 | ||
| C | -2 L | -1 L | 1 L | -2 | -1 L | -5 | ||
| D | 2 H | -1 L | 1 L | 2 | -1 L | -1 |
Given the conditions for each scenario:
- A: R.x+R.y+R.z=1⇒R.y+R.z=−2
- B: R.x+R.y+R.z=1⇒R.y+R.z=2
- C: R.x+R.y+R.z=−3⇒R.y+R.z=−2
- D: R.x+R.y+R.z=−3⇒R.y+R.z=−2
For scenario A, possible values for (R.y,R.z) are:
- (-3, 1)
- (-1, -1)
Case A1: (𝑅.𝑦,𝑅.𝑧)=(−3,1)(R.y,R.z)=(−3,1)
- Since both C and D require 𝑅.𝑦+𝑅.𝑧=−2R.y+R.z=−2, there are no valid combinations that satisfy all given conditions. Hence, this case is not feasible.
Case A2: (𝑅.𝑦,𝑅.𝑧)=(−1,−1)(R.y,R.z)=(−1,−1)
- For B, C, D, valid combinations are:
- (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for B: (3, -1)
- (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for C: (-1, -1)
- (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for D: (-1, -1)
Based on these combinations, the bids should be:
- A: (L, H)
- B: (H, H)
- C: (L, H)
- D: (L, H)
This configuration satisfies all the given conditions and is therefore valid.
| R1 | R2 | R3 | T1 | RX | RY | RX | T2 | |
| A | 2 H | 3 H | 1 L | 6 | 3 H | -1 L | -1 H | 7 |
| B | -2 L | -1 7 | 1 L | -2 | -1 L | 3 H | 3 H | -1 |
| C | -2 L | -1 L | 1 L | -2 | -1 L | -1 L | -1 H | -5 |
| D | 2 L | -1 L | 1 L | 2 | -1 L | -1 L | -1 H | -1 |
The bids by Arun, Bankim, Charu and Dipak, respectively in the first round are HLLH
So, the correct option is (D): Hi, Lo, Lo, Hi.
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Question: 2
In how many rounds did Arun bid Hi?
In how many rounds did Arun bid Hi?
Updated On: Jan 15, 2026
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Solution and Explanation
Lets follow the table :
| R1 | R2 | R3 | T1 | RX | RY | RZ | T2 | |
| A | 2 H | 3 H | 1 L | 6 | 3 H | 7 | ||
| B | -2 L | -1 L | 1 L | -2 | -1 L | -1 | ||
| C | -2 L | -1 L | 1 L | -2 | -1 L | -5 | ||
| D | 2 H | -1 L | 1 L | 2 | -1 L | -1 |
Given the conditions for each scenario:
- A: R.x+R.y+R.z=1⇒R.y+R.z=−2
- B: R.x+R.y+R.z=1⇒R.y+R.z=2
- C: R.x+R.y+R.z=−3⇒R.y+R.z=−2
- D: R.x+R.y+R.z=−3⇒R.y+R.z=−2
For scenario A, possible values for (R.y,R.z) are:
- (-3, 1)
- (-1, -1)
Case A1: (𝑅.𝑦,𝑅.𝑧)=(−3,1)(R.y,R.z)=(−3,1)
- Since both C and D require 𝑅.𝑦+𝑅.𝑧=−2R.y+R.z=−2, there are no valid combinations that satisfy all given conditions. Hence, this case is not feasible.
Case A2: (𝑅.𝑦,𝑅.𝑧)=(−1,−1)(R.y,R.z)=(−1,−1)
- For B, C, D, valid combinations are:
- (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for B: (3, -1)
- (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for C: (-1, -1)
- (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for D: (-1, -1)
Based on these combinations, the bids should be:
- A: (L, H)
- B: (H, H)
- C: (L, H)
- D: (L, H)
This configuration satisfies all the given conditions and is therefore valid.
| R1 | R2 | R3 | T1 | RX | RY | RX | T2 | |
| A | 2 H | 3 H | 1 L | 6 | 3 H | -1 L | -1 H | 7 |
| B | -2 L | -1 7 | 1 L | -2 | -1 L | 3 H | 3 H | -1 |
| C | -2 L | -1 L | 1 L | -2 | -1 L | -1 L | -1 H | -5 |
| D | 2 L | -1 L | 1 L | 2 | -1 L | -1 L | -1 H | -1 |
Arun bid Hi in R1, R2, R, X, R.z
So, the correct answer is 4.
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Question: 3
In how many rounds did Bankim bid Lo?
In how many rounds did Bankim bid Lo?
Updated On: Jan 15, 2026
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Solution and Explanation
Lets follow the table :
| R1 | R2 | R3 | T1 | RX | RY | RZ | T2 | |
| A | 2 H | 3 H | 1 L | 6 | 3 H | 7 | ||
| B | -2 L | -1 L | 1 L | -2 | -1 L | -1 | ||
| C | -2 L | -1 L | 1 L | -2 | -1 L | -5 | ||
| D | 2 H | -1 L | 1 L | 2 | -1 L | -1 |
Given the conditions for each scenario:
- A: R.x+R.y+R.z=1⇒R.y+R.z=−2
- B: R.x+R.y+R.z=1⇒R.y+R.z=2
- C: R.x+R.y+R.z=−3⇒R.y+R.z=−2
- D: R.x+R.y+R.z=−3⇒R.y+R.z=−2
For scenario A, possible values for (R.y,R.z) are:
- (-3, 1)
- (-1, -1)
Case A1: (𝑅.𝑦,𝑅.𝑧)=(−3,1)(R.y,R.z)=(−3,1)
- Since both C and D require 𝑅.𝑦+𝑅.𝑧=−2R.y+R.z=−2, there are no valid combinations that satisfy all given conditions. Hence, this case is not feasible.
Case A2: (𝑅.𝑦,𝑅.𝑧)=(−1,−1)(R.y,R.z)=(−1,−1)
- For B, C, D, valid combinations are:
- (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for B: (3, -1)
- (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for C: (-1, -1)
- (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for D: (-1, -1)
Based on these combinations, the bids should be:
- A: (L, H)
- B: (H, H)
- C: (L, H)
- D: (L, H)
This configuration satisfies all the given conditions and is therefore valid.
| R1 | R2 | R3 | T1 | RX | RY | RX | T2 | |
| A | 2 H | 3 H | 1 L | 6 | 3 H | -1 L | -1 H | 7 |
| B | -2 L | -1 7 | 1 L | -2 | -1 L | 3 H | 3 H | -1 |
| C | -2 L | -1 L | 1 L | -2 | -1 L | -1 L | -1 H | -5 |
| D | 2 L | -1 L | 1 L | 2 | -1 L | -1 L | -1 H | -1 |
Bikram bid Lo in R1, R2, R3, R. X.
So, the correct answer is 4.
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Question: 4
In how many rounds did all four players make identical bids?
In how many rounds did all four players make identical bids?
Updated On: Jan 15, 2026
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Solution and Explanation
Lets follow the table :
| R1 | R2 | R3 | T1 | RX | RY | RZ | T2 | |
| A | 2 H | 3 H | 1 L | 6 | 3 H | 7 | ||
| B | -2 L | -1 L | 1 L | -2 | -1 L | -1 | ||
| C | -2 L | -1 L | 1 L | -2 | -1 L | -5 | ||
| D | 2 H | -1 L | 1 L | 2 | -1 L | -1 |
Given the conditions for each scenario:
- A: R.x+R.y+R.z=1⇒R.y+R.z=−2
- B: R.x+R.y+R.z=1⇒R.y+R.z=2
- C: R.x+R.y+R.z=−3⇒R.y+R.z=−2
- D: R.x+R.y+R.z=−3⇒R.y+R.z=−2
For scenario A, possible values for (R.y,R.z) are:
- (-3, 1)
- (-1, -1)
Case A1: (𝑅.𝑦,𝑅.𝑧)=(−3,1)(R.y,R.z)=(−3,1)
- Since both C and D require 𝑅.𝑦+𝑅.𝑧=−2R.y+R.z=−2, there are no valid combinations that satisfy all given conditions. Hence, this case is not feasible.
Case A2: (𝑅.𝑦,𝑅.𝑧)=(−1,−1)(R.y,R.z)=(−1,−1)
- For B, C, D, valid combinations are:
- (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for B: (3, -1)
- (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for C: (-1, -1)
- (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for D: (-1, -1)
Based on these combinations, the bids should be:
- A: (L, H)
- B: (H, H)
- C: (L, H)
- D: (L, H)
This configuration satisfies all the given conditions and is therefore valid.
| R1 | R2 | R3 | T1 | RX | RY | RX | T2 | |
| A | 2 H | 3 H | 1 L | 6 | 3 H | -1 L | -1 H | 7 |
| B | -2 L | -1 7 | 1 L | -2 | -1 L | 3 H | 3 H | -1 |
| C | -2 L | -1 L | 1 L | -2 | -1 L | -1 L | -1 H | -5 |
| D | 2 L | -1 L | 1 L | 2 | -1 L | -1 L | -1 H | -1 |
All the players made identical bids in R3 and R.z
So, the correct answer is 2 round.
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Question: 5
In how many rounds did Dipak gain exactly 1 point?
In how many rounds did Dipak gain exactly 1 point?
Updated On: Jan 15, 2026
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Solution and Explanation
Lets follow the table :
| R1 | R2 | R3 | T1 | RX | RY | RZ | T2 | |
| A | 2 H | 3 H | 1 L | 6 | 3 H | 7 | ||
| B | -2 L | -1 L | 1 L | -2 | -1 L | -1 | ||
| C | -2 L | -1 L | 1 L | -2 | -1 L | -5 | ||
| D | 2 H | -1 L | 1 L | 2 | -1 L | -1 |
Given the conditions for each scenario:
- A: R.x+R.y+R.z=1⇒R.y+R.z=−2
- B: R.x+R.y+R.z=1⇒R.y+R.z=2
- C: R.x+R.y+R.z=−3⇒R.y+R.z=−2
- D: R.x+R.y+R.z=−3⇒R.y+R.z=−2
For scenario A, possible values for (R.y,R.z) are:
- (-3, 1)
- (-1, -1)
Case A1: (𝑅.𝑦,𝑅.𝑧)=(−3,1)(R.y,R.z)=(−3,1)
- Since both C and D require 𝑅.𝑦+𝑅.𝑧=−2R.y+R.z=−2, there are no valid combinations that satisfy all given conditions. Hence, this case is not feasible.
Case A2: (𝑅.𝑦,𝑅.𝑧)=(−1,−1)(R.y,R.z)=(−1,−1)
- For B, C, D, valid combinations are:
- (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for B: (3, -1)
- (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for C: (-1, -1)
- (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for D: (-1, -1)
Based on these combinations, the bids should be:
- A: (L, H)
- B: (H, H)
- C: (L, H)
- D: (L, H)
This configuration satisfies all the given conditions and is therefore valid.
| R1 | R2 | R3 | T1 | RX | RY | RX | T2 | |
| A | 2 H | 3 H | 1 L | 6 | 3 H | -1 L | -1 H | 7 |
| B | -2 L | -1 7 | 1 L | -2 | -1 L | 3 H | 3 H | -1 |
| C | -2 L | -1 L | 1 L | -2 | -1 L | -1 L | -1 H | -5 |
| D | 2 L | -1 L | 1 L | 2 | -1 L | -1 L | -1 H | -1 |
Deepak got exactly one point in only R3.
So, the correct answer is 1.
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Question: 6
In which of the following rounds, was Arun DEFINITELY the only player to bid Hi?
In which of the following rounds, was Arun DEFINITELY the only player to bid Hi?
Updated On: Jan 15, 2026
- First
- Fourth
- Third
- Second
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The Correct Option is D
Solution and Explanation
Lets follow the table :
| R1 | R2 | R3 | T1 | RX | RY | RZ | T2 | |
| A | 2 H | 3 H | 1 L | 6 | 3 H | 7 | ||
| B | -2 L | -1 L | 1 L | -2 | -1 L | -1 | ||
| C | -2 L | -1 L | 1 L | -2 | -1 L | -5 | ||
| D | 2 H | -1 L | 1 L | 2 | -1 L | -1 |
Given the conditions for each scenario:
- A: R.x+R.y+R.z=1⇒R.y+R.z=−2
- B: R.x+R.y+R.z=1⇒R.y+R.z=2
- C: R.x+R.y+R.z=−3⇒R.y+R.z=−2
- D: R.x+R.y+R.z=−3⇒R.y+R.z=−2
For scenario A, possible values for (R.y,R.z) are:
- (-3, 1)
- (-1, -1)
Case A1: (𝑅.𝑦,𝑅.𝑧)=(−3,1)(R.y,R.z)=(−3,1)
- Since both C and D require 𝑅.𝑦+𝑅.𝑧=−2R.y+R.z=−2, there are no valid combinations that satisfy all given conditions. Hence, this case is not feasible.
Case A2: (𝑅.𝑦,𝑅.𝑧)=(−1,−1)(R.y,R.z)=(−1,−1)
- For B, C, D, valid combinations are:
- (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for B: (3, -1)
- (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for C: (-1, -1)
- (𝑅.𝑦,𝑅.𝑧)(R.y,R.z) for D: (-1, -1)
Based on these combinations, the bids should be:
- A: (L, H)
- B: (H, H)
- C: (L, H)
- D: (L, H)
This configuration satisfies all the given conditions and is therefore valid.
| R1 | R2 | R3 | T1 | RX | RY | RX | T2 | |
| A | 2 H | 3 H | 1 L | 6 | 3 H | -1 L | -1 H | 7 |
| B | -2 L | -1 7 | 1 L | -2 | -1 L | 3 H | 3 H | -1 |
| C | -2 L | -1 L | 1 L | -2 | -1 L | -1 L | -1 H | -5 |
| D | 2 L | -1 L | 1 L | 2 | -1 L | -1 L | -1 H | -1 |
R2 is correct answer
So, the correct answer is (D) : Second.
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