- Gordon did not score consecutively in any two rounds.
- Eric and Fatima both scored in a round.

- Only two players scored in three consecutive rounds. One of them was Chen. No other player scored in any two consecutive rounds
- Joshin scored in Round 7, while Amita scored in Round 10.
- No player scored in all the four rounds.

- 3, 3, 0
- 3, 3, 3
- 3, 6, 3
- 3, 0, 3

- Hansa, Ikea, Joshin
- Bala, Chen, Gordon
- Bala, Ikea, Joshin
- Bala, Hansa, Ikea

- Ikea
- Chen
- Amita
- Joshin

- Amita, Chen, Eric
- Amita, Bala, Chen
- Amita, Eric, Joshin
- Amita, Chen, David

Team playing in different round can be tabulated as given below:

NP : Not played in that round

Player | R1 | R2 | R3 | R4 | R5 | R6 | Points after round 6 | R7 | R8 | R9 | R10 | Points after round 10 |

1 – A | NP | NP | NP | NP | 8 | 18 | ||||||

2 – B | NP | NP | NP | NP | 2 | 5 | ||||||

3 – C | NP | NP | NP | 3 | NP | 6 | ||||||

4 – D | NP | NP | 6 | NP | NP | 6 | ||||||

5 – E | NP | 3 | NP | NP | NP | 10 | ||||||

6 – F | 10 | NP | NP | NP | NP | 10 | ||||||

7 – G | NP | 17 | 0 | NP | NP | NP | 17 | |||||

8 – H | NP | NP | 1 | NP | NP | 4 | ||||||

9 – I | NP | NP | NP | 2 | NP | 17 | ||||||

10 – J | NP | NP | NP | NP | 14 | 17 |

Now as B had scored 2 points till round 6 and he had played only in round 1 and in round 2 till that thus he must have scored 1 â€“ 1 points in each round.

So A can score 1 in round 6 and 7 in round 1 as he had played only in two rounds and had 8 points.

Similarly, Joshin (J) must have scored 7 points each in round 5 and round 6.

Now from the information known about Rounds 7 through 10: from point 1 and 2, C must score 1 points each in round 8, 9 and 10 as hi points has been increased 3 (from 3 to 6) and J must score 3 points in round 7.

Now the only possibility for I is to score 1, 7 and 7 points in round 7, 8 and 9 respectively.

Now A can not score in two consecutive rounds and his points has been increased from 8 to 18 so it is the only possibility that he score 7 points in round 7 and 3 points in round 10. Thus B will score 3 points in round 9 and H in round 8.

Now if we fill these data in the table, the following table can be made

Player | R1 | R2 | R3 | R4 | R5 | R6 | Points after round 6 | R7 | R8 | R9 | R10 | Points after round 10 |

1 – A | 7 | NP | NP | NP | NP | 1 | 8 | 7 | 0 | 0 | 3 | 18 |

2 – B | 1 | 1 | NP | NP | NP | NP | 2 | 0 | 0 | 3 | 0 | 5 |

3 – C | 0/3 | 3/0 | 0 | NP | NP | NP | 3 | NP | 1 | 1 | 1 | 6 |

4 – D | 3/0 | 0/3 | 0 | 3 | NP | NP | 6 | NP | NP | 0 | 0 | 6 |

5 – E | 0 | 0 | 3 | 0 | 0 | NP | 3 | NP | NP | NP | 7 | 10 |

6 – F | 0 | 0 | 7 | 0 | 3 | 0 | 10 | NP | NP | NP | NP | 10 |

7 – G | NP | 7 | 0 | 7 | 0 | 3 | 17 | 0 | NP | NP | NP | 17 |

8 – H | NP | NP | 1 | 0 | 0 | 0 | 1 | 0 | 3 | NP | NP | 4 |

9 – I | NP | NP | NP | 1 | 1 | 0 | 2 | 1 | 7 | 7 | NP | 17 |

10 – J | NP | NP | NP | NP | 7 | 7 | 14 | 3 | 0 | 0 | 0 | 17 |

Now all the question can be answered:

- Score of Chen, David, and Eric respectively after Round 3 =3, 3,3
- option a)
- option a)
- option a)

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