Discussion

Explanation:

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: 
P1 → P2 → P3 → P4 → P5 → P6 → P7 → P8 → P9 → P10

In the first round only 6 players had a valid throw: P1, P3, P5, P6, P7 and P9
∴ Ranking at the end of 1st round.

1    P7    87.2
2    P5    86.4
3    P9    84.1
4    P1    82.9
5    P6    82.5
6    P3    81.5
7    P2    -
8    P4    -
9    P8    -
10  P10    -


P7, P5, P9, P1, P6, P3 P2, P4, P8, P10 

Now, in round 2, only last two throws i.e., of P8 and P10 were 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. One of P8 or P10 is at the sixth positions.  P1 > P7 > P5 > P9.

∴ Their ranks at the end of round 3.

1/2    P1    88.6
2/3    P7    87.2
3/4    P5    86.4
4/5    P9    84.1
6       P8/P10    

The other person between P8 / P10 can go anywhere between rank 1 and 5.

Now let us consider the two players who got a double. Doubles happen in the transition between rounds.

Round 1 → Round 2: Double is not possible as P10 (last to throw in Round 1) did not have a valid throw.

Round 2 → Round 1: Possible if P10 (last to throw in Round 1) reaches Rank 1 in round 2.

Round 3 → Round 4: P8/P10 who is the last among qualifying will be the first to throw. So, here it definitely happens.

Round 4 → Round 5: Not possible since only one player improves his/her rank in round 4.

Round 5 → Round 6: Not possible since only one player improves his/her rank in round 5.

∴ After Round 2 P10 definitely reaches top of the ranking i.e., P10 throws more than 87.2.

∴ Ranks at the end of round 3.

Case 1
1    P1    88.6
2    P10    ?
3    P7    87.2
4    P5    86.4
5    P9    84.1
6    P8    

Case 2
1    P10    ?
2    P1    88.6
3    P7    87.2
4    P5    86.4
5    P9    84.1
6    P8    

P10 and P8 do not win any medals.

Case 2: 3 of P1, P7, P5 and P9 need to score better than P10. Each of these 3 will improve their score by same amount say x.

Possible scores after improvement:
P1: 88.6 + x
P7: 87.2 + x
P5: 86.4 + x
P9: 84.1 + x

No value of x will satisfy the 5th condition given in the question. Hence, this case is rejected.

Case 1: 2 or 3 of P1, P7, P5 and P9 need to score better than P10. Each of these 3 will improve their score by same amount say x.

Possible scores after improvement:
P1: 88.6 + x
P7: 87.2 + x
P5: 86.4 + x
P9: 84.1 + x

If P1 does not win any medal, P7, P5 and P9 will have to improve their scores. But this will not satisfy conditions 3 and 5 simultaneously. Hence P1 has to be one of the medalists.

P10 has scored more than 87.2, hence the three medalists need to score more than 87.2.

The only way this is possible is when P7 improves his score twice by 1.2 while P5 improves his score once by 1.2.

Scores at the end of round 5.
P7: 89.6
P1: 88.6
P5: 87.6

∴ P8 and P10 get the doubles.

Hence, option (d).

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