10 players – P1, P2, … , P10 – competed in an international javelin throw event. The number (after P) of a player reflects his rank at the beginning of the event, with rank 1 going to the topmost player. There were two phases in the event with the first phase consisting of rounds 1, 2, and 3, and the second phase consisting of rounds 4, 5, and 6. A throw is measured in terms of the distance it covers (in meters, up to one decimal point accuracy), only if the throw is a ‘valid’ one. For an invalid throw, the distance is taken as zero. A player’s score at the end of a round is the maximum distance of all his throws up to that round. Players are re-ranked after every round based on their current scores. In case of a tie in scores, the player with a prevailing higher rank retains the higher rank. This ranking determines the order in which the players go for their throws in the next round.
In each of the rounds in the first phase, the players throw in increasing order of their latest rank, i.e. the player ranked 1 at that point throws first, followed by the player ranked 2 at that point and so on. The top six players at the end of the first phase qualify for the second phase. In each of the rounds in the second phase, the players throw in decreasing order of their latest rank i.e. the player ranked 6 at that point throws first, followed by the player ranked 5 at that point and so on. The players ranked 1, 2, and 3 at the end of the sixth round receive gold, silver, and bronze medals respectively.
All the valid throws of the event were of distinct distances (as per stated measurement accuracy). The tables below show distances (in meters) covered by all valid throws in the first and the third round in the event.
The following facts are also known.
i. Among the throws in the second round, only the last two were valid. Both the throws enabled these players to qualify for the second phase, with one of them qualifying with the least score.
None of these players won any medal.
ii. If a player throws first in a round AND he was also the last (among the players in the current round) to throw in the previous round, then the player is said to get a double. Two players got a double.
iii. In each round of the second phase, exactly one player improved his score. Each of these improvements was by the same amount.
iv. The gold and bronze medalists improved their scores in the fifth and the sixth rounds respectively. One medal winner improved his score in the fourth round.
v. The difference between the final scores of the gold medalist and the silver medalist, as well as the difference between the final scores of the silver medalist and the bronze medalist was 1.0 m.
Q. 1 Which two players got the double?
1) P8, P10
2) P2, P4
3) P1, P10
4) P1, P8
Explanation
Correct Answer
1
Sure. Let’s walk through the step-by-step reasoning behind the solution without any emoji or MathJax and in plain, clear text:
1. Understanding the situation
- There are 10 participants: P1 to P10.
- There are 6 rounds of throws.
- After Round 3, the top 6 participants qualify for Rounds 4, 5, and 6.
- The final medals are based on the best throws.
- Only 3 improvements happened in the final three rounds, and each improvement is of the same value (let’s call this x).
2. What is known from early rounds?
- P10 had to throw a double (back-to-back best throws in R2 and R3), so his throw in R2 had to be better than the highest from R1, which was 87.2 by P7.
Therefore, P10’s throw in R2 is greater than 87.2. - P8’s throw in R3 allowed him to beat P6 and P3, but not P1.
Therefore, his throw in R3 is between 82.5 and 82.9. - P1 threw 88.6 in R3. This was the best in R3.
- Since P10 did not win a medal and there were only 3 improvements left, P10 could not have thrown more than P1’s 88.6 in any round.
3. Identifying potential medalists
- We know only 3 improvements occurred (in R4, R5, or R6).
- P9’s throw in R3 was 84.1. Since this is way behind the top throws (P1: 88.6, P5: 86.4, P7: 87.2), P9 is not one of the medalists — he could not catch up with just one improvement.
- Therefore, the 3 medalists must be P1, P5, and P7.
4. How the improvements work
- Since there are 3 improvements and they are of the same value, x, the improvements were distributed like this:
- One athlete improved twice (gold medalist).
- One athlete improved once (silver medalist).
- One athlete improved once (bronze medalist).
- P1 already had the highest throw (88.6) at the start of R4.
If P1 got both improvements, no one could catch him.
Therefore, P1 could not be the gold medalist. - P5 had the lowest of the three throws (86.4).
If P5 got both improvements, he could not surpass both P1 and P7 even with two improvements. - That leaves P7 to be the athlete who got two improvements and became the gold medalist.
5. Determining the value of x
We need the medalist scores to be 1 meter apart:
- P7 (gold): 87.2 + 2x
- P1 (silver): 88.6 + x
- P5 (bronze): 86.4 + x
Let’s set the scores so that the difference between each medalist is 1 meter:
- P7 = P5 + 2 meters
- P7 = P1 + 1 meter
Substitute:
- 87.2 + 2x = 86.4 + x + 2
- 87.2 + 2x = 88.6 + x + 1
Solve the first:
87.2 + 2x = 88.4 + x
2x – x = 88.4 – 87.2
x = 1.2
Verify the second:
87.2 + 2x = 88.6 + x + 1
87.2 + 2x = 89.6 + x
2x – x = 89.6 – 87.2
x = 2.4
Actually, we need to think in terms of one athlete improving twice, one once, and one not improving:
- P7 improves twice: 87.2 + 2x
- P1 improves not at all: 88.6
- P5 improves once: 86.4 + x
We want:
87.2 + 2x = 88.6 + 1
87.2 + 2x = 89.6
2x = 89.6 – 87.2
2x = 2.4
x = 1.2
6. Final scores
- P7 (gold): 87.2 + 2 * 1.2 = 89.6
- P1 (silver): 88.6 (no improvement)
- P5 (bronze): 86.4 + 1.2 = 87.6
7. Medal order
- Gold: P7 with 89.6
- Silver: P1 with 88.6
- Bronze: P5 with 87.6
8. Final table completion
The final table reflects this logic:
- P7 improves in Round 5 and 6
- P5 improves in Round 5
- P1 does not improve
| Player | Round 1 | Round 2 | Round 3 | Round 4 | Round 5 | Round 6 |
|---|---|---|---|---|---|---|
| P1 | 84.1 | 82.9 | 88.6 | 88.6 | 88.6 | 88.6 |
| P2 | – | 86.4 | 87.2 | 87.2 | 87.2 | 87.2 |
| P3 | 81.5 | 81.5 | 81.5 | 81.5 | 81.5 | 81.5 |
| P4 | 82.9 | 82.9 | 84.1 | 84.1 | 84.1 | 84.1 |
| P5 | 86.4 | 86.4 | 86.4 | 86.4 | 87.6 | 87.6 |
| P6 | 82.5 | 82.5 | 82.5 | 82.5 | 82.5 | 82.5 |
| P7 | 87.2 | 87.2 | 87.2 | 88.4 | 89.6 | 89.6 |
| P8 | – | – | 82.5 < y < 82.9 | 82.5 < y < 82.9 | 82.5 < y < 82.9 | 82.5 < y < 82.9 |
| P9 | 84.1 | 84.1 | 84.1 | 84.1 | 84.1 | 84.1 |
| P10 | 82.9 | >87.2 | 87.2 < w < 87.6 | 87.2 < w < 87.6 | 87.2 < w < 87.6 | 87.2 < w < 87.6 |
Summary of logic
- P10’s throws were high early but not medal-worthy.
- P9 was too far behind to medal.
- P1 started best, but couldn’t get two improvements or he’d win gold easily.
- P5 had the weakest starting position among top three so couldn’t be gold.
- P7’s middle position meant he could win with two improvements.
- The improvements were +1.2 meters each, giving medals to P7, P1, and P5 in that order.
Q. 2 Who won the silver medal?
1) P7
2) P9
3) P1
4) P5
Explanation
Correct Answer
3
Q. 3 Who threw the last javelin in the event?
1) P10
2) P9
3) P1
4) P7
Explanation
Correct Answer
4
Q. 4 What was the final score (in m) of the silver-medalist?
1) 87.2
2) 88.6
3) 89.6
4) 88.4
Explanation
Correct Answer
2
