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CAT 2021 Slot 3 DILR · Set 1 of 4Logical ReasoningHard DifficultyQ1–Q6 · 6 MCQ

CAT 2021 Slot 3 DILR Set 1: Arrangements (6 Questions) with Solutions

The Arrangements set from the official CAT 2021 Slot 3 DILR paper: the caselet, all 6 questions, the official answer key and a step-by-step solution for each. 100% free, no sign-up.

The Set: Read the Caselet First

CAT 2021 Slot 3 DILR · Questions 1–6
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 around 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.
Distances covered by all the valid throws in the first round
PlayerDistance (in m)
P182.9
P381.5
P586.4
P682.5
P787.2
P984.1
Distances covered by all the valid throws in the third round
PlayerDistance (in m)
P188.6
P379.0
P981.4
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 1.01.0 m.

In the exam this set competed with 3 others for 40 minutes. Budget about 10 minutes if you attempt it.

Q1. Arrangements

MCQ · +3 / −1
Which two players got the double?
  1. A) P1, P10
  2. B) P2, P4
  3. C) P8, P10
  4. D) P1, P8
Correct Answer
C) P8, P10
Step-by-step solution
📌 Core Concept
A double needs a player to be last in one round and first in the next. Combining the round-1 data with the phase-1 and phase-2 facts shows which rounds can produce one.
🔢 Step-by-Step Solution
1
Round 1 gives P2, P4, P8 and P10 a score of 0, so they rank 7 to 10 in that order. Round 2 therefore opens with P7 (87.2), and P10 throws last. P7 was not last in round 1, so round 2 has no double.
2
Fact (i): P8 and P10 are the only valid throws in round 2, and both qualify. Round 3 adds nothing for them, so both must beat P6's 82.5 to reach the top six.
3
Rounds 4 and 6 only matter if the first thrower was the previous round's last thrower. Rounds 5 and 6 cannot give a double. P1 (88.6) is rank 1 after rounds 3 and 4, so P1 throws last in rounds 4 and 5. P1 is never ranked 6 and so never throws first in the next round.
4
Two doubles are needed, so they must fall in rounds 3 and 4.
5
Round 3: P10 threw last in round 2, so P10 must be rank 1 after round 2. This needs a score above 87.2. P10 gets the double.
6
Round 4: the first thrower is the lowest of the six after round 3, and must have been lowest in rank among the six after round 2. P9 (84.1) cannot be that player, because P1 (82.9) was below P9 after round 2. So it is P8, with a score between 82.5 and 82.9.
Why Not the Others
P1 can never throw first in a later round.
P2 and P4 are out after phase 1.
P9 sat above P1 after round 2.
Correct Answer: C

Q2. Arrangements

MCQ · +3 / −1
Who won the silver medal?
  1. A) P9
  2. B) P5
  3. C) P7
  4. D) P1
Correct Answer
D) P1
Step-by-step solution
📌 Core Concept
Every phase-2 improvement is the same size dd, and the final scores of gold, silver and bronze differ by exactly 1.0. Testing who could have improved fixes the medalists.
🔢 Step-by-Step Solution
1
After round 3 the scores are P1 88.6, P7 87.2, P5 86.4 and P9 84.1. P9's 81.4 does not beat 84.1. P8 and P10 win no medal.
2
Fact (iii) allows three improvements, one in each of rounds 4, 5 and 6, all by dd. Gold improved in round 5 and bronze in round 6. The round-4 improver is a medalist. P8 and P10 therefore never improve.
3
If silver improved in round 4, then gold, silver and bronze each improve once by dd. Their starting scores would also be 1.0 apart. No two of 88.6, 87.2, 86.4 and 84.1 differ by 1.0, so this fails.
4
So silver did not improve, and the final scores are GG, G−1G-1 and G−2G-2. Silver has to be one of the round-3 scores.
5
Silver P1 at 88.6 gives gold 89.6 and bronze 87.6. Gold P7 improving twice (rounds 4 and 5) gives 87.2+2d=89.687.2+2d=89.6, so d=1.2d=1.2. Bronze P5 then gives 86.4+1.2=87.686.4+1.2=87.6 in round 6. This works.
6
Silver P7, P5 or P9 fails. Silver P7 would need gold of 88.2, below P1's 88.6. The other cases give a bronze or gold start that is not a score.
Why Not the Others
P9, P5 and P7 cannot be silver, as step 6 shows.
Correct Answer: D

Q3. Arrangements

MCQ · +3 / −1
Who threw the last javelin in the event?
  1. A) P7
  2. B) P1
  3. C) P10
  4. D) P9
Correct Answer
A) P7
Step-by-step solution
📌 Core Concept
In phase 2 the order is decreasing rank, so the last thrower of round 6 is whoever is ranked 1 after round 5.
🔢 Step-by-Step Solution
1
Facts (iii) to (v) give final scores of gold 89.6 (P7), silver 88.6 (P1) and bronze 87.6 (P5). Each improvement is d=1.2d=1.2.
2
P7 started phase 2 on 87.2 and improved in rounds 4 and 5: 87.2+1.2+1.2=89.687.2+1.2+1.2=89.6.
3
P5 improved only in round 6, so after round 5 P5 is still on 86.4. P1 is on 88.6, and P8 and P10 are below 87.6.
4
After round 5, P7 (89.6) is rank 1, ahead of P1 (88.6).
5
Round 6 runs from rank 6 down to rank 1, so P7 throws last. This is the final javelin of the event.
Why Not the Others
P1 is rank 2 and throws second to last.
P10 is ranked below the leaders and throws earlier.
P9 sits on 84.1 and throws early.
Correct Answer: A

Q4. Arrangements

MCQ · +3 / −1
What was the final score (in m) of the silver-medalist?
  1. A) 87.2
  2. B) 89.6
  3. C) 88.6
  4. D) 88.4
Correct Answer
C) 88.6
Step-by-step solution
📌 Core Concept
The final scores of gold, silver and bronze are 1.0 apart, and all phase-2 improvements are equal. Together these identify the silver score.
🔢 Step-by-Step Solution
1
After round 3 the scores are P1 88.6, P7 87.2, P5 86.4 and P9 84.1. P8 and P10 are non-medalists and never improve.
2
Gold improved in round 5 and bronze in round 6. The round-4 improver is a medalist, so there are three improvements of dd in total.
3
Silver cannot improve. If it did, all three medalists would improve once, and their starting scores would need to be 1.0 apart. No such pair exists among 88.6, 87.2, 86.4 and 84.1.
4
So silver's final score equals a round-3 score, and gold and bronze are 1.0 above and below it.
5
Silver 88.6 gives gold 89.6 and bronze 87.6. Gold P7 improving twice gives 87.2+2d=89.687.2+2d=89.6, so d=1.2d=1.2. Bronze P5 improving once gives 86.4+1.2=87.686.4+1.2=87.6. Both fit.
6
Silver 87.2, 86.4 or 84.1 fails. For example, silver 87.2 needs gold of 88.2, which is below P1's 88.6.
7
The silver medalist is P1, who never improved, with a final score of 88.6.
Quick Check
The scores 89.6, 88.6 and 87.6 are spaced 1.0 apart, and d=1.2d=1.2 is used in rounds 4, 5 and 6.
Correct Answer: C

Q5. Arrangements

MCQ · +3 / −1
Which of the following can be the final score (in m) of P8?
  1. A) 85.1
  2. B) 0
  3. C) 81.9
  4. D) 82.7
Correct Answer
D) 82.7
Step-by-step solution
📌 Core Concept
P8's only valid throw was in round 2. The double in round 4 and the qualification condition pin its score to a narrow range.
🔢 Step-by-Step Solution
1
Fact (i) says P8 threw one of the last two valid throws of round 2 and qualified. P8's final score equals that throw, because P8 is not a medalist and never improves in phase 2.
2
P8 must reach the top six. After round 3 the others are P1 88.6, P7 87.2, P5 86.4, P9 84.1, P6 82.5 and P3 81.5. P8 and P10 must both beat P6, so P8 is above 82.5. A score of 0 or 81.9 fails this.
3
Fact (ii) allows two doubles. Rounds 2, 5 and 6 give none, so they fall in rounds 3 and 4. P10 takes the round-3 double, so P8 must take the round-4 double.
4
P8 would be first in round 4 as the lowest of the six after round 3. It would also be last in round 3 among those six, so lowest after round 2. P1 stood at 82.9 after round 2, so P8 is below 82.9.
5
So P8 lies strictly between 82.5 and 82.9. Only 82.7 fits.
Why Not the Others
85.1 is above P1's 82.9 and P9's 84.1, so P8 would not be lowest.
0 and 81.9 do not beat P6's 82.5, so P8 would miss the top six.
Correct Answer: D

Q6. Arrangements

MCQ · +3 / −1
By how much did the gold medalist improve his score (in m) in the second phase?
  1. A) 2.0
  2. B) 1.2
  3. C) 1.0
  4. D) 2.4
Correct Answer
D) 2.4
Step-by-step solution
📌 Core Concept
Facts (iii) to (v) fix who won each medal and the common improvement dd. The gold medalist's total gain is then a multiple of dd.
🔢 Step-by-Step Solution
1
After round 3 the scores are P1 88.6, P7 87.2, P5 86.4 and P9 84.1. P8 and P10 win no medal and never improve.
2
Gold improved in round 5 and bronze in round 6. A medalist improved in round 4. Silver cannot have improved, because no two round-3 scores are 1.0 apart (all three would then need starting scores 1.0 apart).
3
So the round-4 improver is gold or bronze. Silver is unchanged, with gold 1.0 above and bronze 1.0 below it.
4
Silver P1 on 88.6 gives gold 89.6 and bronze 87.6. This is the only workable case.
5
Gold P7 improving in rounds 4 and 5 gives 87.2+2d=89.687.2+2d=89.6, so d=1.2d=1.2. Bronze P5 improving once gives 86.4+1.2=87.686.4+1.2=87.6.
6
Gold improved twice by 1.2, so the total improvement in phase 2 is 1.2+1.2=2.41.2+1.2=2.4 m.
Quick Check
The gain is 89.6−87.2=2.489.6-87.2=2.4 m.
Correct Answer: 2.4
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CAT 2021 Slot 3 DILR at a Glance

Sets in this section4
Questions in this section20
Section time40 minutes
Convenor IIMIIM Ahmedabad

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