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CAT 2023 Slot 2 DILR · Set 2 of 4Data InterpretationModerate DifficultyQ6–Q10 · 3 MCQ + 2 TITA

CAT 2023 Slot 2 DILR Set 2: Tables (5 Questions) with Solutions

The Tables set from the official CAT 2023 Slot 2 DILR paper: the caselet, all 5 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 2023 Slot 2 DILR · Questions 6–10
Three participants – Akhil, Bimal and Chatur participate in a random draw competition for five days. Every day, each participant randomly picks up a ball numbered between 1 and 9, both inclusive. The number on the ball determines his score on that day. The total score of a participant is the sum of his scores attained in the five days. The total score of a day is the sum of participants' scores on that day. The 2-day average on a day, except on Day 1, is the average of the total scores of that day and of the previous day. For example, if the total scores of Day 1 and Day 2 are 25 and 20, then the 2-day average on Day 2 is calculated as 22.5. Table 1 gives the 2-day averages for Days 2 through 5.
Table 1: 2-day averages for Days 2 through 5
Day 2Day 3Day 4Day 5
1515.51617
Participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. If there is a tie, all participants with the tied score are awarded the best available rank. For example, if on a day Akhil, Bimal, and Chatur score 8, 7 and 7 respectively, then their ranks will be 1, 2 and 2 respectively on that day. These ranks are given in Table 2.
Table 2: Ranks of participants on each day
Day 1Day 2Day 3Day 4Day 5
Akhil12233
Bimal23211
Chatur31122
The following information is also known.
1
Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil's score on Day 4.
2
The total score on Day 3 is the same as the total score on Day 4.
3
Bimal's scores are the same on Day 1 and Day 3.

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

Q6. Tables

MCQ · +3 / −1
What is Akhil's score on Day 1?
  1. A) 7
  2. B) 5
  3. C) 6
  4. D) 8
Correct Answer
A) 7
Step-by-step solution
📌 Core Concept
The 2-day averages fix every daily total. The ranks and clues then fix individual scores.
🔢 Step-by-Step Solution
1
Let the daily totals be T1T_1 to T5T_5. The averages give T1+T2=30T_1+T_2=30, T2+T3=31T_2+T_3=31, T3+T4=32T_3+T_4=32 and T4+T5=34T_4+T_5=34.
2
Clue 2 says T3=T4T_3=T_4. From the first three equations, T3=T1+1T_3=T_1+1 and T4=31−T1T_4=31-T_1. Setting these equal gives T1=15T_1=15, so T2=15T_2=15, T3=16T_3=16, T4=16T_4=16 and T5=18T_5=18.
3
Table 2 ranks Akhil and Bimal equal on Day 3. Clue 3 says Bimal scores the same bb on Days 1 and 3, so Akhil also scores bb on Day 3. Chatur is rank 1 on Day 3, so his score c3>bc_3>b. Then 2b+c3=162b+c_3=16.
4
Chatur scores only multiples of 3, and his minimum occurs only on Day 1. Hence c3≥6c_3\ge 6. If c3=6c_3=6, then b=5b=5. If c3=9c_3=9, then 2b=72b=7, which is impossible. So c3=6c_3=6 and b=5b=5.
5
Chatur's Day 1 score is his unique minimum and is a multiple of 3 below Bimal's 5, so it is 3.
6
On Day 1 the total is 15 and the ranks are Akhil 1, Bimal 2, Chatur 3. So A1=15−5−3=7A_1=15-5-3=7, which is above Bimal's 5 as the ranking requires.
Why Not the Others
5, 6 and 8 give a Day 1 total different from 15 once Bimal has 5 and Chatur has 3.
Quick Check
7+5+3=15=T17+5+3=15=T_1.
Correct Answer: A

Q7. Tables

MCQ · +3 / −1
Who attains the maximum total score?
  1. A) Akhil
  2. B) Cannot be determined
  3. C) Bimal
  4. D) Chatur
Correct Answer
D) Chatur
Step-by-step solution
📌 Core Concept
Fix every score that is forced. Then compare the best possible totals of Akhil and Bimal with Chatur's total.
🔢 Step-by-Step Solution
1
The daily totals are 15, 15, 16, 16 and 18. Bimal scores 5 on Days 1 and 3, Akhil scores 7 on Day 1 and 5 on Day 3, and Chatur scores 3 on Day 1 and 6 on Day 3.
2
Day 4 ranks are Bimal 1, Chatur 2, Akhil 3. Akhil's Day 4 score equals Chatur's minimum, 3. Chatur's Day 4 score is 6 or 9. If it were 9, Bimal would need more than 9, which is impossible. So Chatur scores 6 and Bimal scores 16−3−6=716-3-6=7.
3
Chatur's Day 2 score is the unique highest in the competition, so it exceeds Bimal's 7. It is a multiple of 3, so it is 9. Day 5 works the same way: Chatur has 6, Bimal exceeds 6 but stays below 9, and Akhil is below 6.
4
Chatur's total is 3+9+6+6+6=303+9+6+6+6=30.
5
On Day 2, Chatur has 9 and the other two sum to 6, with Akhil above Bimal. So Bimal scores at most 2 and Akhil at most 5. Day 5 gives A5+B5=12A_5+B_5=12 with A5≤5A_5\le5 and B5≤8B_5\le8.
6
Bimal's maximum is 5+2+5+7+8=275+2+5+7+8=27. Akhil's maximum is 7+5+5+3+5=257+5+5+3+5=25. Both are below 30.
Why Not the Others
Akhil and Bimal cannot reach 30 under any allowed split.
The total is determinable, so "Cannot be determined" is wrong.
Correct Answer: D

Q8. Arrangements

TITA · +3 / 0
What is the minimum possible total score of Bimal?
Correct Answer
25
Step-by-step solution
📌 Core Concept
The forced scores leave only Bimal's Day 2 and Day 5 scores open. Minimise each within the rank and highest-score constraints.
🔢 Step-by-Step Solution
1
The daily totals are 15, 15, 16, 16 and 18. Bimal scores 5 on Days 1 and 3 and 7 on Day 4. Chatur scores 3, 9, 6, 6 and 6 on Days 1 to 5.
2
Day 2: Chatur has 9, so Akhil and Bimal sum to 6. Ranks put Akhil above Bimal, so Bimal's options are 1 or 2, giving (A2,B2)=(5,1)(A_2,B_2)=(5,1) or (4,2)(4,2). The minimum for Bimal is 1.
3
Day 5: the total is 18 and Chatur has 6, so A5+B5=12A_5+B_5=12. Ranks put Akhil below Chatur, so A5≤5A_5\le5. Hence B5≥7B_5\ge7, and the minimum is 7 with A5=5A_5=5.
4
Minimum total for Bimal is 5+1+5+7+7=255+1+5+7+7=25.
Quick Check
These scores keep every rank in Table 2, and Day 2 sums to 5+1+9=155+1+9=15.
Correct Answer: 25

Q9. Arrangements

MCQ · +3 / −1
If the total score of Bimal is a multiple of 3, what is the score of Akhil on Day 2?
  1. A) 5
  2. B) 4
  3. C) Cannot be determined
  4. D) 6
Correct Answer
B) 4
Step-by-step solution
📌 Core Concept
Bimal's open scores are B2∈{1,2}B_2\in\{1,2\} and B5∈{7,8}B_5\in\{7,8\}. A divisibility condition on his total selects one pair.
🔢 Step-by-Step Solution
1
Bimal scores 5, 5 and 7 on Days 1, 3 and 4, a sum of 17. His total is 17+B2+B517+B_2+B_5.
2
Day 2: Chatur scores 9, and Akhil and Bimal sum to 6 with Akhil above Bimal. So B2B_2 is 1 or 2, with A2=5A_2=5 or 4 respectively.
3
Day 5: A5+B5=12A_5+B_5=12 with A5≤5A_5\le5, so B5≥7B_5\ge7. Chatur's 9 on Day 2 is the unique highest score, so B5<9B_5<9. Hence B5B_5 is 7 or 8.
4
The four possible totals are 25 (1,7)(1,7), 26 (1,8)(1,8), 26 (2,7)(2,7) and 27 (2,8)(2,8). Only 27 is a multiple of 3, so B2=2B_2=2.
5
Then A2=6−2=4A_2=6-2=4.
Why Not the Others
5 would need B2=1B_2=1, which gives a total of 25 or 26, not a multiple of 3.
6 would equal Chatur's score minus 3 and leave B2=0B_2=0, which is not a valid score.
The value is fixed, so "Cannot be determined" is wrong.
Correct Answer: B

Q10. Tables

TITA · +3 / 0
If Akhil attains a total score of 24, then what is the total score of Bimal?
Correct Answer
26
Step-by-step solution
📌 Core Concept
Akhil's total of 24 restricts his two open scores. Both consistent cases give Bimal the same total.
🔢 Step-by-Step Solution
1
Akhil's fixed scores are 7, 5 and 3 on Days 1, 3 and 4, a sum of 15. So A2+A5=24−15=9A_2+A_5=24-15=9.
2
Day 2 gives A2∈{4,5}A_2\in\{4,5\}, with B2=6−A2B_2=6-A_2. Day 5 gives A5≤5A_5\le5 and B5=12−A5B_5=12-A_5, so A5∈{4,5}A_5\in\{4,5\} since B5<9B_5<9.
3
A sum of 9 needs one score to be 4 and the other 5.
4.
Case 1
A2=4A_2=4, A5=5A_5=5. Then B2=2B_2=2 and B5=7B_5=7, so Bimal totals 5+2+5+7+7=265+2+5+7+7=26.
5.
Case 2
A2=5A_2=5, A5=4A_5=4. Then B2=1B_2=1 and B5=8B_5=8, so Bimal totals 5+1+5+7+8=265+1+5+7+8=26.
6
Both cases give 26.
Quick Check
The three totals are 24+26+30=8024+26+30=80. The daily totals are 15+15+16+16+18=8015+15+16+16+18=80.
Correct Answer: 26
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CAT 2023 Slot 2 DILR at a Glance

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

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