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CAT 2022 Slot 3 DILR · Set 3 of 4Logical ReasoningEasy DifficultyQ11–Q15 · 5 MCQ

CAT 2022 Slot 3 DILR Set 3: Arrangements (5 Questions) with Solutions

The Arrangements set from the official CAT 2022 Slot 3 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 2022 Slot 3 DILR · Questions 11–15
There are only four neighbourhoods in a city - Levmisto, Tyhrmisto, Pesmisto and Kitmisto. During the onset of a pandemic, the number of new cases of a disease in each of these neighbourhoods was recorded over a period of five days. On each day, the number of new cases recorded in any of the neighbourhoods was either 0, 1, 2 or 3.
The following facts are also known:
1
There was at least one new case in every neighbourhood on Day 1.
2
On each of the five days, there were more new cases in Kitmisto than in Pesmisto.
3
The number of new cases in the city in a day kept increasing during the five-day period. The number of new cases on Day 3 was exactly one more than that on Day 2.
4
The maximum number of new cases in a day in Pesmisto was 2, and this happened only once during the five-day period.
5
Kitmisto is the only place to have 3 new cases on Day 2.
6
The total numbers of new cases in Levmisto, Tyhrmisto, Pesmisto and Kitmisto over the five-day period were 12, 12, 5 and 14 respectively.

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

Q11. Arrangements

MCQ · +3 / −1
What BEST can be concluded about the total number of new cases in the city on Day 2?
  1. A) Exactly 7
  2. B) Either 6 or 7
  3. C) Either 7 or 8
  4. D) Exactly 8
Correct Answer
D) Exactly 8
Step-by-step solution
📌 Core Concept
Fix the daily city totals using the strict increase, the cap of 3 per cell, and the total of 43 cases. Then read off Day 2.
🔢 Step-by-Step Solution
1
Pesmisto's total is 5 and its maximum is 2, reached only once. So its five values are one 2, three 1s and one 0 (any other split needs another 2 or a 3).
2
The five daily totals add up to 12 + 12 + 5 + 14 = 43. Call them d1<d2<d3<d4<d5d_1<d_2<d_3<d_4<d_5, with d3=d2+1d_3=d_2+1.
3
Pesmisto never has 3, so the most any day can total is 3+3+3+2=113+3+3+2=11. Hence d5≤11d_5\le 11 and d4≤10d_4\le 10.
4
Clue 1 gives d1≥4d_1\ge 4. Since d1<d2d_1<d_2, we get d2≥5d_2\ge 5.
5
Try d2=7d_2=7. Then d3=8d_3=8 and d4+d5=43−d1−15=28−d1d_4+d_5=43-d_1-15=28-d_1. With d1≤6d_1\le 6 this is at least 22, but d4+d5≤10+11=21d_4+d_5\le 10+11=21. This is impossible. Smaller d2d_2 is worse, since it forces d4+d5d_4+d_5 even higher.
6
Try d2=9d_2=9. Then d3=10d_3=10, so d4≥11d_4\ge 11. That contradicts d4≤10d_4\le 10.
7
Only d2=8d_2=8 is left: d3=9d_3=9, and d1+d4+d5=26d_1+d_4+d_5=26 with d4≤10d_4\le10, d5≤11d_5\le11, d1≥4d_1\ge4. So d4=10d_4=10, d5=11d_5=11, d1=5d_1=5, and the totals are 5, 8, 9, 10, 11.
Quick Check
5 + 8 + 9 + 10 + 11 = 43. ✓
Correct Answer: D

Q12. Arrangements

MCQ · +3 / −1
What BEST can be concluded about the number of new cases in Levmisto on Day 3?
  1. A) Exactly 2
  2. B) Exactly 3
  3. C) Either 0 or 1
  4. D) Either 2 or 3
Correct Answer
B) Exactly 3
Step-by-step solution
📌 Core Concept
Work out the daily totals (5, 8, 9, 10, 11), then fill the grid from the days with the least slack.
🔢 Step-by-Step Solution
1
Pesmisto's values are one 2, three 1s and one 0. The daily city totals are 5, 8, 9, 10, 11, since the totals and Pesmisto's pattern force this.
2
Day 5 totals 11, which needs 3+3+3+23+3+3+2. So Pesmisto has its only 2 on Day 5, and every other neighbourhood has 3.
3
Day 4 totals 10, and Pesmisto is at most 1 that day because its 2 is already used. So Levmisto, Tyhrmisto and Kitmisto are all 3, and Pesmisto is 1.
4
Day 1 needs at least 1 in each place. So Pesmisto's Day 1 is 1, and its 0 falls on Day 2 or Day 3.
5
Clue 5 gives Kitmisto 3 on Day 2, so the other three places have at most 2 that day. Day 2 totals 8, so the other three sum to 5.
6
If Pesmisto were 0 on Day 2, Levmisto and Tyhrmisto would need to sum to 5. They can only reach 2+2=42+2=4, so this fails. Pesmisto is 1 on Day 2 and Levmisto = Tyhrmisto = 2.
7
Pesmisto's 0 is therefore on Day 3. Day 3 totals 9 with Pesmisto at 0, so Levmisto, Tyhrmisto and Kitmisto sum to 9. Each is 3.
Levmisto had exactly 3 new cases on Day 3.
Quick Check
Kitmisto's Day 1 value is then 14−3−3−3−3=214-3-3-3-3=2, which exceeds Pesmisto's 1. ✓
Correct Answer: B

Q13. Arrangements

MCQ · +3 / −1
On which days did Pesmisto not have any new case?
  1. A) Only Day 3
  2. B) Both Day 2 and Day 3
  3. C) Only Day 2
  4. D) Both Day 2 and 4
Correct Answer
A) Only Day 3
Step-by-step solution
📌 Core Concept
Pesmisto's pattern is one 2, three 1s and one 0. The task is to find which day holds the 0.
🔢 Step-by-Step Solution
1
Pesmisto totals 5 with a maximum of 2 occurring once. So its values are one 2, three 1s and one 0.
2
The daily city totals are forced to be 5, 8, 9, 10, 11. Day 5 totals 11, which needs 3, 3, 3 and Pesmisto at 2. So Pesmisto's only 2 is on Day 5.
3
Day 4 totals 10. Pesmisto is at most 1 that day, so the other three places are all 3 and Pesmisto is 1.
4
Clue 1 gives Pesmisto at least 1 on Day 1, so its Day 1 value is 1. The single 0 is on Day 2 or Day 3.
5
Clue 5 caps Levmisto and Tyhrmisto at 2 on Day 2, and Kitmisto is 3. With a Day 2 total of 8, a 0 for Pesmisto would need Levmisto + Tyhrmisto = 5, which is more than 2+2=42+2=4. So Pesmisto is not 0 on Day 2.
6
The 0 is therefore on Day 3, and Pesmisto has cases on Days 1, 2, 4 and 5.
Why Not the Others
B, C and D all include Day 2, which step 5 rules out.
Correct Answer: A

Q14. Arrangements

MCQ · +3 / −1
Which of the two statements below is/are necessarily false?
Statement A: There were 2 new cases in Tyhrmisto on Day 3.
Statement B: There were no new cases in Pesmisto on Day 2.
  1. A) Statement B only
  2. B) Both Statement A and Statement B
  3. C) Neither Statement A nor Statement B
  4. D) Statement A only
Correct Answer
B) Both Statement A and Statement B
Step-by-step solution
📌 Core Concept
Once the grid is fixed, check each statement against the Day 2 and Day 3 values.
🔢 Step-by-Step Solution
1
The daily totals are 5, 8, 9, 10, 11. Day 5 is 3+3+3+23+3+3+2, with Pesmisto's only 2 there. Day 4 is 3+3+3+13+3+3+1.
2
Day 2 totals 8. Kitmisto is 3 (Clue 5), and Levmisto and Tyhrmisto are each at most 2. Pesmisto = 0 would need Levmisto + Tyhrmisto = 5, which is impossible. So Pesmisto is 1 and Levmisto = Tyhrmisto = 2 on Day 2.
3
Pesmisto's 0 therefore falls on Day 3. Day 3 totals 9, so Levmisto, Tyhrmisto and Kitmisto are all 3.
4
Statement A says Tyhrmisto had 2 on Day 3. It actually had 3, so A is false.
5
Statement B says Pesmisto had no cases on Day 2. It actually had 1, so B is false.
Quick Check
Day 1 is Levmisto 1, Tyhrmisto 1, Pesmisto 1, Kitmisto 2, total 5. Both Levmisto and Tyhrmisto then total 12. ✓
Correct Answer: B

Q15. Arrangements

MCQ · +3 / −1
On how many days did Levmisto and Tyhrmisto have the same number of new cases?
  1. A) 2
  2. B) 5
  3. C) 3
  4. D) 4
Correct Answer
B) 5
Step-by-step solution
📌 Core Concept
Build the full grid. Levmisto and Tyhrmisto turn out to be identical on every day.
🔢 Step-by-Step Solution
1
The daily city totals are 5, 8, 9, 10, 11. Pesmisto has one 2, three 1s and one 0.
2
Day 5 (11): Levmisto, Tyhrmisto and Kitmisto are 3, and Pesmisto is 2.
3
Day 4 (10): Pesmisto is at most 1, so Levmisto, Tyhrmisto and Kitmisto are 3 and Pesmisto is 1.
4
Day 2 (8): Kitmisto is 3, and Levmisto and Tyhrmisto are at most 2. Pesmisto = 0 would need them to sum to 5, so Pesmisto is 1 and Levmisto = Tyhrmisto = 2.
5
Day 3 (9): Pesmisto has the remaining 0, so Levmisto, Tyhrmisto and Kitmisto are 3 each.
6
Day 1 (5): Pesmisto is 1. Kitmisto's total of 14 gives 14−3−3−3−3=214-3-3-3-3=2 on Day 1. So Levmisto + Tyhrmisto = 5−1−2=25-1-2=2, and Clue 1 makes each of them 1.
7
Levmisto is 1, 2, 3, 3, 3 and Tyhrmisto is 1, 2, 3, 3, 3, each totalling 12. They are equal on all 5 days.
Correct Answer: B
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CAT 2022 Slot 3 DILR at a Glance

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

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