CAT 2022 Slot 3 DILR Set 4 Q16–Q20

A DILR caselet from the Data Interpretation & Logical Reasoning (DILR) section of CAT 2022 Slot 3, with 5 questions (Grids & Matrices). Answers are CAT's official key.

The set

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.

Question 16

What BEST can be concluded about the total number of new cases in the city on Day 2?
  1. A
    Either 6 or 7
  2. B
    Exactly 8
  3. C
    Exactly 7
  4. D
    Either 7 or 8

Answer: (B) Exactly 8 Charts & Tables MCQ

Solution

  1. Pesmisto total 5 with a single 2: one 2, three 1s, one 0. Kitmisto total 14: four 3s and one 2.
  2. Daily totals rise strictly, sum to 43, with Day 3 = Day 2 + 1. Pesmisto is at most 2, so a day's total is at most 11. Day 1 is at least 4, and Day 2 is at most 9 (only Kitmisto has 3). The only sequence that works is 5, 8, 9, 10, 11.
  3. Day 5 = 11 needs Pesmisto 2 and the other three at 3. Day 4 = 10 needs Pesmisto 1 and the others at 3. Day 2 = 8 with Kitmisto 3 gives Levmisto = Tyhrmisto = 2 and Pesmisto 1.
  4. The city had exactly 8 new cases on Day 2.

Question 17

What BEST can be concluded about the number of new cases in Levmisto on Day 3?
  1. A
    Exactly 3
  2. B
    Either 2 or 3
  3. C
    Exactly 2
  4. D
    Either 0 or 1

Answer: (A) Exactly 3 Grids & Matrices MCQ

Solution

  1. Pesmisto total 5 with a single 2: one 2, three 1s, one 0. Kitmisto total 14: four 3s and one 2.
  2. Daily totals rise strictly, sum to 43, with Day 3 = Day 2 + 1. Pesmisto is at most 2, so a day's total is at most 11. Day 1 is at least 4, and Day 2 is at most 9 (only Kitmisto has 3). The only sequence that works is 5, 8, 9, 10, 11.
  3. Day 5 = 11 needs Pesmisto 2 and the other three at 3. Day 4 = 10 needs Pesmisto 1 and the others at 3. Day 2 = 8 with Kitmisto 3 gives Levmisto = Tyhrmisto = 2 and Pesmisto 1.
  4. Day 1 = 5 with everyone at least 1 forces Kitmisto 2, Levmisto = Tyhrmisto = Pesmisto = 1. So Pesmisto has 0 on Day 3 and Kitmisto has 3, and the totals of 12 give Levmisto = Tyhrmisto = 3 on Day 3 (Day 3 total \(3+3+0+3=9\)).
  5. Levmisto had exactly 3 new cases on Day 3.

Question 18

On which day(s) did Pesmisto not have any new case?
  1. A
    Only Day 3
  2. B
    Only Day 2
  3. C
    Both Day 2 and Day 4
  4. D
    Both Day 2 and Day 3

Answer: (A) Only Day 3 Grids & Matrices MCQ

Solution

  1. Pesmisto total 5 with a single 2: one 2, three 1s, one 0. Kitmisto total 14: four 3s and one 2.
  2. Daily totals rise strictly, sum to 43, with Day 3 = Day 2 + 1. Pesmisto is at most 2, so a day's total is at most 11. Day 1 is at least 4, and Day 2 is at most 9 (only Kitmisto has 3). The only sequence that works is 5, 8, 9, 10, 11.
  3. Day 5 = 11 needs Pesmisto 2 and the other three at 3. Day 4 = 10 needs Pesmisto 1 and the others at 3. Day 2 = 8 with Kitmisto 3 gives Levmisto = Tyhrmisto = 2 and Pesmisto 1.
  4. Day 1 = 5 with everyone at least 1 forces Kitmisto 2, Levmisto = Tyhrmisto = Pesmisto = 1. So Pesmisto has 0 on Day 3 and Kitmisto has 3, and the totals of 12 give Levmisto = Tyhrmisto = 3 on Day 3 (Day 3 total \(3+3+0+3=9\)).
  5. Pesmisto had no case only on Day 3.

Question 19

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
    Statement A only
  4. D
    Neither Statement A nor Statement B

Answer: (B) Both Statement A and Statement B Grids & Matrices MCQ

Solution

  1. Pesmisto total 5 with a single 2: one 2, three 1s, one 0. Kitmisto total 14: four 3s and one 2.
  2. Daily totals rise strictly, sum to 43, with Day 3 = Day 2 + 1. Pesmisto is at most 2, so a day's total is at most 11. Day 1 is at least 4, and Day 2 is at most 9 (only Kitmisto has 3). The only sequence that works is 5, 8, 9, 10, 11.
  3. Day 5 = 11 needs Pesmisto 2 and the other three at 3. Day 4 = 10 needs Pesmisto 1 and the others at 3. Day 2 = 8 with Kitmisto 3 gives Levmisto = Tyhrmisto = 2 and Pesmisto 1.
  4. Day 1 = 5 with everyone at least 1 forces Kitmisto 2, Levmisto = Tyhrmisto = Pesmisto = 1. So Pesmisto has 0 on Day 3 and Kitmisto has 3, and the totals of 12 give Levmisto = Tyhrmisto = 3 on Day 3 (Day 3 total \(3+3+0+3=9\)).
  5. Tyhrmisto had 3 on Day 3 (not 2) and Pesmisto had 1 on Day 2 (not 0), so both statements are false.

Question 20

On how many days did Levmisto and Tyhrmisto have the same number of new cases?
  1. A
    4
  2. B
    5
  3. C
    3
  4. D
    2

Answer: (B) 5 Grids & Matrices MCQ

Solution

  1. Pesmisto total 5 with a single 2: one 2, three 1s, one 0. Kitmisto total 14: four 3s and one 2.
  2. Daily totals rise strictly, sum to 43, with Day 3 = Day 2 + 1. Pesmisto is at most 2, so a day's total is at most 11. Day 1 is at least 4, and Day 2 is at most 9 (only Kitmisto has 3). The only sequence that works is 5, 8, 9, 10, 11.
  3. Day 5 = 11 needs Pesmisto 2 and the other three at 3. Day 4 = 10 needs Pesmisto 1 and the others at 3. Day 2 = 8 with Kitmisto 3 gives Levmisto = Tyhrmisto = 2 and Pesmisto 1.
  4. Day 1 = 5 with everyone at least 1 forces Kitmisto 2, Levmisto = Tyhrmisto = Pesmisto = 1. So Pesmisto has 0 on Day 3 and Kitmisto has 3, and the totals of 12 give Levmisto = Tyhrmisto = 3 on Day 3 (Day 3 total \(3+3+0+3=9\)).
  5. Levmisto and Tyhrmisto: (1,1), (2,2), (3,3), (3,3), (3,3). They are equal on all 5 days.

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