CAT 2026 Slot 3 DILR: Set 2: DILR – District-wise Pandemic Cases with Solutions

4 questions: Q5, Q6, Q7, Q8. Solve each one here: tap the answer, press Check, then open the solution.

Logical Reasoning

The set

There are only four districts in a city — Alden, Brim, Cova and Dune. During the onset of a pandemic, the number of new cases of a disease in each of these districts was recorded over a period of five days. On each day, the number of new cases recorded in any of the districts was either 0, 1, 2 or 3.

The following facts are also known:
1. There was at least one new case in every district on Day 1.
2. On each of the five days, there were more new cases in Dune than in Cova.
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 Cova was 2, and this happened only once during the five-day period.
5. Dune is the only place to have 3 new cases on Day 2.
6. The total numbers of new cases in Alden, Brim, Cova and Dune over the five-day period were 12, 12, 5 and 14 respectively.

Question 5Integer distribution

MCQ · +3 / −1 · medium

On how many days did Alden and Brim have the same number of new cases?

Answer and solution

Correct answer: A

Solution

Answer: A — 5

1. The daily city total is at least 5 on Day 1: each district has at least 1 and Dune must exceed Cova. Any daily total is at most 11: three districts can have 3 each and Cova at most 2.

2. The five strictly increasing integer city totals sum to 12+12+5+14=43. With Day 3=Day 2+1, the only possible sequence between 5 and 11 is 5,8,9,10,11. This can be checked by listing increasing five-number subsets of 5,…,11.

3. Day 1 total 5 forces Alden=Brim=Cova=1 and Dune=2. Dune total 14 then forces 3 on each later day.

4. Alden has 11 remaining cases over four days. On Day 2 it can have at most 2, because only Dune can have 3 then. The maximum possible remaining sum is 2+3+3+3=11, so all these maxima are forced. The same argument applies to Brim.

5. Subtract Alden, Brim and Dune from daily totals to obtain Cova=1,1,0,1,2. This has one maximum of 2, total 5, and is below Dune on every day.

6. Final grid (Alden, Brim, Cova, Dune): Day 1 (1,1,1,2); Day 2 (2,2,1,3); Day 3 (3,3,0,3); Day 4 (3,3,1,3); Day 5 (3,3,2,3). Independent exhaustive enumeration of all 0–3 district sequences gives exactly one valid grid.

7. Alden and Brim match on all five days. Answer A (5).

G Strategy shortcut

Method: Use total 43 and daily bounds 5–11 to force 5,8,9,10,11. Day 1 is 1,1,1,2; Alden/Brim each need 11 more, forcing 2,3,3,3. Obtain Cova by subtraction.
Verification: 7. Alden and Brim match on all five days. Answer A (5).

G Strategy: Method: Use total 43 and daily bounds 5–11 to force 5,8,9,10,11. Day 1 is 1,1,1,2; Alden/Brim each need 11 more, forcing 2,3,3,3. Obtain Cova by subtraction. Verification: 7. Alden and Brim match on all five days. Answer A (5).

Question 6Integer distribution

MCQ · +3 / −1 · medium

Which of the two statements below is/are necessarily false?
Statement A: There were 2 new cases in Brim on Day 3.
Statement B: There were no new cases in Cova on Day 2.

Answer and solution

Correct answer: D

Solution

Answer: D — Both Statement A and Statement B

1. The daily city total is at least 5 on Day 1: each district has at least 1 and Dune must exceed Cova. Any daily total is at most 11: three districts can have 3 each and Cova at most 2.

2. The five strictly increasing integer city totals sum to 12+12+5+14=43. With Day 3=Day 2+1, the only possible sequence between 5 and 11 is 5,8,9,10,11. This can be checked by listing increasing five-number subsets of 5,…,11.

3. Day 1 total 5 forces Alden=Brim=Cova=1 and Dune=2. Dune total 14 then forces 3 on each later day.

4. Alden has 11 remaining cases over four days. On Day 2 it can have at most 2, because only Dune can have 3 then. The maximum possible remaining sum is 2+3+3+3=11, so all these maxima are forced. The same argument applies to Brim.

5. Subtract Alden, Brim and Dune from daily totals to obtain Cova=1,1,0,1,2. This has one maximum of 2, total 5, and is below Dune on every day.

6. Final grid (Alden, Brim, Cova, Dune): Day 1 (1,1,1,2); Day 2 (2,2,1,3); Day 3 (3,3,0,3); Day 4 (3,3,1,3); Day 5 (3,3,2,3). Independent exhaustive enumeration of all 0–3 district sequences gives exactly one valid grid.

7. Brim has 3 on Day 3 and Cova has 1 on Day 2. Both proposed statements are false. Answer D.

G Strategy shortcut

Method: Use total 43 and daily bounds 5–11 to force 5,8,9,10,11. Day 1 is 1,1,1,2; Alden/Brim each need 11 more, forcing 2,3,3,3. Obtain Cova by subtraction.
Verification: 7. Brim has 3 on Day 3 and Cova has 1 on Day 2. Both proposed statements are false. Answer D.

G Strategy: Method: Use total 43 and daily bounds 5–11 to force 5,8,9,10,11. Day 1 is 1,1,1,2; Alden/Brim each need 11 more, forcing 2,3,3,3. Obtain Cova by subtraction. Verification: 7. Brim has 3 on Day 3 and Cova has 1 on Day 2. Both proposed statements are false. Answer D.

Question 7Integer distribution

MCQ · +3 / −1 · medium

On which day(s) did Cova not have any new case?

Answer and solution

Correct answer: B

Solution

Answer: B — Only Day 3

1. The daily city total is at least 5 on Day 1: each district has at least 1 and Dune must exceed Cova. Any daily total is at most 11: three districts can have 3 each and Cova at most 2.

2. The five strictly increasing integer city totals sum to 12+12+5+14=43. With Day 3=Day 2+1, the only possible sequence between 5 and 11 is 5,8,9,10,11. This can be checked by listing increasing five-number subsets of 5,…,11.

3. Day 1 total 5 forces Alden=Brim=Cova=1 and Dune=2. Dune total 14 then forces 3 on each later day.

4. Alden has 11 remaining cases over four days. On Day 2 it can have at most 2, because only Dune can have 3 then. The maximum possible remaining sum is 2+3+3+3=11, so all these maxima are forced. The same argument applies to Brim.

5. Subtract Alden, Brim and Dune from daily totals to obtain Cova=1,1,0,1,2. This has one maximum of 2, total 5, and is below Dune on every day.

6. Final grid (Alden, Brim, Cova, Dune): Day 1 (1,1,1,2); Day 2 (2,2,1,3); Day 3 (3,3,0,3); Day 4 (3,3,1,3); Day 5 (3,3,2,3). Independent exhaustive enumeration of all 0–3 district sequences gives exactly one valid grid.

7. Cova has zero only on Day 3. Answer B.

G Strategy shortcut

Method: Use total 43 and daily bounds 5–11 to force 5,8,9,10,11. Day 1 is 1,1,1,2; Alden/Brim each need 11 more, forcing 2,3,3,3. Obtain Cova by subtraction.
Verification: 7. Cova has zero only on Day 3. Answer B.

Question 8Integer distribution

MCQ · +3 / −1 · medium

What can be concluded about the number of new cases in Alden on Day 3?

Answer and solution

Correct answer: C

Solution

Answer: C — Exactly 3

1. The daily city total is at least 5 on Day 1: each district has at least 1 and Dune must exceed Cova. Any daily total is at most 11: three districts can have 3 each and Cova at most 2.

2. The five strictly increasing integer city totals sum to 12+12+5+14=43. With Day 3=Day 2+1, the only possible sequence between 5 and 11 is 5,8,9,10,11. This can be checked by listing increasing five-number subsets of 5,…,11.

3. Day 1 total 5 forces Alden=Brim=Cova=1 and Dune=2. Dune total 14 then forces 3 on each later day.

4. Alden has 11 remaining cases over four days. On Day 2 it can have at most 2, because only Dune can have 3 then. The maximum possible remaining sum is 2+3+3+3=11, so all these maxima are forced. The same argument applies to Brim.

5. Subtract Alden, Brim and Dune from daily totals to obtain Cova=1,1,0,1,2. This has one maximum of 2, total 5, and is below Dune on every day.

6. Final grid (Alden, Brim, Cova, Dune): Day 1 (1,1,1,2); Day 2 (2,2,1,3); Day 3 (3,3,0,3); Day 4 (3,3,1,3); Day 5 (3,3,2,3). Independent exhaustive enumeration of all 0–3 district sequences gives exactly one valid grid.

7. Alden has exactly 3 on Day 3. Answer C.

G Strategy shortcut

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