CAT 2022 Slot 3 · DILR · Set 2: New Cases in Four Neighbourhoods

The full set and all 5 questions (Q6–Q10) from CAT 2022 Slot 3, with answers and solutions.

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 6 · Caselet DI

What BEST can be concluded about the total number of new cases in the city on Day 2?
  1. A
    Either 7 or 8
  2. B
    Exactly 7
  3. C
    Either 6 or 7
  4. D
    Exactly 8
Pick an option, then check.
Answer & solution

Answer: Option D

Solving the set

Step 1 – Kitmisto (total 14, at most 3 a day) has 3 on four days and 2 on one day; it has 3 on Day 2 (fact 5).

Step 2 – Pesmisto (total 5) has exactly one 2 and at most 1 on every other day (fact 4).

Step 3 – Largest possible daily totals. Pesmisto reaches 2 only once, so the most any day can have is 3 + 3 + 2 + 3 = 11. The city totals strictly increase, so Day 5 ≤ 11, Day 4 ≤ 10, Day 3 ≤ 9 and Day 2 = Day 3 − 1 ≤ 8.

Step 4 – The five daily totals add up to 12 + 12 + 5 + 14 = 43. Write them as d1, d2, d2 + 1, d4, d5: d1 + 2·d2 = 42 − d4 − d5 ≥ 42 − 21 = 21. If d2 ≤ 7, then d1 ≥ 7 ≥ d2, which breaks 'increasing'. So d2 = 8, and then d1 = 26 − d4 − d5 forces d4 = 10, d5 = 11, d1 = 5. Daily totals: 5, 8, 9, 10, 11.

Step 5 – Day by day.

• Day 5 = 11 is the maximum possible: Levmisto 3, Tyhrmisto 3, Pesmisto 2 (its only 2), Kitmisto 3.

• Day 1 = 5 with everyone ≥ 1 and Kitmisto > Pesmisto: Kitmisto 2, the others 1.

• Day 2 = 8 with Kitmisto 3 (only 3 that day): the other three share 5 with each ≤ 2 and Pesmisto ≤ 1 ⇒ Levmisto 2, Tyhrmisto 2, Pesmisto 1.

• Levmisto and Tyhrmisto need 12 − 1 − 2 − 3 = 6 more on Days 3 and 4 ⇒ 3 and 3 each. Kitmisto needs 14 − 2 − 3 − 3 = 6 ⇒ 3 and 3.

• Day 3 = 9 ⇒ Pesmisto 0; Day 4 = 10 ⇒ Pesmisto 1. Pesmisto's total 1 + 1 + 0 + 1 + 2 = 5 ✓.

Neighbourhood

Day 1

Day 2

Day 3

Day 4

Day 5

Total

Levmisto

1

2

3

3

3

12

Tyhrmisto

1

2

3

3

3

12

Pesmisto

1

1

0

1

2

5

Kitmisto

2

3

3

3

3

14

City

5

8

9

10

11

43

This question

From the table, Day 2 total = 8.

Hence, the answer is option 4.

Question 7 · Caselet DI

What BEST can be concluded about the number of new cases in Levmisto on Day 3?
  1. A
    Either 2 or 3
  2. B
    Exactly 2
  3. C
    Exactly 3
  4. D
    Either 0 or 1
Pick an option, then check.
Answer & solution

Answer: Option C

Solving the set

Step 1 – Kitmisto (total 14, at most 3 a day) has 3 on four days and 2 on one day; it has 3 on Day 2 (fact 5).

Step 2 – Pesmisto (total 5) has exactly one 2 and at most 1 on every other day (fact 4).

Step 3 – Largest possible daily totals. Pesmisto reaches 2 only once, so the most any day can have is 3 + 3 + 2 + 3 = 11. The city totals strictly increase, so Day 5 ≤ 11, Day 4 ≤ 10, Day 3 ≤ 9 and Day 2 = Day 3 − 1 ≤ 8.

Step 4 – The five daily totals add up to 12 + 12 + 5 + 14 = 43. Write them as d1, d2, d2 + 1, d4, d5: d1 + 2·d2 = 42 − d4 − d5 ≥ 42 − 21 = 21. If d2 ≤ 7, then d1 ≥ 7 ≥ d2, which breaks 'increasing'. So d2 = 8, and then d1 = 26 − d4 − d5 forces d4 = 10, d5 = 11, d1 = 5. Daily totals: 5, 8, 9, 10, 11.

Step 5 – Day by day.

• Day 5 = 11 is the maximum possible: Levmisto 3, Tyhrmisto 3, Pesmisto 2 (its only 2), Kitmisto 3.

• Day 1 = 5 with everyone ≥ 1 and Kitmisto > Pesmisto: Kitmisto 2, the others 1.

• Day 2 = 8 with Kitmisto 3 (only 3 that day): the other three share 5 with each ≤ 2 and Pesmisto ≤ 1 ⇒ Levmisto 2, Tyhrmisto 2, Pesmisto 1.

• Levmisto and Tyhrmisto need 12 − 1 − 2 − 3 = 6 more on Days 3 and 4 ⇒ 3 and 3 each. Kitmisto needs 14 − 2 − 3 − 3 = 6 ⇒ 3 and 3.

• Day 3 = 9 ⇒ Pesmisto 0; Day 4 = 10 ⇒ Pesmisto 1. Pesmisto's total 1 + 1 + 0 + 1 + 2 = 5 ✓.

Neighbourhood

Day 1

Day 2

Day 3

Day 4

Day 5

Total

Levmisto

1

2

3

3

3

12

Tyhrmisto

1

2

3

3

3

12

Pesmisto

1

1

0

1

2

5

Kitmisto

2

3

3

3

3

14

City

5

8

9

10

11

43

This question

From Step 5, Levmisto had 3 cases on Day 3.

Hence, the answer is option 3.

Question 8 · Caselet DI

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 3
  4. D
    Both Day 2 and Day 4
Pick an option, then check.
Answer & solution

Answer: Option A

Solving the set

Step 1 – Kitmisto (total 14, at most 3 a day) has 3 on four days and 2 on one day; it has 3 on Day 2 (fact 5).

Step 2 – Pesmisto (total 5) has exactly one 2 and at most 1 on every other day (fact 4).

Step 3 – Largest possible daily totals. Pesmisto reaches 2 only once, so the most any day can have is 3 + 3 + 2 + 3 = 11. The city totals strictly increase, so Day 5 ≤ 11, Day 4 ≤ 10, Day 3 ≤ 9 and Day 2 = Day 3 − 1 ≤ 8.

Step 4 – The five daily totals add up to 12 + 12 + 5 + 14 = 43. Write them as d1, d2, d2 + 1, d4, d5: d1 + 2·d2 = 42 − d4 − d5 ≥ 42 − 21 = 21. If d2 ≤ 7, then d1 ≥ 7 ≥ d2, which breaks 'increasing'. So d2 = 8, and then d1 = 26 − d4 − d5 forces d4 = 10, d5 = 11, d1 = 5. Daily totals: 5, 8, 9, 10, 11.

Step 5 – Day by day.

• Day 5 = 11 is the maximum possible: Levmisto 3, Tyhrmisto 3, Pesmisto 2 (its only 2), Kitmisto 3.

• Day 1 = 5 with everyone ≥ 1 and Kitmisto > Pesmisto: Kitmisto 2, the others 1.

• Day 2 = 8 with Kitmisto 3 (only 3 that day): the other three share 5 with each ≤ 2 and Pesmisto ≤ 1 ⇒ Levmisto 2, Tyhrmisto 2, Pesmisto 1.

• Levmisto and Tyhrmisto need 12 − 1 − 2 − 3 = 6 more on Days 3 and 4 ⇒ 3 and 3 each. Kitmisto needs 14 − 2 − 3 − 3 = 6 ⇒ 3 and 3.

• Day 3 = 9 ⇒ Pesmisto 0; Day 4 = 10 ⇒ Pesmisto 1. Pesmisto's total 1 + 1 + 0 + 1 + 2 = 5 ✓.

Neighbourhood

Day 1

Day 2

Day 3

Day 4

Day 5

Total

Levmisto

1

2

3

3

3

12

Tyhrmisto

1

2

3

3

3

12

Pesmisto

1

1

0

1

2

5

Kitmisto

2

3

3

3

3

14

City

5

8

9

10

11

43

This question

From Step 5, Pesmisto had 0 cases only on Day 3.

Hence, the answer is option 1.

Question 9 · Caselet DI

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 A only
  2. B
    Neither Statement A nor Statement B
  3. C
    Statement B only
  4. D
    Both Statement A and Statement B
Pick an option, then check.
Answer & solution

Answer: Option D

Solving the set

Step 1 – Kitmisto (total 14, at most 3 a day) has 3 on four days and 2 on one day; it has 3 on Day 2 (fact 5).

Step 2 – Pesmisto (total 5) has exactly one 2 and at most 1 on every other day (fact 4).

Step 3 – Largest possible daily totals. Pesmisto reaches 2 only once, so the most any day can have is 3 + 3 + 2 + 3 = 11. The city totals strictly increase, so Day 5 ≤ 11, Day 4 ≤ 10, Day 3 ≤ 9 and Day 2 = Day 3 − 1 ≤ 8.

Step 4 – The five daily totals add up to 12 + 12 + 5 + 14 = 43. Write them as d1, d2, d2 + 1, d4, d5: d1 + 2·d2 = 42 − d4 − d5 ≥ 42 − 21 = 21. If d2 ≤ 7, then d1 ≥ 7 ≥ d2, which breaks 'increasing'. So d2 = 8, and then d1 = 26 − d4 − d5 forces d4 = 10, d5 = 11, d1 = 5. Daily totals: 5, 8, 9, 10, 11.

Step 5 – Day by day.

• Day 5 = 11 is the maximum possible: Levmisto 3, Tyhrmisto 3, Pesmisto 2 (its only 2), Kitmisto 3.

• Day 1 = 5 with everyone ≥ 1 and Kitmisto > Pesmisto: Kitmisto 2, the others 1.

• Day 2 = 8 with Kitmisto 3 (only 3 that day): the other three share 5 with each ≤ 2 and Pesmisto ≤ 1 ⇒ Levmisto 2, Tyhrmisto 2, Pesmisto 1.

• Levmisto and Tyhrmisto need 12 − 1 − 2 − 3 = 6 more on Days 3 and 4 ⇒ 3 and 3 each. Kitmisto needs 14 − 2 − 3 − 3 = 6 ⇒ 3 and 3.

• Day 3 = 9 ⇒ Pesmisto 0; Day 4 = 10 ⇒ Pesmisto 1. Pesmisto's total 1 + 1 + 0 + 1 + 2 = 5 ✓.

Neighbourhood

Day 1

Day 2

Day 3

Day 4

Day 5

Total

Levmisto

1

2

3

3

3

12

Tyhrmisto

1

2

3

3

3

12

Pesmisto

1

1

0

1

2

5

Kitmisto

2

3

3

3

3

14

City

5

8

9

10

11

43

This question

Statement A: Tyhrmisto had 3 (not 2) on Day 3 – false. Statement B: Pesmisto had 1 (not 0) on Day 2 – false. Both are necessarily false.

Hence, the answer is option 4.

Question 10 · Caselet DI

On how many days did Levmisto and Tyhrmisto have the same number of new cases?
  1. A
    2
  2. B
    3
  3. C
    4
  4. D
    5
Pick an option, then check.
Answer & solution

Answer: Option D

Solving the set

Step 1 – Kitmisto (total 14, at most 3 a day) has 3 on four days and 2 on one day; it has 3 on Day 2 (fact 5).

Step 2 – Pesmisto (total 5) has exactly one 2 and at most 1 on every other day (fact 4).

Step 3 – Largest possible daily totals. Pesmisto reaches 2 only once, so the most any day can have is 3 + 3 + 2 + 3 = 11. The city totals strictly increase, so Day 5 ≤ 11, Day 4 ≤ 10, Day 3 ≤ 9 and Day 2 = Day 3 − 1 ≤ 8.

Step 4 – The five daily totals add up to 12 + 12 + 5 + 14 = 43. Write them as d1, d2, d2 + 1, d4, d5: d1 + 2·d2 = 42 − d4 − d5 ≥ 42 − 21 = 21. If d2 ≤ 7, then d1 ≥ 7 ≥ d2, which breaks 'increasing'. So d2 = 8, and then d1 = 26 − d4 − d5 forces d4 = 10, d5 = 11, d1 = 5. Daily totals: 5, 8, 9, 10, 11.

Step 5 – Day by day.

• Day 5 = 11 is the maximum possible: Levmisto 3, Tyhrmisto 3, Pesmisto 2 (its only 2), Kitmisto 3.

• Day 1 = 5 with everyone ≥ 1 and Kitmisto > Pesmisto: Kitmisto 2, the others 1.

• Day 2 = 8 with Kitmisto 3 (only 3 that day): the other three share 5 with each ≤ 2 and Pesmisto ≤ 1 ⇒ Levmisto 2, Tyhrmisto 2, Pesmisto 1.

• Levmisto and Tyhrmisto need 12 − 1 − 2 − 3 = 6 more on Days 3 and 4 ⇒ 3 and 3 each. Kitmisto needs 14 − 2 − 3 − 3 = 6 ⇒ 3 and 3.

• Day 3 = 9 ⇒ Pesmisto 0; Day 4 = 10 ⇒ Pesmisto 1. Pesmisto's total 1 + 1 + 0 + 1 + 2 = 5 ✓.

Neighbourhood

Day 1

Day 2

Day 3

Day 4

Day 5

Total

Levmisto

1

2

3

3

3

12

Tyhrmisto

1

2

3

3

3

12

Pesmisto

1

1

0

1

2

5

Kitmisto

2

3

3

3

3

14

City

5

8

9

10

11

43

This question

Levmisto and Tyhrmisto both had 1, 2, 3, 3, 3 – equal on all five days.

Hence, the answer is option 4.

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