CAT 2022 Slot 3 · DILR · Question 6
CAT 2022 Slot 3 previous year question on Caselet DI, with the answer and a step-by-step solution.
Set 2: New Cases in Four Neighbourhoods — this question belongs to a set. Read the full set with all 5 questions →
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.
- AEither 7 or 8
- BExactly 7
- CEither 6 or 7
- DExactly 8
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.
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