CAT 2022 Slot 3 — DILR Question 6
Answer the next 5 questions based on the information given below:
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.
What BEST can be concluded about the total number of new cases in the city on Day 2?
Answer & solution
- A
Exactly 7
- B
Either 7 or 8
- C
Either 6 or 7
Exactly 8
We can make the following table:
1. There was at least one new case in every neighborhood 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.
5. Kitmisto is the only place to have 3 new cases on Day 2.
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From (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.
Total cases during the 5 days = 12 + 12 + 5 + 14 = 43
On day 5, neighborhoods L, T and K can have maximum 3 cases each, while P can have maximum 2 cases. Hence, maximum possible total cases on day 5 = 3 + 2 + 3 + 3 = 11.
⇒ Maximum possible cases on day 4 = 10.
∴ Maximum possible cases on day 4 + day 5 = 10 + 11 = 21.
Since on day 1 there will be at least 5 cases in the city and every day total cases increase, hence on day 2 total cases must be 6 or more.
Case 1: Total cases on Day 2 = 6
⇒ Total cases on Day 3 = 7
⇒ Total cases on Day 1 = 5
∴ Total cases on day 4 + day 5 = 43 – (5 + 6 + 7) = 25
This is not possible as maximum cases possible on day 4 + day 5 is 21.
Case 2: Total cases on Day 2 = 7
⇒ Total cases on Day 3 = 8
⇒ Total cases on Day 1 = 5 or 6
∴ Total cases on day 4 + day 5 = 43 – (5/6 + 7 + 8) = 23 or 22
This is not possible as maximum cases possible on day 4 + day 5 is 21.
Case 3: Total cases on Day 2 = 8
⇒ Total cases on Day 3 = 9
Since total cases on day 4 cannot be more than 10, hence total cases on day 4 = 10.
Since total cases on day 5 cannot be more than 11, hence total cases on day 5 = 11.
⇒ Total cases on Day 1 = 43 – (8 + 9 + 10 + 11) = 5
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Only possibility for total cases of 11 on day 5 is: L – 3, T – 3, P – 2, K – 3.
Only possibility for total cases of 11 on day 5 is: L – 3, T – 3, P – 1, K – 3.
[Note: There is only 1 day when Pesmisto had 2 cases. On remaining days P will have less than 2 cases.]
Total cases in Kitmisto is 14. This is possible when there are 3 cases on 4 days each and 2 cases on the remaining 5th day.
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Levmisto and Tyhrmisto have total 12 cases.
Total cases on day 2 + day 3 for these two neighborhoods = 12 – (1 + 3 + 3) = 5
5 cases in 2 days are possible when there are 2 and 3 cases in these 2 days.
On day 2 only Kitmisto had 3 cases, hence we can fill the table accordingly.
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Now the remaining 2 slots can be filled.
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∴ The total number of new cases in the city on Day 2 = 8
Hence, option (d).