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CAT 2022 Slot 3 Paper

Instructions [30 - 34 ]
There are only four neighborhoods in a city - Levmisto, Tyhrmisto, Pesmisto and Kitmisto. During the onset of a pandemic, the number ofnew cases of a disease in each of these neighborhoods was recorded over a period of five days. On each day, the number of new casesrecorded in any of the neighborhoods was either 0, 1, 2 or 3.
The following facts are also known:
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 exactlyone more than that on Day 2.
4.The maximum number of new cases in a day in Pesmisto was 2, and this happened onlyonce 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 14respectively.

Question : 32

Total: 66

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

Only Day 3
Only Day 2
Both Day 2 and Day 3
Both Day 2 and Day 4

Validate

Solution:

From statement 6 , it can be concluded that the total number of new cases is equal to 12+12+5+14=43.
Also, since the total number of cases in Kitmisto is 14 , it can be concluded that the number of cases each day is either 2 or 3 , where 3 cases are recorded on 4 days and 2 cases are recorded on 1 day.
From statement 4 , it can be concluded that the number of new cases for Pesmisto will be 0,1,1,1, and 2 , in any order(Since the total number of cases is 5 , and the maximum number of new cases is 2 ).
In statement 3 , it is given that the number of new cases kept increasing during the 5 -day period.
Now, as it is already known that the maximum number of cases for Pesmisto is 2, the maximum total number of cases in a day(or on day 5) will be less than 12.
Let us consider the maximum number of cases on Day 5 as 10 .
Thus the maximum number of cases possible for the remaining days will be 9,8,7, and 6 . So, the total number of maximum cases possible for this case will be 40 (less than 43 ).
Thus, the number of cases on Day 5 will be 11(i.e., b/w 10 and 11)
Now, if the number of cases on Day 4 is 9 , the maximum number of cases possible for the remaining days will be 8,7, and 6.
Thus, the maximum number of cases, in this case, will be 41 (less than 43).
So, the number of cases on day 4 will be 10 .
Now, if the number of cases on Day 3 is 8 , the number of cases on day 2 will be 7 , and the maximum possible number of cases on Day 1 will be 6 .
Thus, the number of cases, in this case, will be 42(less than 43 ).
Thus, the number of cases on day 3 will be 9 , the number of cases on day 2 will be 8 , and the number of cases on day 1 will be 5 .

Neighbourhood	Day 1	Day 2	Day 3	Day 4	Day 5	Total
Levmisto						12
Tyhrmisto						12
Pesmisto						5
Kitmisto						14
Total	5	8	9	10	11	43

Since all the neighborhood has at least one case on Day 1 , the only possible combination will be 1,1,1, and 2 for Levmisto, Tyhrmisto, Pesmisto and Kitmisto, respectively.
Now, for the other 4 days, the number of cases in Kitmisto will be 3 .
For day 5 , the number of cases will be 3,3,2, and 3 for Levmisto, Tyhrmisto, Pesmisto and Kitmisto, respectively(since the maximum number of cases in Pesmisto is 2 ).
And since Pesmisto only has the maximum number of cases on one day, the number of cases on day 4 will be 3 , 3,1 , and 3 for Levmisto, Tyhrmisto, Pesmisto and Kitmisto, respectively.
On day 2 , since Kitmisto is the only neighbourhood to have 3 cases, the number of cases on day 2 will be 2, 2, 1, and 3 for Levmisto, Tyhrmisto, Pesmisto and Kitmisto, respectively.
Now, on day 3 , the number of cases will be 3,3,0, and 3 for Levmisto, Tyhrmisto, Pesmisto and Kitmisto, respectively.
Thus, the final table will be as follows:

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
Total	5	8	9	10	11	43

From the final table, it can be concluded that on Day 3 , the number of cases will be zero for Pesmisto.
Thus, the correct option is A.

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