Question

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

• A. Exactly 2
• B. Either 2 or 3
• C. Either 0 or 1
• D. Exactly 3
• A. Both Day 2 and Day 3
• B. Only Day 3
• C. Both Day 2 and Day 4
• D. Only Day 2
• A. Neither Statement A nor Statement B
• B. Statement B only
• C. Both Statement A and Statement B
• D. Statement A only
• A. 4
• B. 5
• C. 2
• D. 3
• A. Either 7 or 8
• B. Either 6 or 7
• C. Exactly 7
• D. Exactly 8

Answer 1

The Correct Option is D

To determine the number of new cases in Levmisto on Day 3, we'll use the data given in the comprehension:

• On Day 1, each neighborhood had at least one new case, and the total must be such that it allows sequential increases over five days.
• On Day 2, Kitmisto has 3 new cases. Since Kitmisto always has more new cases than Pesmisto, and the sum for Day 2 must be less than Day 3, let us explore the possibilities logically considering all constraints:

We are given that Levmisto had a total of 12 cases over five days. The number of new cases on each day could initially be:

• Day 1: Can be set to have moderate cases like 2 (ensuring gradual increase over days).
• Day 2: Incremental but still low enough for subsequent days to increase.

Let's distribute based on total increasing daily totals and satisfying all constraints:

• Assume Day 1 total is 6 (a reasonably possible minimum to start from given constraints and total ongoing increase).
• Day 2 total, let's try 7, given that adding one more on Day 3 compared to Day 2 is feasible, making Day 3 sum 8.
• Ensuring Levmisto fits total cases, Kitmisto always has more cases than Pesmisto (especially on Day 2 with Pesmisto having 1 in place and Kitmisto already given 3).

Assign options:

Day	Levmisto	Tyhrmisto	Pesmisto	Kitmisto
Day 1	2	1	1	2
Day 2	2	0	1	3
Total on Day 3	Cannot be less than 13 from Day 1 increasing assumption overall (needs trial of options)

• Following constraints, Levmisto hits 3 cases exactly, fits increase total needing balance of days as logical sums. Remaining calculation yields:
• The only feasible case for Day 3 supports the exact count matching the expected overall number.

Conclusion: Levmisto had Exactly 3 new cases on Day 3, logically fitting all given conditions.

Answer 2

Step 1: Analyze the given conditions

• Kitmisto has more new cases than Pesmisto on every day.
• Pesmisto had a total of 5 cases over 5 days.
• Pesmisto had at most 2 cases on any day, and that only once.
• Kitmisto had exactly 3 cases on Day 2.
• Day 3's total new cases = Day 2's total + 1 (i.e., a strict increase).

Step 2: Understand Pesmisto's possible case distribution

Since Pesmisto must have fewer cases than Kitmisto on every day, and Kitmisto had 3 cases on Day 2, the maximum Pesmisto can have on Day 2 is:

\[ \text{Pesmisto on Day 2} \leq 2 \]

And if Pesmisto had 2 cases on Day 2, then it cannot have 2 on any other day due to the one-time maximum rule.

Step 3: Consider the total and Day 3 condition

The total cases in Pesmisto over 5 days is 5. Suppose Pesmisto had 2 cases on Day 2. That leaves:

\[ 5 - 2 = 3 \text{ cases to distribute across 4 other days} \]

To maintain increasing total cases from Day 2 to Day 3, the other neighborhoods (Levmisto, Tyhrmisto, Kitmisto) must account for that 1 case difference. Therefore, it's most efficient if Pesmisto had 0 cases on Day 3.

Step 4: Confirm feasibility

Let’s try a sample distribution:

• Day 1: Pesmisto = 1
• Day 2: Pesmisto = 2
• Day 3: Pesmisto = 0
• Day 4: Pesmisto = 1
• Day 5: Pesmisto = 1

This satisfies all constraints:

• Total = 1 + 2 + 0 + 1 + 1 = 5
• Only one day with 2 cases (Day 2)
• Kitmisto can remain greater than Pesmisto on each day
• Day 3 allows for +1 increase if other neighborhoods increase

✅ Therefore, the only valid day on which Pesmisto had no new cases is:
Day 3

Answer 3

Given Facts

• Every neighborhood had at least one new case on Day 1.
• Kitmisto > Pesmisto for new cases on every day.
• Total new cases on Day 3 = Day 2 total + 1.
• Pesmisto never had more than 2 cases in a day.
• Kitmisto had exactly 3 new cases on Day 2.
• Total new cases over 5 days:
  • Levmisto = 12
  • Tyhrmisto = 12
  • Pesmisto = 5
  • Kitmisto = 14

Statement A:

"Tyhrmisto had 2 new cases on Day 3."

Let’s assume this is true and examine feasibility:

• Day 2 total cases = \( x \)
• Day 3 total cases = \( x + 1 \)
• If Tyhrmisto = 2 on Day 3, the sum of the other three neighborhoods must match rest of the total.
• Given the tight constraints on Pesmisto and fixed values on Kitmisto (3 on Day 2), this leaves minimal room for combinations that maintain the +1 increase.

 

Conclusion: Statement A is necessarily false.

Statement B:

"Pesmisto had 0 new cases on Day 2."

But Kitmisto had 3 cases on Day 2 (Fact 5), and Fact 2 requires Kitmisto > Pesmisto on every day.
Therefore, Pesmisto must have had at least 1 case on Day 2 to satisfy: \[ \text{Kitmisto} = 3 > \text{Pesmisto} \]

Conclusion: Statement B is necessarily false.

Final Verdict

Both Statement A and Statement B are necessarily false.
The constraints are too strict to allow either condition to be satisfied while maintaining all known data.

Answer 4

To determine the number of days Levmisto and Tyhrmisto had the same number of new cases, we need to deduce the distribution of cases using the constraints provided. 

• Day 1: Since there is at least one case in every neighborhood, let's assume minimal values for total increment on ensuing days.
  We need permutation for initial assignments considering the total daily increment.
• Day 2 to Day 5: Using the fact that Kitmisto had more cases than Pesmisto and had exactly 3 new cases on Day 2, we can assign values incrementally to meet the total sum of new cases while adhering to daily city increases.
• We analyze how the sum of cases evolves, respecting conditions for Pesmisto's max value and other neighborhoods reaching their totals.
• Day-by-Day Assignment: Use backward induction to test each number assignment for congruence in totals. Step through permutations until Levmisto = Tyhrmisto aligns five times across the days.

Day	Levmisto	Tyhrmisto	Pesmisto	Kitmisto	Total
1	2	1	1	2	6
2	2	2	0	3	7
3	3	3	1	3	10
4	3	3	1	3	10
5	2	3	2	3	10

• Answer: Levmisto and Tyhrmisto had the same number of new cases on:
• Days 2, 3, 4, 5 (as calculated from projected values).

Result: They had the same number of new cases on 5 days.

Answer 5

To determine the total number of new cases in the city on Day 2, we analyze the given constraints and logic:

1. Kitmisto had exactly 3 cases on Day 2.
2. It is stated that Kitmisto always has more cases than Pesmisto on any day.
3. Therefore, Pesmisto must have had 0, 1, or 2 cases on Day 2.
4. Given that Pesmisto can have 2 cases at most and that only once across 5 days, it is valid to assume:
   • Pesmisto = 2 on Day 2.
5. So far, total = Kitmisto (3) + Pesmisto (2) = 5 cases.
6. To preserve the increasing trend, and knowing that Day 3 must have one more case than Day 2, we need:
   • Day 2 total = 8
   • Day 3 total = 9 (as required by the increase condition)
7. Now, the remaining number of cases for Day 2 must come from Levmisto and Tyhrmisto:
   • Remaining cases = \( 8 - 5 = 3 \)
   • Possible distributions: Levmisto = 1, Tyhrmisto = 2 or Levmisto = 2, Tyhrmisto = 1
   • Both values lie within typical allowed day-wise limits (e.g., 0, 1, 2)
8. Total cases across 5 days = 12 + 12 + 5 + 14 = 43 (given)

✅ Therefore, the total number of new cases on Day 2 is: 8 cases.