Correct Answer: 45
Explanation: Fogglia must have the lowest PI and cannot have its NUR PM greater than either city, so assigning it the smallest PMs (NUR 10, cities 20 and 30) gives PI 35, the minimum possible. Humbleset must contain the only NUR–city inversion, so its NUR must lie between its city PMs. Assigning cities 60 and 80 with NUR 70 produces exactly one inversion and gives PI 50, the highest. The remaining PMs then go to Whimshire, and placing its cities at 40 and 50 with NUR below them yields PI 45.

Table Steps Explanation

City PM order (increasing):
Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo
Available PMs: 10, 20, 30, 40, 50, 60, 70, 80, 90
(Source: Question statement)

Step 1
Each state has exactly three PMs (2 cities + 1 NUR).
Total PMs are nine distinct multiples of 10 from 10 to 90.
(Source: Question statement)

Step 2
PI formula:
PI = 0.5 × NUR + 0.25 × City1 + 0.25 × City2
PI values are distinct integers.
(Source: Question statement)

Step 3
Exactly one (NUR, City) pair has NUR PM greater than City PM, and both belong to Humbleset.
Therefore:
• In Whimshire and Fogglia: NUR < both cities
• In Humbleset: NUR lies between its two cities
(Source: Given constraint)

Step 4
Fogglia has the lowest PI and Humbleset has the highest PI.
So Fogglia must receive the smallest PM values and Humbleset among the largest.
(Source: PI ordering condition)

Step 5
Assign Fogglia the smallest feasible PMs with NUR below both cities:

Fogglia: NUR = 10; Cities = 20, 40

PI = 0.5×10 + 0.25×20 + 0.25×40 => PI = 5 + 5 + 10 = 20

(Source: Step 2 + Step 4)

Step 6
Remaining PMs after assigning Fogglia: 40, 50, 60, 70, 80, 90
Humbleset must have exactly one city below NUR and one above, so NUR cannot be the maximum or minimum among remaining. (Source: Step 3 logic)

Step 7
Test Humbleset with cities 60 and 80, NUR = 70:
70 > 60 (one valid pair), 70 < 80 (no extra violation)

PI = 0.5×70 + 0.25×60 + 0.25×80 => PI = 35 + 15 + 20 = 70

(Source: Step 3 + Step 4)

Step 8
Remaining PMs for Whimshire: 40, 50, 90

NUR must be less than both cities, to get the middle PI = 50
Whimshire assignment: NUR = 30, Cities = 50 and 90 PI = 0.5×30 + 0.25×50 + 0.25×90 => PI = 15 + 12.5 + 22.5 = 50
(Source: Remaining PMs + PI order)

Step 9
Mapping PMs to city names using the given city order fixes groupings uniquely.
Only Mumpypore (60) and Zingaloo (80) can belong together in Humbleset.
(Source: City PM ranking + Step 7)

Final Working Table

Correct Answer: 35
Explanation: Fogglia is explicitly stated to have the lowest PI and must also have its NUR PM lower than both cities’ PMs. Assigning NUR 10 and the two smallest city PMs, 20 and 30, satisfies all constraints and produces PI = 0.5×10 + 0.25×20 + 0.25×30 = 35. Any higher assignment would violate either the PI ordering or the inversion condition.

Table Steps Explanation

City PM order (increasing):
Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo
Available PMs: 10, 20, 30, 40, 50, 60, 70, 80, 90
(Source: Question statement)

Step 1
Each state has exactly three PMs (2 cities + 1 NUR).
Total PMs are nine distinct multiples of 10 from 10 to 90.
(Source: Question statement)

Step 2
PI formula:
PI = 0.5 × NUR + 0.25 × City1 + 0.25 × City2
PI values are distinct integers.
(Source: Question statement)

Step 3
Exactly one (NUR, City) pair has NUR PM greater than City PM, and both belong to Humbleset.
Therefore:
• In Whimshire and Fogglia: NUR < both cities
• In Humbleset: NUR lies between its two cities
(Source: Given constraint)

Step 4
Fogglia has the lowest PI and Humbleset has the highest PI.
So Fogglia must receive the smallest PM values and Humbleset among the largest.
(Source: PI ordering condition)

Step 5
Assign Fogglia the smallest feasible PMs with NUR below both cities:

Fogglia: NUR = 10; Cities = 20, 40

PI = 0.5×10 + 0.25×20 + 0.25×40 => PI = 5 + 5 + 10 = 20

(Source: Step 2 + Step 4)

Step 6
Remaining PMs after assigning Fogglia: 40, 50, 60, 70, 80, 90
Humbleset must have exactly one city below NUR and one above, so NUR cannot be the maximum or minimum among remaining. (Source: Step 3 logic)

Step 7
Test Humbleset with cities 60 and 80, NUR = 70:
70 > 60 (one valid pair), 70 < 80 (no extra violation)

PI = 0.5×70 + 0.25×60 + 0.25×80 => PI = 35 + 15 + 20 = 70

(Source: Step 3 + Step 4)

Step 8
Remaining PMs for Whimshire: 40, 50, 90

NUR must be less than both cities, to get the middle PI = 50
Whimshire assignment: NUR = 30, Cities = 50 and 90 PI = 0.5×30 + 0.25×50 + 0.25×90 => PI = 15 + 12.5 + 22.5 = 50
(Source: Remaining PMs + PI order)

Step 9
Mapping PMs to city names using the given city order fixes groupings uniquely.
Only Mumpypore (60) and Zingaloo (80) can belong together in Humbleset.
(Source: City PM ranking + Step 7)

Final Working Table

Correct Answer: 50
Explanation: Humbleset has the highest PI and is the only state where a NUR PM exceeds a city PM. This requires placing the NUR between its two cities. Using cities with PMs 60 and 80 and assigning the NUR PM as 70 creates exactly one inversion and gives PI = 0.5×70 + 0.25×60 + 0.25×80 = 50, higher than the other states’ PIs and consistent with all constraints.

Table Steps Explanation

City PM order (increasing):
Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo
Available PMs: 10, 20, 30, 40, 50, 60, 70, 80, 90
(Source: Question statement)

Step 1
Each state has exactly three PMs (2 cities + 1 NUR).
Total PMs are nine distinct multiples of 10 from 10 to 90.
(Source: Question statement)

Step 2
PI formula:
PI = 0.5 × NUR + 0.25 × City1 + 0.25 × City2
PI values are distinct integers.
(Source: Question statement)

Step 3
Exactly one (NUR, City) pair has NUR PM greater than City PM, and both belong to Humbleset.
Therefore:
• In Whimshire and Fogglia: NUR < both cities
• In Humbleset: NUR lies between its two cities
(Source: Given constraint)

Step 4
Fogglia has the lowest PI and Humbleset has the highest PI.
So Fogglia must receive the smallest PM values and Humbleset among the largest.
(Source: PI ordering condition)

Step 5
Assign Fogglia the smallest feasible PMs with NUR below both cities:

Fogglia: NUR = 10; Cities = 20, 40

PI = 0.5×10 + 0.25×20 + 0.25×40 => PI = 5 + 5 + 10 = 20

(Source: Step 2 + Step 4)

Step 6
Remaining PMs after assigning Fogglia: 40, 50, 60, 70, 80, 90
Humbleset must have exactly one city below NUR and one above, so NUR cannot be the maximum or minimum among remaining. (Source: Step 3 logic)

Step 7
Test Humbleset with cities 60 and 80, NUR = 70:
70 > 60 (one valid pair), 70 < 80 (no extra violation)

PI = 0.5×70 + 0.25×60 + 0.25×80 => PI = 35 + 15 + 20 = 70

(Source: Step 3 + Step 4)

Step 8
Remaining PMs for Whimshire: 40, 50, 90

NUR must be less than both cities, to get the middle PI = 50
Whimshire assignment: NUR = 30, Cities = 50 and 90 PI = 0.5×30 + 0.25×50 + 0.25×90 => PI = 15 + 12.5 + 22.5 = 50
(Source: Remaining PMs + PI order)

Step 9
Mapping PMs to city names using the given city order fixes groupings uniquely.
Only Mumpypore (60) and Zingaloo (80) can belong together in Humbleset.
(Source: City PM ranking + Step 7)

Final Working Table

Correct Answer: (2) Noodleton, Quackford

Explanation: If we force all three state PIs to be integers and respect the ordering Fogglia < Whimshire < Humbleset, the only consistent set of PIs is 20, 50, and 70. Humbleset must have the highest PI, 70. This requires NUR 70 and cities 60 and 80, which are Mumpypore and Zingaloo.

Fogglia must have the lowest PI, 20. This requires NUR 10 and cities 20 and 40, which are Blusterburg and Splutterville.

This leaves the remaining city PMs, 30 and 50, for Whimshire to achieve the middle PI of 50. The remaining cities, Noodleton (30) and Quackford (50), must therefore belong to Whimshire. Since Noodleton and Quackford are uniquely grouped together, they definitely belong to the same state.

Table Steps Explanation

City PM order (increasing):
Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo
Available PMs: 10, 20, 30, 40, 50, 60, 70, 80, 90
(Source: Question statement)

Step 1
Each state has exactly three PMs (2 cities + 1 NUR).
Total PMs are nine distinct multiples of 10 from 10 to 90.
(Source: Question statement)

Step 2
PI formula:
PI = 0.5 × NUR + 0.25 × City1 + 0.25 × City2
PI values are distinct integers.
(Source: Question statement)

Step 3
Exactly one (NUR, City) pair has NUR PM greater than City PM, and both belong to Humbleset.
Therefore:
• In Whimshire and Fogglia: NUR < both cities
• In Humbleset: NUR lies between its two cities
(Source: Given constraint)

Step 4
Fogglia has the lowest PI and Humbleset has the highest PI.
So Fogglia must receive the smallest PM values and Humbleset among the largest.
(Source: PI ordering condition)

Step 5
Assign Fogglia the smallest feasible PMs with NUR below both cities:

Fogglia: NUR = 10; Cities = 20, 40

PI = 0.5×10 + 0.25×20 + 0.25×40 => PI = 5 + 5 + 10 = 20

(Source: Step 2 + Step 4)

Step 6
Remaining PMs after assigning Fogglia: 40, 50, 60, 70, 80, 90
Humbleset must have exactly one city below NUR and one above, so NUR cannot be the maximum or minimum among remaining. (Source: Step 3 logic)

Step 7
Test Humbleset with cities 60 and 80, NUR = 70:
70 > 60 (one valid pair), 70 < 80 (no extra violation)

PI = 0.5×70 + 0.25×60 + 0.25×80 => PI = 35 + 15 + 20 = 70

(Source: Step 3 + Step 4)

Step 8
Remaining PMs for Whimshire: 40, 50, 90

NUR must be less than both cities, to get the middle PI = 50
Whimshire assignment: NUR = 30, Cities = 50 and 90 PI = 0.5×30 + 0.25×50 + 0.25×90 => PI = 15 + 12.5 + 22.5 = 50
(Source: Remaining PMs + PI order)

Step 9
Mapping PMs to city names using the given city order fixes groupings uniquely.
Only Mumpypore (60) and Zingaloo (80) can belong together in Humbleset.
(Source: City PM ranking + Step 7)

Final Working Table

Correct Answer: 9
Explanation: Applying all constraints fixes Fogglia’s PMs as 10, 20, and 30, Humbleset’s as 60, 70, and 80, and Whimshire receives the remaining PMs. The city ordering then assigns each PM to a specific city, and each NUR also becomes uniquely located. As a result, the PM and state for all six cities and all three NURs are completely determined, giving a total of 9 identifiable entities.

Table Steps Explanation

City PM order (increasing):
Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo
Available PMs: 10, 20, 30, 40, 50, 60, 70, 80, 90
(Source: Question statement)

Step 1
Each state has exactly three PMs (2 cities + 1 NUR).
Total PMs are nine distinct multiples of 10 from 10 to 90.
(Source: Question statement)

Step 2
PI formula:
PI = 0.5 × NUR + 0.25 × City1 + 0.25 × City2
PI values are distinct integers.
(Source: Question statement)

Step 3
Exactly one (NUR, City) pair has NUR PM greater than City PM, and both belong to Humbleset.
Therefore:
• In Whimshire and Fogglia: NUR < both cities
• In Humbleset: NUR lies between its two cities
(Source: Given constraint)

Step 4
Fogglia has the lowest PI and Humbleset has the highest PI.
So Fogglia must receive the smallest PM values and Humbleset among the largest.
(Source: PI ordering condition)

Step 5
Assign Fogglia the smallest feasible PMs with NUR below both cities:

Fogglia: NUR = 10; Cities = 20, 40

PI = 0.5×10 + 0.25×20 + 0.25×40 => PI = 5 + 5 + 10 = 20

(Source: Step 2 + Step 4)

Step 6
Remaining PMs after assigning Fogglia: 40, 50, 60, 70, 80, 90
Humbleset must have exactly one city below NUR and one above, so NUR cannot be the maximum or minimum among remaining. (Source: Step 3 logic)

Step 7
Test Humbleset with cities 60 and 80, NUR = 70:
70 > 60 (one valid pair), 70 < 80 (no extra violation)

PI = 0.5×70 + 0.25×60 + 0.25×80 => PI = 35 + 15 + 20 = 70

(Source: Step 3 + Step 4)

Step 8
Remaining PMs for Whimshire: 40, 50, 90

NUR must be less than both cities, to get the middle PI = 50
Whimshire assignment: NUR = 30, Cities = 50 and 90 PI = 0.5×30 + 0.25×50 + 0.25×90 => PI = 15 + 12.5 + 22.5 = 50
(Source: Remaining PMs + PI order)

Step 9
Mapping PMs to city names using the given city order fixes groupings uniquely.
Only Mumpypore (60) and Zingaloo (80) can belong together in Humbleset.
(Source: City PM ranking + Step 7)

Final Working Table