The two most populous cities and the non-urban region (NUR) of each of three states, Whimshire, Fogglia, and Humbleset, are assigned Pollution Measures (PMs).

These nine PMs are all distinct multiples of 10, ranging from 10 to 90. The six cities in increasing order of their PMs are: Blusterburg, Noodleton, Splutterville, Quackford, Mumpypore, Zingaloo.

The Pollution Index (PI) of a state is a weighted average of the PMs of its NUR and cities, with a weight of 50% for the NUR, and 25% each for its two cities.

There is only one pair of an NUR and a city (considering all cities and all NURs) where the PM of the NUR is greater than that of the city. That NUR and the city both belong to Humbleset.
The PIs of all three states are distinct integers, with Humbleset and Fogglia having the highest and the lowest PI respectively.

What is the PI of Fogglia?

The correct answer is 35.

Step 1: Understand the Pollution Measures (PMs) for Non-Urban Regions (NURs) and Cities
We are given that there are three states (Whimshire, Fogglia, and Humbleset), each containing one NUR and two cities. This means there are 3 NURs and 6 cities in total, making 9 entities. Their PMs are distinct multiples of 10 ranging from 10 to 90. Therefore, the nine PM values are 10, 20, 30, 40, 50, 60, 70, 80, and 90.

Step 2: Use the "Only One Pair" Rule to Determine the PMs of NURs and Cities
The problem states: "There is only one pair of an NUR and a city... where the PM of the NUR is greater than that of the city." If we compare the 3 NURs against the 6 cities, exactly one NUR has a higher PM than exactly one city. For this to happen, the remaining two NURs must have lower PMs than all the cities, and the third NUR must be higher than only the city with the lowest PM.

Therefore, the two lowest values must be NURs, followed by the lowest city, and then the third NUR. This gives us the exact assignments for the 9 PM values in increasing order:

$\text{NURs:}10,20,40$

$\text{Cities:}30,50,60,70,80,90$

This perfectly creates exactly one pair where the NUR is greater than the city: the NUR with a PM of 40 and the city with a PM of 30.

Step 3: Analyze Humbleset's Composition
The problem states that the unique pair (NUR greater than city) belongs entirely to Humbleset. Therefore, Humbleset consists of the NUR with a PM of 40 and at least one city with a PM of 30.

The Pollution Index (PI) is calculated using the weights: 50% for the NUR, and 25% for each of the two cities. The formula for PI is:

$\text{PI}=\frac{2\times \text{NUR}+C_1+C_2}{4}$

Since the PI must be an integer, the numerator must be a multiple of 4.

For Humbleset, we have NUR = 40 and C₁ = 30. Plugging these into our formula:

${\text{PI}}_{\text{Humbleset}}=\frac{2\times 40+30+C_2}{4}=\frac{110+C_2}{4}$

To make the numerator a multiple of 4, the possible remaining city values for C₂ (from 50, 60, 70, 80, 90) can be tested. Since we know Humbleset has the highest PI of all states, we can test assigning it the highest possible city, C₂ = 90. This gives:

${\text{PI}}_{\text{Humbleset}}=\frac{110+90}{4}=50$

Step 4: Determine Fogglia and Whimshire's Attributes
With Humbleset's cities confirmed as 30 and 90, the remaining unassigned cities are 50, 60, 70, and 80. The remaining unassigned NURs are 10 and 20.

We are told Fogglia has the lowest PI. To achieve the lowest possible integer PI, let's assign it the lowest available NUR (10) and test the lowest combination of remaining cities that yields a multiple of 4.

For Fogglia, NUR = 10:

${\text{PI}}_{\text{Fogglia}}=\frac{2\times 10+C_1+C_2}{4}=\frac{20+C_1+C_2}{4}$

For the numerator to be divisible by 4, the sum of the two cities must be a multiple of 4. From the remaining cities (50, 60, 70, 80), the pair 50 and 70 sums to 120. Using these cities for Fogglia gives:

${\text{PI}}_{\text{Fogglia}}=\frac{20+120}{4}=35$

By process of elimination, Whimshire receives the remaining NUR (20) and the remaining cities (60 and 80):

${\text{PI}}_{\text{Whimshire}}=\frac{2\times 20+60+80}{4}=\frac{40+140}{4}=45$

Step 5: Verify all the constraints
The three PIs are 35, 45, and 50. These are all distinct integers. Humbleset has the highest PI (50), and Fogglia has the lowest PI (35). This satisfies all conditions in the problem perfectly.

Therefore, the Pollution Index of Fogglia is 35.