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 Whimshire?

The correct answer is 45.

Step 1: Determining the PMs of NURs and Cities

The nine Pollution Measures (PMs) are all distinct multiples of 10 ranging from 10 to 90: 10, 20, 30, 40, 50, 60, 70, 80, and 90.

There are 6 cities and 3 Non-Urban Regions (NURs). The problem states there is exactly one pair of an NUR and a city where the PM of the NUR is greater than the city.

If the 3 NURs were assigned the three lowest PMs (10, 20, and 30), the cities would take the six highest values (40 to 90). In this scenario, no NUR would be greater than any city.

To have exactly one instance where an NUR is greater than a city, the highest NUR must be 40, and the lowest city must be 30. The remaining two NURs must be 10 and 20, meaning they are smaller than all cities.

Thus, the NURs are 10, 20, and 40.

The six cities, ordered by increasing PM (Blusterburg, Noodleton, Splutterville, Quackford, Mumpypore, Zingaloo), are naturally assigned as follows:

Blusterburg = 30, Noodleton = 50, Splutterville = 60, Quackford = 70, Mumpypore = 80, and Zingaloo = 90.

The single valid pair where NUR > City is 40 > 30.

Step 2: Analyzing Humbleset

The problem states that this single pair (the NUR of 40 and the city of 30) belongs entirely to Humbleset. So, Humbleset has an NUR of 40 and one of its cities is Blusterburg (30).

Let Humbleset's second city be denoted as:

$C_2$

The Pollution Index (PI) is calculated as:

$PI=0.5\times NUR+0.25\times C_1+0.25\times C_2$

Which simplifies algebraically to:

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

For Humbleset, substituting the known values gives:

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

Because the PI must be an integer, the numerator must be divisible by 4. Looking at the remaining available cities (50, 60, 70, 80, 90), the only values that make the sum a multiple of 4 are 50, 70, and 90.

Step 3: Finding the Valid PIs

Let's test the possibilities for Humbleset's second city:

Case 1: The second city is 50.

Humbleset's PI would be:

$\frac{160}{4}=40$

The remaining cities (60, 70, 80, 90) must be paired so their sum is a multiple of 4. The unique pairs are (60, 80) summing to 140, and (70, 90) summing to 160. Pairing these with the remaining NURs (10, 20) yields PIs of 40, 45, and 50 in various combinations. However, this produces duplicate PIs (e.g., 40, 40, 50 or 40, 45, 45), violating the condition that all PIs are distinct.

Case 2: The second city is 70.

Humbleset's PI would be:

$\frac{180}{4}=45$

The remaining cities (50, 60, 80, 90) must form pairs summing to a multiple of 4. The only valid pairs are (60, 80) and (50, 90), both summing to 140. This would identically force the other two states to have matching PIs, which is invalid.

Case 3: The second city is 90.

Humbleset's PI is:

$\frac{200}{4}=50$

The remaining cities are 50, 60, 70, and 80. The valid pairs are (60, 80) summing to 140, and (50, 70) summing to 120. Assigning the remaining NURs (10 and 20):

If NUR 20 is paired with cities 60 and 80, the PI is:

$\frac{40+140}{4}=45$

If NUR 10 is paired with cities 50 and 70, the PI is:

$\frac{20+120}{4}=35$

This gracefully results in three perfectly distinct integer PIs: 50, 45, and 35.

Step 4: Final Conclusion

The problem requires that Humbleset has the highest PI and Fogglia has the lowest PI. Checking our only valid scenario:

Humbleset's PI is 50 (the highest).

Fogglia's PI must be 35 (the lowest).

Therefore, Whimshire must have the remaining middle PI, which is 45.