First, I'll list out all the given information and constraints to understand the problem better.
Clue Breakdown:
1
States and Their Components:
Three states: Whimshire, Fogglia, Humbleset.
Each has two cities and one NUR (Non-Urban Region).
Total entities: 3 states × (2 cities + 1 NUR) = 9 PMs.
All PMs are distinct multiples of 10, ranging from 10 to 90.
So, possible PMs: 10, 20, 30, 40, 50, 60, 70, 80, 90.
3
Cities in Order of PMs:
Blusterburg, Noodleton, Splutterville, Quackford, Mumpypore, Zingaloo.
Their PMs are in increasing order: Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo.
4
Pollution Index (PI) Calculation:
PI of a state = 50% × NUR PM + 25% × City1 PM + 25% × City2 PM.
Since all PMs are multiples of 10, PI will be a weighted average, possibly a multiple of 5 or 10.
Only one pair where NUR PM > City PM.
Both the NUR and the city in this pair belong to Humbleset.
All three states have distinct integer PIs.
Humbleset has the highest PI.
Fogglia has the lowest PI.
Step-by-Step Deduction Grid:
I'll create a table to organize the information. Let's denote the states as W (Whimshire), F (Fogglia), H (Humbleset). Each has two cities and one NUR.
| State | Cities | NUR | PMs (Cities) | PMs (NUR) | PI Calculation | PI Value |
|---|
| W | C1, C2 | N1 | | | | |
| F | C3, C4 | N2 | | | | |
| H | C5, C6 | N3 | | | | |
But wait, the cities are given in order: Blusterburg, Noodleton, Splutterville, Quackford, Mumpypore, Zingaloo. So, each state has two cities, and the NURs are separate.
Let me list the cities with their possible PMs:
Cities in order of PMs:
Wait, but the PMs go up to 90, so maybe the cities have higher PMs? Wait, no, the cities are six in total, each with distinct PMs from 10 to 90, multiples of 10. So, the cities have PMs: 10, 20, 30, 40, 50, 60, and the NURs have 70, 80, 90.
Wait, no, total PMs are 9: 10,20,30,40,50,60,70,80,90. Each state has two cities and one NUR, so 3 states × 3 = 9.
So, cities have 6 PMs: 10,20,30,40,50,60.
NURs have 3 PMs: 70,80,90.
Wait, but the cities are six, so their PMs are 10,20,30,40,50,60.
NURs are three: 70,80,90.
But the problem says all nine PMs are distinct multiples of 10 from 10 to 90. So yes, cities have 6 PMs, NURs have 3.
Now, the cities are ordered by PMs: Blusterburg (B) < Noodleton (N) < Splutterville (S) < Quackford (Q) < Mumpypore (M) < Zingaloo (Z).
So, B=10, N=20, S=30, Q=40, M=50, Z=60.
Wait, but the cities are six, so their PMs are 10,20,30,40,50,60.
NURs are 70,80,90.
Now, the PI for each state is 50% NUR + 25% city1 + 25% city2.
So, PI = 0.5NUR + 0.25C1 + 0.25*C2.
Since all PMs are multiples of 10, let's represent them as 10a, where a is 1 to 9.
So, NURs are 70 (a=7), 80 (a=8), 90 (a=9).
Cities are 10 (a=1), 20 (a=2), 30 (a=3), 40 (a=4), 50 (a=5), 60 (a=6).
Now, the cities are assigned to the states. Each state has two cities.
Also, the NURs are assigned to the states.
Additionally, only one pair where NUR PM > City PM, and both belong to Humbleset.
So, in Humbleset, one NUR PM is greater than one of its cities.
Now, the PIs of the states are distinct integers, with H having the highest, F the lowest, and W in the middle.
Our goal is to find the PI of Whimshire.
Let me start by assigning the NURs and cities to the states.
First, let's note that the NURs are 70,80,90.
Since H has the highest PI, and PI is 50% NUR +
