Comprehension
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 dis tinct 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 Pls of all three states are distinct integers, with Humbleset and Fogglia having the highest and the lowest PI respectively.
Question: 1

What is the PI of Whimshire?

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In complex assignment-based DILR sets, start with the most restrictive condition. Here, the integer PI requirement (leading to the same-parity city pairs) and the unique NUR-city size relationship were the keys to unlocking the puzzle. Build a solution step-by-step and verify all conditions as you go.
Updated On: Jul 4, 2026
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Correct Answer: 45

Solution and Explanation

Step 1: Whimshire's NUR = 20, cities = 60 and 80 (from the passage's deduction).
Step 2: Apply the weights: NUR contributes \(0.5\times20=10\); the two cities together contribute \(0.25\times(60+80)=35\).
Step 3: Add the contributions.
\[ \boxed{10+35=45} \]
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Question: 2

What is the PI of Fogglia?

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Once you have solved a DILR set and found a unique solution, the subsequent questions are typically straightforward lookups or simple calculations based on that solution. Trust your initial detailed work, but keep the derived table or structure handy to answer questions quickly and accurately.
Updated On: Jul 4, 2026
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Correct Answer: 35

Solution and Explanation

Step 1: Fogglia's NUR = 10, cities = 50 and 70.
Step 2: NUR's contribution \(=0.5\times10=5\); cities' combined contribution \(=0.25\times(50+70)=30\).
Step 3: Sum the two parts.
\[ \boxed{5+30=35} \]
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Question: 3

What is the PI of Humbleset?

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For complex logic puzzles, carefully re-read any rules that seem ambiguous. A single word can change the entire logic. Here, understanding that the "only one pair" rule was global, not state-specific, was the crucial step to finding the correct solution that matches the answer key.
Updated On: Jul 4, 2026
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Correct Answer: 50

Solution and Explanation

Step 1: Humbleset's NUR = 40, cities = 30 and 90.
Step 2: NUR's share \(=0.5\times40=20\); cities' combined share \(=0.25\times(30+90)=30\).
Step 3: Add them up.
\[ \boxed{20+30=50} \]
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Question: 4

Which pair of cities definitely belong to the same state?

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For "definitely true" questions in a grouping or assignment set, identify the constraints that force certain items to be together. In this case, the parity rule was the key constraint that created fixed city pairings, making the answer certain.
Updated On: Jul 2, 2026
  • Noodleton, Quackford
  • Splutterville, Quackford
  • Mumpypore, Zingaloo
  • Blusterburg, Mumpypore
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The Correct Option is A

Solution and Explanation

Approach: Instead of building the full ladder first, attack the answer choices using the single most powerful filter — the integer-PI / same-parity rule — and the fact that Blusterburg is forced into Humbleset.

Step 1 — Forced facts: Nine PMs are the distinct multiples of 10 from 10 to 90 (6 cities + 3 NURs). For only one NUR to exceed any city, two NURs must be the two smallest values (10, 20) and the third NUR (40) tops only the smallest city. That smallest city is Blusterburg, and “both in Humbleset” puts Blusterburg in Humbleset with NUR 40. The remaining city PMs in order are $N=50,\ S=60,\ Q=70,\ M=80,\ Z=90$.

Step 2 — Parity test for each state: A state has an integer PI only if its two cities have the same tens-parity, since $0.25(\text{city}_1+\text{city}_2)$ must be whole. Tens digits: $B=3,\ N=5,\ S=6,\ Q=7,\ M=8,\ Z=9$. The only even ones are $S=60$ and $M=80$, so $S$ and $M$ are unavoidably together (Splutterville & Mumpypore). The four odd-tens cities are $B,N,Q,Z$; with $B$ fixed in Humbleset, its partner is one of $N,Q,Z$, and the leftover two odd cities form the last state.

Step 3 — Test the options against this: Splutterville & Quackford (option 2) and Blusterburg & Mumpypore (option 4) mix even and odd tens — impossible, non-integer PI. Mumpypore & Zingaloo (option 3): Mumpypore is locked with Splutterville, so it cannot pair with Zingaloo. That leaves Noodleton & Quackford (option 1), and indeed $50$ and $70$ share odd parity.

Step 4 — Confirm: Choosing Zingaloo $=90$ as Humbleset’s second city gives PIs Humbleset $50$, Whimshire $45$ (S,M,NUR 20), Fogglia $35$ (N,Q,NUR 10) — distinct integers, Humbleset highest, Fogglia lowest. So Noodleton and Quackford (both Fogglia) are the definite same-state pair.

Answer: Noodleton, Quackford.
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Question: 5

For how many of the cities and NURs is it possible to identify their PM and the state they belong to?

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In complex DILR arrangement sets, the goal is often to see if the rules force a single outcome. If you can build a complete table or assignment that follows every rule, and you can demonstrate through logic (e.g., by eliminating other possibilities) that this is the only such arrangement, then all elements are "definitely" identified.
Updated On: Jul 4, 2026
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Correct Answer: 9

Solution and Explanation

Go through the nine entities one by one: the three NURs are forced to exactly 10, 20 and 40, each tied to a specific state by the violation-pair clue and the PI ranking; Blusterburg is forced to 30 and to Humbleset by that same clue; and the other five cities are forced to 50-90 in their stated order, with their state-split forced uniquely once "distinct integer PIs, Humbleset highest, Fogglia lowest" is imposed. Nothing is left as a genuine either/or choice, so the answer covers every city and NUR.
\[ \boxed{9} \]
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