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.
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?
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
Approach Solution - 1
Approach: Once the unique grid is fixed, Whimshire's PI is a direct plug-in. Whimshire holds NUR PM 20 and cities Splutterville (60) and Mumpypore (80).
Step 1 \(-\) recall the assignment. From the master grid, Whimshire's NUR PM is 20, and its two cities carry PMs 60 and 80.
Step 2 \(-\) apply the weighted formula. \[ \text{PI}(\text{Whimshire}) = \frac{2 \times 20 + 60 + 80}{4} = \frac{40 + 140}{4} = \frac{180}{4} = 45. \]
Final answer: 45
Step 1 \(-\) recall the assignment. From the master grid, Whimshire's NUR PM is 20, and its two cities carry PMs 60 and 80.
Step 2 \(-\) apply the weighted formula. \[ \text{PI}(\text{Whimshire}) = \frac{2 \times 20 + 60 + 80}{4} = \frac{40 + 140}{4} = \frac{180}{4} = 45. \]
Final answer: 45
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Approach Solution -2
Direct calculation. From the deduced assignment, Whimshire's NUR has PM 20, and its two cities are Splutterville (60) and Mumpypore (80). The Pollution Index weights the NUR at 50% and each city at 25%:
\[ PI_{\text{Whimshire}} = 0.5(20) + 0.25(60) + 0.25(80) = 10 + 15 + 20 = 45. \]
Answer: 45.
\[ PI_{\text{Whimshire}} = 0.5(20) + 0.25(60) + 0.25(80) = 10 + 15 + 20 = 45. \]
Answer: 45.
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Question: 2
What is the PI of Fogglia?
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
Approach Solution - 1
Approach: Fogglia is the lowest-PI state in the fixed grid; plug its PMs straight in. Fogglia holds NUR PM 10 and cities Noodleton (50) and Quackford (70).
Step 1 \(-\) recall the assignment. Fogglia's NUR PM is 10, and its two cities carry PMs 50 and 70 (the smallest NUR, consistent with Fogglia having the lowest PI).
Step 2 \(-\) apply the formula. \[ \text{PI}(\text{Fogglia}) = \frac{2 \times 10 + 50 + 70}{4} = \frac{20 + 120}{4} = \frac{140}{4} = 35. \]
Final answer: 35
Step 1 \(-\) recall the assignment. Fogglia's NUR PM is 10, and its two cities carry PMs 50 and 70 (the smallest NUR, consistent with Fogglia having the lowest PI).
Step 2 \(-\) apply the formula. \[ \text{PI}(\text{Fogglia}) = \frac{2 \times 10 + 50 + 70}{4} = \frac{20 + 120}{4} = \frac{140}{4} = 35. \]
Final answer: 35
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Approach Solution -2
Direct calculation. Fogglia's NUR has PM 10, and its two cities are Noodleton (50) and Quackford (70). So:
\[ PI_{\text{Fogglia}} = 0.5(10)+0.25(50)+0.25(70)=5+12.5+17.5=35. \]
Answer: 35.
\[ PI_{\text{Fogglia}} = 0.5(10)+0.25(50)+0.25(70)=5+12.5+17.5=35. \]
Answer: 35.
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Question: 3
What is the PI of Humbleset?
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
Approach Solution - 1
Approach: Humbleset is the highest-PI state and the home of the lone dominance pair; plug in its PMs. Humbleset holds NUR PM 40 and cities Blusterburg (30) and Zingaloo (90).
Step 1 \(-\) recall the assignment. Humbleset's NUR PM is 40 \(-\) the one NUR that beats a city (Blusterburg, PM 30). Its second city is Zingaloo, PM 90.
Step 2 \(-\) apply the formula. \[ \text{PI}(\text{Humbleset}) = \frac{2 \times 40 + 30 + 90}{4} = \frac{80 + 120}{4} = \frac{200}{4} = 50. \]
Step 3 \(-\) sanity check. PI order is \(50 > 45 > 35\), so Humbleset is highest and Fogglia lowest \(-\) exactly as required.
Final answer: 50
Step 1 \(-\) recall the assignment. Humbleset's NUR PM is 40 \(-\) the one NUR that beats a city (Blusterburg, PM 30). Its second city is Zingaloo, PM 90.
Step 2 \(-\) apply the formula. \[ \text{PI}(\text{Humbleset}) = \frac{2 \times 40 + 30 + 90}{4} = \frac{80 + 120}{4} = \frac{200}{4} = 50. \]
Step 3 \(-\) sanity check. PI order is \(50 > 45 > 35\), so Humbleset is highest and Fogglia lowest \(-\) exactly as required.
Final answer: 50
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Approach Solution -2
Direct calculation. Humbleset's NUR has PM 40, and its two cities are Blusterburg (30) and Zingaloo (90). So:
\[ PI_{\text{Humbleset}}=0.5(40)+0.25(30)+0.25(90)=20+7.5+22.5=50. \]
This is indeed the highest of the three PIs (45 and 35 for Whimshire and Fogglia), consistent with the passage.
Answer: 50.
\[ PI_{\text{Humbleset}}=0.5(40)+0.25(30)+0.25(90)=20+7.5+22.5=50. \]
This is indeed the highest of the three PIs (45 and 35 for Whimshire and Fogglia), consistent with the passage.
Answer: 50.
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Question: 4
Which pair of cities definitely belong to the same state?
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
Approach Solution - 1
Approach: The whole set hinges on one tight clue — across all NUR–city comparisons, only ONE NUR is bigger than a city, and both sit in Humbleset. That forces almost every NUR to be tiny, which pins the entire grid. Build the PM ladder first, then split into states using the “integer PI” condition.
Step 1 — What the data gives: Three states (Whimshire, Fogglia, Humbleset), each with 2 cities and 1 NUR — nine PMs in all, the distinct multiples of 10 from 10 to 90. Cities in increasing PM: Blusterburg \(<\) Noodleton \(<\) Splutterville \(<\) Quackford \(<\) Mumpypore \(<\) Zingaloo. PI of a state \(= 0.5\,\text{NUR} + 0.25\,\text{city}_1 + 0.25\,\text{city}_2\). Exactly one (NUR, city) pair has NUR \(>\) city, both in Humbleset. PIs are distinct integers; Humbleset highest, Fogglia lowest.
Step 2 — Use the “only one NUR beats a city” clue: For only one NUR-above-city pair to exist, two of the three NURs must be smaller than every city, and the third NUR may exceed just one city. So the two smallest PMs (10, 20) are NURs, and the remaining special NUR sits just above the smallest city. Ladder of all nine PMs becomes: \(\text{NUR}=10,\ \text{NUR}=20,\ B=30,\ \text{NUR}=40,\ N=50,\ S=60,\ Q=70,\ M=80,\ Z=90.\) The NUR \(=40\) beats only Blusterburg \(=30\) — that lone pair lies in Humbleset, so Blusterburg is a Humbleset city and 40 is Humbleset’s NUR.
Step 3 — Integer-PI splits the rest: A state’s PI is an integer only when its two cities have PMs of the same “tens parity” (so \(0.25(\text{city}_1+\text{city}_2)\) is whole). Among the cities, only Splutterville \(=60\) and Mumpypore \(=80\) are even-tens, so they must share a state. Humbleset’s second city (an odd-tens PM) and the integer/ordering conditions then force a unique fit.
Step 4 — Lock the grid: Testing the odd choices for Humbleset’s second city, only Zingaloo \(=90\) keeps Humbleset highest and Fogglia lowest with distinct integer PIs:
Humbleset = {Blusterburg 30, Zingaloo 90, NUR 40}, PI \(= 20+7.5+22.5 = 50\) (highest).
Whimshire = {Splutterville 60, Mumpypore 80, NUR 20}, PI \(= 10+15+20 = 45\).
Fogglia = {Noodleton 50, Quackford 70, NUR 10}, PI \(= 5+12.5+17.5 = 35\) (lowest).
Step 5 — Read the pairs: Same-state city pairs are (Blusterburg, Zingaloo), (Splutterville, Mumpypore) and (Noodleton, Quackford). Among the options, only Noodleton & Quackford appear — both in Fogglia.
Answer: Noodleton, Quackford.
Step 1 — What the data gives: Three states (Whimshire, Fogglia, Humbleset), each with 2 cities and 1 NUR — nine PMs in all, the distinct multiples of 10 from 10 to 90. Cities in increasing PM: Blusterburg \(<\) Noodleton \(<\) Splutterville \(<\) Quackford \(<\) Mumpypore \(<\) Zingaloo. PI of a state \(= 0.5\,\text{NUR} + 0.25\,\text{city}_1 + 0.25\,\text{city}_2\). Exactly one (NUR, city) pair has NUR \(>\) city, both in Humbleset. PIs are distinct integers; Humbleset highest, Fogglia lowest.
Step 2 — Use the “only one NUR beats a city” clue: For only one NUR-above-city pair to exist, two of the three NURs must be smaller than every city, and the third NUR may exceed just one city. So the two smallest PMs (10, 20) are NURs, and the remaining special NUR sits just above the smallest city. Ladder of all nine PMs becomes: \(\text{NUR}=10,\ \text{NUR}=20,\ B=30,\ \text{NUR}=40,\ N=50,\ S=60,\ Q=70,\ M=80,\ Z=90.\) The NUR \(=40\) beats only Blusterburg \(=30\) — that lone pair lies in Humbleset, so Blusterburg is a Humbleset city and 40 is Humbleset’s NUR.
Step 3 — Integer-PI splits the rest: A state’s PI is an integer only when its two cities have PMs of the same “tens parity” (so \(0.25(\text{city}_1+\text{city}_2)\) is whole). Among the cities, only Splutterville \(=60\) and Mumpypore \(=80\) are even-tens, so they must share a state. Humbleset’s second city (an odd-tens PM) and the integer/ordering conditions then force a unique fit.
Step 4 — Lock the grid: Testing the odd choices for Humbleset’s second city, only Zingaloo \(=90\) keeps Humbleset highest and Fogglia lowest with distinct integer PIs:
Humbleset = {Blusterburg 30, Zingaloo 90, NUR 40}, PI \(= 20+7.5+22.5 = 50\) (highest).
Whimshire = {Splutterville 60, Mumpypore 80, NUR 20}, PI \(= 10+15+20 = 45\).
Fogglia = {Noodleton 50, Quackford 70, NUR 10}, PI \(= 5+12.5+17.5 = 35\) (lowest).
Step 5 — Read the pairs: Same-state city pairs are (Blusterburg, Zingaloo), (Splutterville, Mumpypore) and (Noodleton, Quackford). Among the options, only Noodleton & Quackford appear — both in Fogglia.
Answer: Noodleton, Quackford.
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Approach Solution -2
Step 1: Understanding the Question:
The question asks which of the given pairs of cities must be in the same state. This requires us to check the city pairings in our logically derived unique solution.
Step 2: Analyzing the Logically Necessary Pairings:
The "integer PI" condition forces the two cities in any state to have PMs with same-parity tens-digits.
The PM values for the six cities in our unique solution are {10, 40, 50, 60, 70, 90}.
- Odd PMs: {10, 50, 70, 90}
- Even PMs: {40, 60}
To create three pairs with same-parity partners, the two even-PM cities, Noodleton (40) and Quackford (60), must be paired together.
The four odd-PM cities must form the other two pairs: {Blusterburg (10), Zingaloo (90)} and {Splutterville (50), Mumpypore (70)}.
Therefore, the pairing of Noodleton and Quackford is a logical necessity.
Step 3: Evaluating the Options Based on the Derived Solution:
- (A) Noodleton, Quackford: This pair belongs to Fogglia in our solution. This pairing is logically forced by the parity rule. This is correct.
- (B) Splutterville, Quackford: Splutterville (PM=50) and Quackford (PM=60) have different parity and cannot be in the same state.
- (C) Mumpypore, Zingaloo: Mumpypore is in Whimshire and Zingaloo is in Humbleset.
- (D) Blusterburg, Mumpypore: Blusterburg is in Humbleset and Mumpypore is in Whimshire.
Step 4: Final Answer:
The pair of cities that definitely belong to the same state is Noodleton and Quackford.
The question asks which of the given pairs of cities must be in the same state. This requires us to check the city pairings in our logically derived unique solution.
Step 2: Analyzing the Logically Necessary Pairings:
The "integer PI" condition forces the two cities in any state to have PMs with same-parity tens-digits.
The PM values for the six cities in our unique solution are {10, 40, 50, 60, 70, 90}.
- Odd PMs: {10, 50, 70, 90}
- Even PMs: {40, 60}
To create three pairs with same-parity partners, the two even-PM cities, Noodleton (40) and Quackford (60), must be paired together.
The four odd-PM cities must form the other two pairs: {Blusterburg (10), Zingaloo (90)} and {Splutterville (50), Mumpypore (70)}.
Therefore, the pairing of Noodleton and Quackford is a logical necessity.
Step 3: Evaluating the Options Based on the Derived Solution:
- (A) Noodleton, Quackford: This pair belongs to Fogglia in our solution. This pairing is logically forced by the parity rule. This is correct.
- (B) Splutterville, Quackford: Splutterville (PM=50) and Quackford (PM=60) have different parity and cannot be in the same state.
- (C) Mumpypore, Zingaloo: Mumpypore is in Whimshire and Zingaloo is in Humbleset.
- (D) Blusterburg, Mumpypore: Blusterburg is in Humbleset and Mumpypore is in Whimshire.
Step 4: Final Answer:
The pair of cities that definitely belong to the same state is Noodleton and 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?
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
Approach Solution - 1
Approach: “How many can we identify” is really asking — is the grid unique? If the constraints pin down one and only one arrangement, every entity is identified.
Step 1 — Set the ladder: Nine PMs are the distinct multiples of 10 from 10 to 90 (6 cities, 3 NURs). \(\text{PI} = 0.5\,\text{NUR} + 0.25\,\text{city}_1 + 0.25\,\text{city}_2\). The clue “exactly one NUR exceeds a city, both in Humbleset” forces two NURs to be the smallest values and the third to top just one city. So: NUR \(=10\), NUR \(=20\), Blusterburg \(=30\), NUR \(=40\), Noodleton \(=50\), Splutterville \(=60\), Quackford \(=70\), Mumpypore \(=80\), Zingaloo \(=90\). The NUR \(=40\) beats only Blusterburg, so Blusterburg and NUR 40 are Humbleset’s.
Step 2 — Integer PI fixes the split: Integer PI needs the two cities of a state to share tens-parity. Only Splutterville (60) and Mumpypore (80) are even-tens, so they sit together. Humbleset’s second city is odd-tens; testing Noodleton(50)/Quackford(70)/Zingaloo(90), only Zingaloo keeps Humbleset highest and Fogglia lowest with distinct integer PIs.
Step 3 — The unique grid:
Humbleset: Blusterburg 30, Zingaloo 90, NUR 40, PI \(=50\) (highest).
Whimshire: Splutterville 60, Mumpypore 80, NUR 20, PI \(=45\).
Fogglia: Noodleton 50, Quackford 70, NUR 10, PI \(=35\) (lowest).
Step 4 — Count: Every PM and every state assignment is forced — nothing is left ambiguous. So all 6 cities and all 3 NURs, i.e. \(6+3 = 9\) entities, are fully identified.
Answer: 9.
Step 1 — Set the ladder: Nine PMs are the distinct multiples of 10 from 10 to 90 (6 cities, 3 NURs). \(\text{PI} = 0.5\,\text{NUR} + 0.25\,\text{city}_1 + 0.25\,\text{city}_2\). The clue “exactly one NUR exceeds a city, both in Humbleset” forces two NURs to be the smallest values and the third to top just one city. So: NUR \(=10\), NUR \(=20\), Blusterburg \(=30\), NUR \(=40\), Noodleton \(=50\), Splutterville \(=60\), Quackford \(=70\), Mumpypore \(=80\), Zingaloo \(=90\). The NUR \(=40\) beats only Blusterburg, so Blusterburg and NUR 40 are Humbleset’s.
Step 2 — Integer PI fixes the split: Integer PI needs the two cities of a state to share tens-parity. Only Splutterville (60) and Mumpypore (80) are even-tens, so they sit together. Humbleset’s second city is odd-tens; testing Noodleton(50)/Quackford(70)/Zingaloo(90), only Zingaloo keeps Humbleset highest and Fogglia lowest with distinct integer PIs.
Step 3 — The unique grid:
Humbleset: Blusterburg 30, Zingaloo 90, NUR 40, PI \(=50\) (highest).
Whimshire: Splutterville 60, Mumpypore 80, NUR 20, PI \(=45\).
Fogglia: Noodleton 50, Quackford 70, NUR 10, PI \(=35\) (lowest).
Step 4 — Count: Every PM and every state assignment is forced — nothing is left ambiguous. So all 6 cities and all 3 NURs, i.e. \(6+3 = 9\) entities, are fully identified.
Answer: 9.
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Approach Solution -2
Checking every entity. The chain of deductions pins down every one of the nine PMs to one specific value AND one specific state, with no leftover ambiguity at any step: the two "always-lower" NURs are forced to exactly 10 and 20 (Fogglia and Whimshire respectively, once the PI ordering is applied), Blusterburg is forced to 30 and to Humbleset, Humbleset's own NUR is forced to 40, and the remaining five cities are forced into their fixed increasing order (50 through 90) and then into a unique 2-2-2 split across the three states by the integer-PI and highest/lowest conditions. Since every one of these nine assignments comes out uniquely rather than as an either/or choice, all nine can be identified.
Answer: 9 (all of them).
Answer: 9 (all of them).
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