Solution

We can draw table for the 3 regions having 2 cities and 1 NUR of its own. We will also give variables to all the cities and NURs to understand the calculation of PIs.

	Whimshire(W)	Fogglia(F)	Humbleset(H)
City1	City1(W)=a	City1(F)=d	City1(H)=g
City2	City2(W)=b	City2(F)=e	City2(H)=h
NUR	NUR(W)=c	NUR(F)=f	NUR(H)=i
PI (integers)	-	Lowest PI	Highest PI
	PI(W)=(a25%)+(b25%)+(c*50%) PI(W)= + + = +	PI(F)=(d25%)+(e25%)+(f*50%) PI(F)= + + = +	PI(H)=(g25%)+(h25%)+(i*50%) PI(H)= + + = +

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(H).

This means among all the 6 cities and 3 NURs, PM of NUR-H is greater than exactly PM of City1- H only. This is true only for NUR-H. For other NURS, they must be lower in value than City1-H and NUR-H. Otherwise, if NUR-W/NUR-F are greater than NUR-H and City1-H, the stated condition will be true for other NURs: NUR-W/NUR-F will also be greater than City1-H, which is not what we want.

Case 1: Noodleton is a city of Humbleset

If Noodleton belongs to Humbleset, then the PI of Humbleset becomes

27.5+12.5=40

From earlier deductions, the combined PI of one of the other two states (excluding Humbleset) is 35.

Depending on whether the remaining NUR contribution is 5 or 10, the PI of that state becomes 40 or 45.

In both possibilities:

Either PI values are not unique, or
Humbleset does not have the highest PI.

Hence, Noodleton cannot be a city of Humbleset.

Case 2: Quackford is a city of Humbleset

If Quackford belongs to Humbleset, then the PI of Humbleset becomes

27.5+17.5=45

The combined PI of the other two states is again 35.

If the NUR contribution is:

5, the state PI becomes 40
10, the state PI becomes 45

Since PI values must be distinct, the case where the PI of other state becomes 45 is rejected.

Thus, the NUR contribution must be 5, fixing that state’s PI at 40.

Now:

Humbleset PI = 45
One other state PI = 40

The remaining state’s PI becomes:

12.5+22.5+10=45

This leads to two states having PI = 45, which violates uniqueness.

Hence, Quackford cannot be a city of Humbleset.

Case 3: Zingaloo is a city of Humbleset

If Zingaloo belongs to Humbleset, then the PI of Humbleset becomes

7.5+22.5+20=50

The combined PI of one of the other two states is 35.

Subcase 3A: One state has PI = 40 (35+5)

Humbleset PI = 50,

and the remaining state PI becomes: 12.5+17.5+10=40, which violates uniqueness.

Subcase 3B: One state has PI = 45 (35+10)

Humbleset PI = 50,

Then the remaining state PI becomes: 12.5+17.5+5=35, which satisfies all the conditions. Hence, Zingaloo is the second city of Humbleset. Since PI(F) is the lowest, PI(F)=35, PI(H)=50 which is the highest, and remaining PI(W)=45. This will help fill the cities for specific states

Q4. Noodleton, Quackford are the cities from the same state. So, the option(d) is the answer.

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