CAT 2025 Slot 2 · DILR · Set 5: Pollution Index of Three States

Only one (NUR, city) pair has the NUR above the city. So two NURs are below all six cities, taking PMs 10 and 20. The third NUR (Humbleset's) is above exactly one city, which must be the lowest city, Blusterburg. So Blusterburg = 30 and Humbleset's NUR = 40, and the other cities are Noodleton 50, Splutterville 60, Quackford 70, Mumpypore 80, Zingaloo 90. Blusterburg is in Humbleset.

PI = (2 × NUR + city1 + city2) ÷ 4. With 2 × NUR a multiple of 20, the PI is an integer only if city1 + city2 is a multiple of 4.

Humbleset: 30 + x must be a multiple of 4, so x = 50, 70 or 90, giving PI = (80 + 30 + x) ÷ 4 = 40, 45 or 50.

x = 50: the remaining cities pair as (60, 80) and (70, 90); the other PIs are at least (20 + 140) ÷ 4 = 40, so Humbleset is not strictly highest.

x = 70: the remaining cities pair as (50, 90) and (60, 80), both summing to 140; the PIs are (20 + 140) ÷ 4 = 40 and (40 + 140) ÷ 4 = 45, tying with Humbleset's 45.

x = 90: Humbleset PI = 50. The remaining cities 50, 60, 70, 80 must pair as (50, 70) and (60, 80). NUR 10 with (50, 70) gives (20 + 120) ÷ 4 = 35; NUR 20 with (60, 80) gives (40 + 140) ÷ 4 = 45. (The swap gives 40 and 40, not distinct.)

So the PIs are Humbleset 50, Whimshire 45, Fogglia 35 (lowest).

Shortcut: fix the PM ladder first (10, 20, 30, 40, then 50 to 90), then test Humbleset's second city against the 'sum divisible by 4' rule.

Hence, the answer is 45.

The PMs are fixed: NURs 10 and 20, Blusterburg 30, Humbleset's NUR 40, and Noodleton 50 to Zingaloo 90.

Humbleset must be Blusterburg + Zingaloo + NUR 40, giving PI = (80 + 30 + 90) ÷ 4 = 50; any other second city produces a tie or makes Humbleset not the highest.

The remaining cities pair as Noodleton–Quackford (50 + 70 = 120) and Splutterville–Mumpypore (60 + 80 = 140), since each pair's sum must be a multiple of 4.

Fogglia has the lowest PI, so it takes NUR 10 with the smaller pair: (20 + 120) ÷ 4 = 35.

Shortcut: with Humbleset at 50, the only distinct split of the other two states is 35 and 45; Fogglia takes 35.

Hence, the answer is 35.

Humbleset has NUR 40 and Blusterburg 30, plus one city x from 50 to 90. PI = (80 + 30 + x) ÷ 4 = (110 + x) ÷ 4, so x must make 110 + x a multiple of 4, i.e. x = 50, 70 or 90.

x = 50 gives 40, but the other states would be at least 40, so Humbleset would not be the highest.

x = 70 gives 45, but the other states would be 40 and 45, a tie.

x = 90 gives 50, with the others at 45 and 35. This works.

Shortcut: Humbleset must be the highest, so try the largest admissible city, Zingaloo, first: (110 + 90) ÷ 4 = 50.

Hence, the answer is 50.

The arrangement is unique:

Humbleset: NUR 40, Blusterburg 30, Zingaloo 90 (PI 50).

Whimshire: NUR 20, Splutterville 60, Mumpypore 80 (PI 45).

Fogglia: NUR 10, Noodleton 50, Quackford 70 (PI 35).

Option 1: Mumpypore (Whimshire) and Zingaloo (Humbleset) – different states.

Option 2: Splutterville (Whimshire) and Quackford (Fogglia) – different states.

Option 3: Blusterburg (Humbleset) and Mumpypore (Whimshire) – different states.

Option 4: Noodleton and Quackford are both in Fogglia.

Shortcut: from 50, 60, 70, 80 the only pairing with both sums divisible by 4 is (50, 70) and (60, 80), so Noodleton and Quackford must be together.

Hence, the answer is option 4.

The PM ladder is forced: the two lowest NURs are 10 and 20, Blusterburg is 30, Humbleset's NUR is 40, and the cities Noodleton to Zingaloo are 50 to 90. Every PM is known.

The integer and ordering conditions force Humbleset = NUR 40 + Blusterburg + Zingaloo. The other cities must pair as Noodleton–Quackford and Splutterville–Mumpypore, and the only distinct assignment is Fogglia = NUR 10 + Noodleton + Quackford (35) and Whimshire = NUR 20 + Splutterville + Mumpypore (45).

All 3 NURs and all 6 cities have a known PM and a known state: 9 in total.

Shortcut: once the arrangement turns out to be unique, all nine places are fully identified.

Hence, the answer is 9.