CAT 2025 Slot 2 · DILR · Question 18

CAT 2025 Slot 2 previous year question on Caselet Based DI, with the answer and a step-by-step solution.

Caselet Based DIDATA INTERPRETATION -High LevelType in the answer (TITA)Medium

Set 5: Pollution Index of Three States — this question belongs to a set. Read the full set with all 5 questions →

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 distinct 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 PIs of all three states are distinct integers, with Humbleset and Fogglia having the highest and the lowest PI respectively.

What is the PI of Whimshire?

Answer & solution

Answer: 45

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.

CAT 2025 Slot 2 · DILR · Question 18

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