We track two linked tables. Table A lists the six cities with their Pollution Measure (PM) and the state each belongs to. Table B lists the three states with the PM of their non-urban region (NUR) and their Pollution Index (PI). | City | PM | State | |------|-----|-------| | Blusterburg | | | | Noodleton | | | | Splutterville | | | | Quackford | | | | Mumpypore | | | | Zingaloo | | | | State | NUR PM | PI | |-------|--------|-----| | Whimshire | | | | Fogglia | | | | Humbleset | | | There are $9$ entities ($6$ cities $+\ 3$ NURs) sharing $9$ distinct PMs: $\{10,20,30,40,50,60,70,80,90\}$. The six cities take six of these values; the three NURs take the remaining three. The given PM order of the cities is $\text{Blusterburg} < \text{Noodleton} < \text{Splutterville} < \text{Quackford} < \text{Mumpypore} < \text{Zingaloo}$. Each PI is $\text{PI} = 0.5\times(\text{NUR PM}) + 0.25\times(\text{city}_1) + 0.25\times(\text{city}_2)$. Goal: find every PM, every state assignment, and the three PIs. —-- | City | PM | State | |------|-----|-------| | Blusterburg | 30 | | | Noodleton | 50 | | | Splutterville | 60 | | | Quackford | 70 | | | Mumpypore | 80 | | | Zingaloo | 90 | | | State | NUR PM | PI | |-------|--------|-----| | Whimshire | | | | Fogglia | | | | Humbleset | | | All six city PMs are filled from the single NUR$>$city rule. Let the cities in increasing order be $c_1$ city equals, for each NUR, the count of cities below it, summed up. This total must be $1$.

If two or more NURs were above the smallest city $c_1$, the total would be $\ge 2$. If no NUR were above $c_1$, the total would be $0$. So exactly one NUR lies above $c_1$, and it must lie below $c_2$ (otherwise it would exceed $\ge 2$ cities). The ordering of all nine values is therefore forced: $n_1$ city is $n_3=40$ over $c_1=30$, i.e. over Blusterburg. This NUR and this city both belong to Humbleset. So Humbleset's NUR PM $=40$, and Blusterburg ($30$) is one of Humbleset's two cities. The remaining NUR PMs $\{10,20\}$ belong to Whimshire and Fogglia, in some order. —-- | City | PM | State | |------|-----|-------| | Blusterburg | 30 | Humbleset | | Noodleton | 50 | | | Splutterville | 60 | W/F | | Quackford | 70 | | | Mumpypore | 80 | W/F | | Zingaloo | 90 | | | State | NUR PM | PI | |-------|--------|-----| | Whimshire | | | | Fogglia | | | | Humbleset | 40 | | Splutterville and Mumpypore are pinned to the same state, which is either Whimshire or Fogglia (marked W/F). Since $0.5\times\text{NUR}$ is an integer for every $\text{NUR}\in\{10,20,40\}$, a PI is an integer only when $0.25\times(\text{city}_1+\text{city}_2)$ is an integer, i.e. the two city PMs of a state sum to a multiple of $4$.

City PMs modulo $4$: $60\equiv 0,\ 80\equiv 0$, and $30,50,70,90\equiv 2$. A pair sums to a multiple of $4$ only if both are $\equiv 0$ or both are $\equiv 2$. Only $60$ and $80$ are $\equiv 0$, so they must be paired together. Thus one state's cities are Splutterville ($60$) and Mumpypore ($80$). Humbleset already contains $30$ (which is $\equiv 2$), so this is not Humbleset. Hence Splutterville and Mumpypore share Whimshire or Fogglia, and Humbleset's second city is one of $\{50,70,90\}$. —-- | City | PM | State | |------|-----|-------| | Blusterburg | 30 | Humbleset | | Noodleton | 50 | | | Splutterville | 60 | W/F | | Quackford | 70 | | | Mumpypore | 80 | W/F | | Zingaloo | 90 | Humbleset | | State | NUR PM | PI | |-------|--------|-----| | Whimshire | | | | Fogglia | | | | Humbleset | 40 | 50 | Zingaloo is fixed as Humbleset's second city, giving Humbleset PI $=50$. Let Humbleset $=\{30,x\}$ with $x\in\{50,70,90\}$: $\text{PI}_H = 20 + 0.25(30+x) = 27.5 + 0.25x$, giving $40,\ 45,\ 50$ for $x=50,70,90$. The $\{60,80\}$ state has PI $= 35 + 0.5\times\text{NUR} \in \{40,45\}$.

Humbleset must have the strictly highest PI. If $x=50$: $\text{PI}_H=40$, but the $\{60,80\}$ state can reach $45$, so Humbleset is not highest. Reject. If $x=70$: $\text{PI}_H=45$; the leftover cities form $\{50,90\}$ (sum $140$) and $\{60,80\}$ (sum $140$), both giving $35+0.5\times\text{NUR}$. With NURs $10,20$ one of these states becomes $45$, tying Humbleset. Reject. So $x=90$: Humbleset $=\{30,90\}=$ Blusterburg, Zingaloo, with $\text{PI}_H = 27.5 + 22.5 = 50$. —-- | City | PM | State | |------|-----|-------| | Blusterburg | 30 | Humbleset | | Noodleton | 50 | Fogglia | | Splutterville | 60 | Whimshire | | Quackford | 70 | Fogglia | | Mumpypore | 80 | Whimshire | | Zingaloo | 90 | Humbleset | | State | NUR PM | PI | |-------|--------|-----| | Whimshire | 20 | 45 | | Fogglia | 10 | 35 | | Humbleset | 40 | 50 | Whimshire and Fogglia are completed. The four remaining cities split as $\{60,80\}$ (Splutterville, Mumpypore) and $\{50,70\}$ (Noodleton, Quackford), with NURs $\{10,20\}$. $\{60,80\}: 35 + 0.5\times\text{NUR} \Rightarrow 40\ (\text{NUR }10)\ \text{or}\ 45\ (\text{NUR }20)$. $\{50,70\}: 30 + 0.5\times\text{NUR} \Rightarrow 35\ (\text{NUR }10)\ \text{or}\ 40\ (\text{NUR }20)$. The combination $\{60,80\}{+}10=40$ and $\{50,70\}{+}20=40$ gives equal PIs, which is rejected (PIs must be distinct). So $\{60,80\}$ takes NUR $20$ (PI $45$) and $\{50,70\}$ takes NUR $10$ (PI $35$).

Fogglia has the lowest PI, so Fogglia $=\{50,70\}$: Noodleton ($50$), Quackford ($70$), NUR $10$, PI $35$. Whimshire $=\{60,80\}$: Splutterville ($60$), Mumpypore ($80$), NUR $20$, PI $45$. —-- | City | PM | State | |------|-----|-------| | Blusterburg | 30 | Humbleset | | Noodleton | 50 | Fogglia | | Splutterville | 60 | Whimshire | | Quackford | 70 | Fogglia | | Mumpypore | 80 | Whimshire | | Zingaloo | 90 | Humbleset | | State | NUR PM | PI | |-------|--------|-----| | Whimshire | 20 | 45 | | Fogglia | 10 | 35 | | Humbleset | 40 | 50 | Every PM, state and PI is now uniquely determined. Verification of the constraints: The only NUR $>$ city pair is $40 > 30$ (Humbleset's NUR over Blusterburg) — exactly one such pair, and both lie in Humbleset. PIs are $50$ (Humbleset), $45$ (Whimshire), $35$ (Fogglia) — distinct integers, with Humbleset highest and Fogglia lowest. All nine entities (six cities and three NURs) have a known PM and a known state.

CAT (DILR) PYQs

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