CAT 2023 Slot 1 Solved Paper

It is given that some of the houses are occupied. The remaining ones are vacant and are the only ones available for sale.
The base price of a vacant house is Rs. 10 lakhs if the house does not have a parking space, and Rs. 12 lakhs if it does. The quoted price (in lakhs of Rs.) of a vacant house is calculated as (base price) $+5\times$ (road adjacency value) $+3\times$ (neighbor count).
It is also known that the maximum quoted price of a house in Block XX is Rs. 24 lakhs
Hence, there can be two cases for the maximum quoted price of a house in block XX.
Case 1: House with parking space:
$\Rightarrow 12+5\mathit{a}+3\mathit{b}=24\Rightarrow 5\mathit{a}+3\mathit{b}=12$ ( $\mathit{a}=$ road adjacency value, $\mathit{b}=$ neighbor count $)$
The only value for which the equation satisfies is $(\mathit{a}=0$, and $\mathit{b}=4$ ). But the value of $\mathit{b}$ can't be 4 because the maximum neighbor count can be at most 3 .
Hence, case 1 is invalid.
Case 2: House without parking space:
$\Rightarrow 10+5\mathit{a}+3\mathit{b}=24\Rightarrow 5\mathit{a}+3\mathit{b}=14$
$\Rightarrow \left(\mathit{a},\mathit{b}\right)=\left(1,3\right)$
Hence, the house must have 3 neighbors and 1 road connected to it. Hence, the only possible case is B2. Therefore, the neighbor houses of $\mathrm{B}2$, which are ( $\mathrm{B}1,\mathrm{A}2$, and $\mathrm{C}2$ ) are also occupied.
It is known that Row 1 has two occupied houses, one in each block. Since B1 is already occupied, it implies A1,
and C1 are vacant.


Hence, the configuration of block XX is given below: (Where U = Unoccupied/ Vacant, and U = Occupied)Now for block YY, we know that both houses in Column E are vacant. Each of Column-D and Column-F has at least one occupied house. There is only one house with parking space in Block YY.

It is also known that the minimum quoted price of a house in block YY is Rs. 15 lakhs, and one such house is in Column E.
Case 1: The minimum quoted house is E2:
We know that the road adjacency of E2 is 1, hence we can calculate whether the house has parking space or not, and the neighbor count (b)
If the house has parking space, then: $12+5*1+3*\mathit{b}=15\Rightarrow 3\mathit{b}=-2$ (which is not possible)
Hence, the house has no parking space $\Rightarrow 10+5^{\star }1+3\mathit{b}=15\Rightarrow \mathit{b}=0$
$\mathit{b}=0$ implies all the neighbor house of $\mathit{E}2$ is vacant, which are (E1, D2, and F2).
It is known that each of Column-D and Column-F has at least one occupied house, which implies D1, and F1 must be occupied.
But D1 and F1 can't be occupied together since the total number of occupied houses in Row 1 is 2 (one in each block).
Hence, This case is invalid.
Case 2: The minimum quoted house is E1:
We know that the road adjacency of E1 is 0 , hence we can calculate whether the house has parking space or not, and the neighbor count (b).
i) If the house has no working space, then: $10+5^{\star }0+3\mathit{b}=15\Rightarrow \mathit{b}=5∕3$ (this is not possible since $\mathit{b}$ has to be an integer value)
Hence, the house has parking space $\Rightarrow 12+5\star 0+3\mathit{b}=15\Rightarrow \mathit{b}=1\Rightarrow$ One neighbor house is occupied among D1 and $\mathrm{F}1$.
We already know that E2 is vacant. Among the houses D2, and F2, at least one must be occupied since each of Column-D and Column-F has at least one occupied house.
Therefore, the final diagram is given below:
From the diagram, we can see that B1 is definitely occupied. The rest opinions are not definitely correct.
The correct option is C