Column | House | Occupied |
---|---|---|
A | A1, A2 | No occupancy info |
B | B1, B2 | No occupancy info |
C | C1, C2 | No occupancy info |
D | D1, D2 | At least one |
E | E1, E2 | Both vacant |
F | F1, F2 | At least one |
Given:
Some houses are already taken, but others are empty and available for sale. The base price for an empty house is either Rs. 10 lakhs if it doesn't have a parking spot, and Rs. 12 lakhs if it does have a parking spot.
So, the price of a vacant house is calculated as:
Price = Base price + 5 × (road adjacency value) + 3 × (neighbor count).
Now calculate for the maximum price of a house in Block XX.
Case 1: House with parking lot
Let road adjacency value be $a$, and neighbor count be $b$.
Then,
$12 + 5a + 3b = 24$
$\Rightarrow 5a + 3b = 12$
Try integer solutions:
If $a = 0$, then $b = 4$, which is invalid since the maximum number of neighbors is 3.
So, this case is not possible.
Case 2: House without parking lot
$10 + 5a + 3b = 24$
$\Rightarrow 5a + 3b = 14$
Try integer values: if $(a, b) = (1, 3)$, it satisfies the equation.
So the house must have 3 neighbors and 1 road adjacent.
The only house in Block XX satisfying this is B2.
Its neighbors: B1, A2, and C2 must be occupied.
Now, we know that Row 1 has two occupied houses, one in each block.
Since B1 is occupied, the other occupied house in Row 1 must be in Block YY.
Hence, A1 and C1 must be vacant.
So, the total occupied houses in Block XX are: B1, A2, C2 → 3 houses.
Correct option: (C) 3
The following histogram represents: