Block Diagram

Understanding the set :- We are given 12 houses, whose names are given in the diagram. Also, some of the houses are occupied. The remaining ones are vacant. and the vacant ones are available for sale. And, Road adjacency value of a house = number of its sides touching road. And, The neighbour count of a house = number of sides of that house touching to an occupied houses. Also the question gives us the base price of a vacant house = 10 lkhs(without parking) and 12 lkhs (with parking) price of a vacant house is calculated as = (base price) + 5 × (road adjacency value) + 3 × (Neighbour count). In the set at last we need to find which houses are occupied and which ones are vacant. —-- Block Diagram

Block XX
Column-A	Column-B	Column-C
A1	B1	C1
A2	B2	C2

ROAD

Block YY
Column-D	Column-E	Column-F
D1	E1	F1
D2	E2	F2

ROAD

According to clue 1, The maximum price of a house in Block XX is Rs. 24 lakhs For a maximum of 24 lakhs there are only 2 possibilities :- CASE -1 - House with parking space: => (base price) + 5 × (road adjacency value) + 3 × (Neighbour count). => 12+5a+3b = 24 => 5a+3b = 12 where, a = road adjacency value and, b= neighbor count The only value for which the equation satisfies is (a = 0, and b=4). But the value of b can't be 4 because the maximum neighbor count can be at most 3. Hence, case 1 is invalid —-- Block Diagram

Block XX
Column-A	Column-B	Column-C
A1	B1 (occupied)	C1
A2 (occupied)	B2	C2 (occupied)

ROAD

Block YY
Column-D	Column-E	Column-F
D1	E1	F1
D2	E2	F2

ROAD

Case 2: House without parking space: => 10+5a+3b = 24 => 5a+3b = 14 => (a, b) = (1, 3) 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 B2, which are (B1, A2, and C2) are occupied. —-- Block Diagram

Block XX
Column-A	Column-B	Column-C
A1 (vacant)	B1 (occupied)	C1 (vacant)
A2 (occupied)	B2 (vacant)	C2 (occupied)

ROAD

Block YY
Column-D	Column-E	Column-F
D1	E1	F1
D2	E2	F2

ROAD

Using clue 2 - Row-1 has two occupied houses, one in each block. 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. —-- Block Diagram

Block XX
Column-A	Column-B	Column-C
A1 (vacant)	B1 (occupied)	C1 (vacant)
A2 (occupied)	B2 (vacant)	C2 (occupied)

ROAD

Block YY
Column-D	Column-E	Column-F
D1	E1	F1
D2	E2	F2

ROAD

Now for block YY, using clue 3 and 4 - 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. —-- Block Diagram

Block XX
Column-A	Column-B	Column-C
A1 (vacant)	B1 (occupied)	C1 (vacant)
A2 (occupied)	B2 (vacant)	C2 (occupied)

ROAD

Block YY
Column-D	Column-E	Column-F
D1	E1	F1
D2	E2	F2

ROAD

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*b = 15 => 3b =-2 (which is not possible) Hence, the house has no parking space => 10+5*1+3b = 15 => b = 0 b = 0 implies all the neighbor house of E2 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 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. —-- Block Diagram

Block XX
Column-A	Column-B	Column-C
A1 (vacant)	B1 (occupied)	C1 (vacant)
A2 (occupied)	B2 (vacant)	C2 (occupied)

ROAD

Block YY
Column-D	Column-E	Column-F
D1	E1	F1
D2	E2	F2

ROAD

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*0+3b = 15 => b = 5/3 (this is not possible since b has to be an integer value) Hence, the house has parking space => 12+5*0+3b = 15 => b = 1 => One neighbor house is occupied among D1 and F1 —-- Block Diagram

Block XX
Column-A	Column-B	Column-C
A1 (vacant)	B1 (occupied)	C1 (vacant)
A2 (occupied)	B2 (vacant)	C2 (occupied)

ROAD

Block YY
Column-D	Column-E	Column-F
D1	E1	F1 (occupied)
D2	E2	F2

ROAD

Let's take the case of house D1 being occupied and F1 being empty. In that case, the value of house F1 would be 10(there is no parking space)+ (5*0) + (3*the number of neighbours) Here, even if we take the number of neighbors to be 1, which is the maximum for F1 in this case, the value of F1 would be a maximum of 13. This is lower than the lowest-value house in block YY. Therefore, F1 cannot be empty. —-- Block Diagram

Block XX
Column-A	Column-B	Column-C
A1 (vacant)	B1 (occupied)	C1 (vacant)
A2 (occupied)	B2 (vacant)	C2 (occupied)

ROAD

Block YY
Column-D	Column-E	Column-F
D1 (vacant)	E1 (vacant)	F1 (occupied)
D2 (occupied)	E2 (vacant)	F2

ROAD

Since F1 is occupied and we know that there is only one house occupied in row 1 of each block, D1 becomes unoccupied. D2 becomes occupied because it is given in the question that each of Column-D and Column-F has at least one occupied house. Here, the value of D1 is 18 as D2 is occupied.

CAT (DILR) PYQs

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