CAT 2023 Slot 1 · DILR · Question 8
CAT 2023 Slot 1 previous year question on Caselet DI, with the answer and a step-by-step solution.
Set 2: Houses for Sale in Blocks XX and YY — this question belongs to a set. Read the full set with all 5 questions →
The schematic diagram below shows 12 rectangular houses in a housing complex. House numbers are mentioned in the rectangles representing the houses. The houses are located in six columns - Column-A through Column-F, and two rows - Row-1 and Row-2. The houses are divided into two blocks - Block XX and Block YY. The diagram also shows two roads, one passing in front of the houses in Row-2 and another between the two blocks.
Some of the houses are occupied. The remaining ones are vacant and are the only ones available for sale.
The road adjacency value of a house is the number of its sides adjacent to a road. For example, the road adjacency values of C2, F2, and B1 are 2, 1, and 0, respectively. The neighbour count of a house is the number of sides of that house adjacent to occupied houses in the same block. For example, E1 and C1 can have the maximum possible neighbour counts of 3 and 2, respectively.
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 × (road adjacency value) + 3 × (neighbour count). The following information is also known.
1. The maximum quoted price of a house in Block XX is Rs. 24 lakhs. The minimum quoted price of a house in block YY is Rs. 15 lakhs, and one such house is in Column-E.
2. Row-1 has two occupied houses, one in each block.
3. Both houses in Column-E are vacant. Each of Column-D and Column-F has at least one occupied house.
4. There is only one house with parking space in Block YY.
- AExactly 3
- BEither 3 or 4
- CExactly 2
- DEither 2 or 3
Answer & solution
Answer: Option D
Solving the set
Step 1 – Road adjacency of every house (from the diagram) and the most neighbours each house can have in its own block:
House | A1 | B1 | C1 | A2 | B2 | C2 | D1 | E1 | F1 | D2 | E2 | F2 |
Road adjacency | 0 | 0 | 1 | 1 | 1 | 2 | 1 | 0 | 0 | 2 | 1 | 1 |
Max neighbours | 2 | 3 | 2 | 2 | 3 | 2 | 2 | 3 | 2 | 2 | 3 | 2 |
Step 2 – The Rs. 24 lakh house in Block XX. A vacant house costs base + 5 × road + 3 × neighbours.
Base price | Needed 5 × road + 3 × neighbours | Possible (road, neighbours)? |
12 (parking) | 12 | None (road 0 needs 4 neighbours; road 1 or 2 leaves 7 or 2) |
10 (no parking) | 14 | Road 1 and 3 neighbours |
Among Block XX houses with road value 1 (C1, A2, B2), only B2 can have 3 neighbours. So B2 is vacant without parking, and its neighbours B1, A2 and C2 are occupied.
Step 3 – Row-1 has exactly one occupied house in each block (fact 2). In Block XX it is B1, so A1 and C1 are vacant. Block XX: vacant A1, B2, C1.
Step 4 – The Rs. 15 lakh house in Column E (fact 1). E1 and E2 are both vacant (fact 3).
• E2 = base + 5 + 3 × (occupied among D2, F2). For 15, base = 10 and D2, F2 both vacant. Then Column D and Column F would need D1 and F1 occupied, giving Block YY two occupied houses in Row-1 – not allowed. So E2 is not the 15-lakh house.
• E1 = base + 0 + 3 × (occupied among D1, F1). Exactly one of D1, F1 is occupied (fact 2), so E1 = base + 3. For 15, base = 12: E1 has parking. It is the only house with parking in Block YY (fact 4).
Step 5 – D1 or F1? If D1 were occupied and F1 vacant, F1 (no parking, road 0, neighbours E1 vacant and at most F2) would cost at most 10 + 3 = 13 < 15. Not allowed. So F1 is occupied and D1 is vacant; Column D then needs D2 occupied (fact 3).
Step 6 – F2 can be occupied or vacant: if vacant, it costs 10 + 5 + 3 = 18 ≥ 15, which is allowed. Final position:
House | Status | Quoted price if vacant (Rs. lakhs) |
A1 | Vacant | 10/12 + 0 + 3 × 2 = 16 or 18 |
B1 | Occupied | — |
C1 | Vacant | 10/12 + 5 + 3 × 2 = 21 or 23 |
A2 | Occupied | — |
B2 | Vacant | 10 + 5 + 3 × 3 = 24 (maximum) |
C2 | Occupied | — |
D1 | Vacant | 10 + 5 + 3 × 1 = 18 |
E1 | Vacant (parking) | 12 + 0 + 3 × 1 = 15 (minimum) |
F1 | Occupied | — |
D2 | Occupied | — |
E2 | Vacant | 10 + 5 + 3 × (1 or 2) = 18 or 21 |
F2 | Occupied or vacant | If vacant: 10 + 5 + 3 × 1 = 18 |
This question
Row-2: A2, C2, D2 are occupied; B2 and E2 are vacant; F2 may be occupied or vacant (Step 6).
So the number of vacant houses in Row-2 is 2 or 3.
Hence, the answer is option 4.
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