CAT 2023 Slot 1 · DILR · Set 2: Houses for Sale in Blocks XX and YY
The full set and all 5 questions (Q6–Q10) from CAT 2023 Slot 1, with answers and solutions.
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.
Question 6 · Caselet DI
Answer & solution
Answer: 3
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
From Step 3, the vacant houses in Block XX are A1, B2 and C1.
Hence, the answer is 3.
Question 7 · Caselet DI
- AA1
- BD2
- CB1
- DF2
Answer & solution
Answer: Option B
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
From the final table, B1, A2, C2, F1 and D2 are definitely occupied, A1 is vacant, and F2 may be either.
Among the options, both D2 (option 2) and B1 (option 3) are definitely occupied; the question has two correct options. Option 2 is entered as the key here; option 3 is equally correct.
Hence, the answer is option 2.
Question 8 · Caselet DI
- 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.
Question 9 · Caselet DI
Answer & solution
Answer: 21
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
Column E: E1 costs 15 (Step 4). E2 = 10 + 5 + 3 × (occupied neighbours among D2 and F2). D2 is occupied; F2 may also be occupied, giving 2 neighbours.
Maximum E2 = 10 + 5 + 6 = 21.
Hence, the answer is 21.
Question 10 · Caselet DI
- AE1
- BF2
- CE2
- DF1
Answer & solution
Answer: Option A
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
From Step 4, E1 must have parking to reach exactly Rs. 15 lakhs, and only one house in Block YY has parking.
Hence, the answer is option 1.
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