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.
How many houses are vacant in Block XX?
Solution and Explanation
Provided Information:
Some houses are occupied, while others are vacant and available for purchase.
- Base price: Rs. 10 lakhs (no parking), Rs. 12 lakhs (with parking).
- Final price calculation: Base price + 5 * (road adjacency value) + 3 * (neighbor count).
The most expensive house in Block XX is priced at Rs. 24 lakhs.
Scenario 1: House with parking
Equation: \(12 + 5a + 3b = 24\), simplifying to \(5a + 3b = 12\).
If \(a = 0\), then \(3b = 12\), yielding \(b = 4\). This is not possible as the maximum number of neighbors is 3.
Conclusion: No house with parking meets the criteria.
Scenario 2: House without parking
Equation: \(10 + 5a + 3b = 24\), simplifying to \(5a + 3b = 14\).
The valid solution is \((a, b) = (1, 3)\).
This implies the house has 3 neighbors and 1 road adjacency.
The only house matching these conditions is B2.
Neighbors of B2: B1, A2, C2. All these must be occupied.
However, Row-1 in Block XX has only one occupied house. Since B1 is occupied, A1 and C1 must be vacant.
Occupied houses in Block XX: B1, A2, B2, C2. Total = 4.
This contradicts the earlier deduction that only 3 houses can be occupied. Therefore, one of A2 or C2 must be vacant. However, for B2 to have 3 neighbors, A2, B1, and C2 must all be occupied.
Revised occupied houses in Block XX: 3 (A2, B1, C2)
Vacant houses in Block XX: 3
In Block YY:
- E1 and E2 are both vacant.
- One of them is priced at Rs. 15 lakhs.
Consider E1:
Neighbor count = 1 (either D1 or F1). Road adjacency = 0.
Price = (Rs. 10 or Rs. 12) + 0 + 3 * 1 = Rs. 13 or Rs. 15 lakhs.
Given the minimum price is Rs. 15 lakhs, E1 must have parking. This means the base price is Rs. 12 lakhs, and E1's price is Rs. 15 lakhs. Conclusion: E1 has parking.
Since Row-1 in Block YY has one occupied house and E1 is vacant, D1 or F1 must be occupied.
Assume F1 is vacant:
- F1 has no parking, so base price = Rs. 10.
- Maximum neighbors = 1 (possibly F2). Road adjacency = 0.
- Price = Rs. 10 + 0 + 3 * 1 = Rs. 13 lakhs.
This is invalid as the minimum price is Rs. 15 lakhs.
Therefore, F1 must be occupied, which means D1 is vacant.
Since Column D must have one occupied house, D2 is occupied.
F2 can be either occupied or vacant.
Final Outcome:
Number of vacant houses in Block XX = 3
Answer: 3
Which of the following houses is definitely occupied?
- D2
- A1
- B1
- F2
The Correct Option is C
Solution and Explanation
To identify the definitively occupied house, let's analyze the given conditions:
1. Rows 1 and 2 each contain two occupied houses, one in each block:
This means that Block XX and Block YY each have at least one occupied house in Row 1 and at least one in Row 2.
2. Both houses in Column E are vacant:
This explicitly states that houses E1 and E2 in Block YY are not occupied.
3. Columns D and F each have at least one occupied house:
This confirms that within Block YY, either D1 or D2 is occupied, and either F1 or F2 is occupied.
Now, let's evaluate each option:
- Option 1: A1 — No information directly supports A1's occupancy. Incorrect.
- Option 2: B1 — Located in Column B of Block XX. Condition 1 implies Block XX, Row 1 has an occupied house. B1 is the most plausible candidate given the constraints. Correct answer.
- Option 3: D2 — Located in Column D of Block YY. Condition 3 mandates an occupied house in Column D, but it could be D1 or D2. Certainty about D2 is lacking. Incorrect.
- Option 4: F2 — Located in Column F of Block YY. Condition 3 mandates an occupied house in Column F, but it could be F1 or F2. Certainty about F2 is lacking. Incorrect.
Therefore, the only house that can be definitively confirmed as occupied is: $B1$.
Which of the following options best describes the number of vacant houses in Row-2?
- Either 2 or 3
- Exactly 3
- Exactly 2
- Either 3 or 4
The Correct Option is A
Solution and Explanation
Result: (A) 2 or 3
Justification:
- Given: B2 and E2 are unoccupied.
- Condition: At least one of D2 or F2 is occupied.
Therefore, the potential states for D2 and F2 are:
- Scenario 1: One of D2 or F2 is unoccupied. → Total unoccupied houses in Row 2 = B2 + E2 + (D2 or F2) = 3
- Scenario 2: Both D2 and F2 are occupied. → Total unoccupied houses in Row 2 = B2 + E2 = 2
Consequently, Row 2 can have 2 or 3 unoccupied houses.
What is the maximum possible quoted price (in lakhs of Rs.) for a vacant house in Column-E?
Solution and Explanation
Given: An empty house costs either Rs. 10 lakhs (no parking) or Rs. 12 lakhs (with parking).
In Block YY, both E1 and E2 are vacant, and one of them is priced at Rs. 15 lakhs. Let's analyze E1:
- Neighbor count = 1 (either D1 or F1 is occupied)
- Road adjacency = 0
- Cost of E1 = \((10 \text{ or } 12) + 5 \times 0 + 3 \times 1 = 13 \text{ or } 15\) lakhs
Since Rs. 15 lakhs is the known cost of one vacant house, and 15 is only achievable with a base price of 12 (indicating parking), E1 must include parking.
Now for E2:
- Base price = Rs. 10 lakhs (no parking)
- Road adjacency = 1
- Maximum neighbors = 2 (D2 and F2 occupied; E1 is vacant)
- Price of E2 = \(10 + 5 \times 1 + 3 \times 2 = 21\) lakhs
Therefore, the maximum possible price of E2 is Rs. 21 lakhs.
Which house in Block YY has parking space?
Show Hint
In the solution to the first question of this set, where only one house in YY, E1 has a parking space
So, the correct option is (C): E1
- E2
- F2
- E1
- F1
The Correct Option is C
Solution and Explanation
To identify the house in Block YY with parking, we examine the following data:
- Block YY contains precisely one house with a parking space.
- Column E lists all vacant houses in Block YY.
- Consequently, the house with parking is either E1 or E2.
- As there is no indication that F1 or F2 possess parking, and given the single-house parking constraint for Block YY, E1 is the most probable option.
Accordingly, E1 is determined to be the house in Block YY with parking space.
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