Question

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.
What is the maximum possible quoted price (in lakhs of Rs.) for a vacant house in Column-E? [This question was asked as TITA]

• A. 20 lakhs
• B. 22 lakhs
• C. 23 lakhs
• D. 21 lakhs

Answer 1

The Correct Option is D

To solve the problem, we need to determine the configuration of the 12 houses and then calculate the maximum possible quoted price for a vacant house in Column-E. We'll follow these steps:

1.  Identify the conditions from the problem:
   • Block XX has a maximum quoted price of Rs. 24 lakhs.
   • Block YY has a minimum quoted price of Rs. 15 lakhs, with one such house in Column-E.
   • Row-1 has two occupied houses, one in each block.
   • Both houses in Column-E are vacant.
   • Each of Column-D and Column-F has at least one occupied house.
   • Block YY has only one house with parking space.
2. Assign base prices based on parking space:
   • Without parking: Rs. 10 lakhs
   • With parking: Rs. 12 lakhs
3. Apply the formula for the quoted price:
   $Q = \text{Base Price} + (5 \times \text{Road Adjacency Value}) + (3 \times \text{Neighbour Count})$
4. Determine the maximum possible road adjacency and neighbour count for the house in Column-E, Block YY:
   • Road adjacency: Since both houses in Column-E are vacant, consider maximal adjacency possible. Road adjacency for any house in Column-E, Row-2 would be 2 (one front, one between blocks).
   • Neighbour count: Calculate assuming maximum adjacency with occupied houses. With two vacant houses in Column-E, assume maximal interaction with occupied houses in Row-1 or adjacent columns like Column-D.
5. Calculate the quoted price for a vacant house in Column-E, Block YY:
   • Assume no parking, so base price = Rs. 10 lakhs.
   • Road adjacency value = 2, neighbour count = 2 (assumed from Column-D and Row-1).
   • Quoted price: $Q = 10 + (5 \times 2) + (3 \times 2) = 10 + 10 + 6 = 26$ lakhs
   • But this exceeds Block YY's maximum limit under the condition that one house in Column-E has Rs. 15 lakhs. Reevaluate under constraint.
6. Reevaluate to meet all constraints:
   • Reduce neighbour count or road adjacency to satisfy block constraint.
   • Try: Base price = 10, Road adjacency = 2, Neighbour count = 1:
     $Q = 10 + 10 + 3 = 23$ lakhs
   • Try: Base price = 10, Road adjacency = 2, Neighbour count = 0:
     $Q = 10 + 10 + 0 = 20$ lakhs
   • Try: Base price = 10, Road adjacency = 1, Neighbour count = 2:
     $Q = 10 + 5 + 6 = 21$ lakhs
   • This satisfies all conditions and stays under Block XX’s 24-lakh max. Since the house is in Block YY, this is acceptable.

Therefore, the maximum possible quoted price for a vacant house in Column-E, Block YY is Rs. 21 lakhs.

Answer 2

Given: The price for an empty house is either Rs. 10 lakhs if it doesn't have a parking spot or Rs. 12 lakhs if it does.

In Block YY, both E1 and E2 are vacant, and one of them costs 15 lakhs. Let's focus on E1.

For E1: 
Neighbor count = 1 (exactly one of D1 or F1 is occupied)
Road adjacency = 0
So, the cost of E1 would be: 
$ (10 \text{ or } 12) + 5 \times 0 + 3 \times 1 = 13 \text{ or } 15 $ lakhs.
Since 15 lakhs is the lowest cost for a house in Block YY, E1 must cost 15 lakhs. This implies E1 is the only house in YY with a parking space.

Now, let's calculate the maximum possible price of E2.
E2's base price is 10 lakhs since it cannot have a parking space (only one house in YY, E1, has a parking space).
The road adjacency for E2 is 1, and the maximum neighbor count of E2 will be 2 (both D2 and F2 are occupied, and E1 is vacant).
So, E2's price would be: 
$ 10 + 5 \times 1 + 3 \times 2 = 21 $ lakhs.

Hence, the maximum possible price for E2 is 21 lakhs.

Correct Option: (D) 21 lakhs.