CAT 2023 Slot 1 — DILR Question 14

Linear ArrangementEasy

Passage / Data

Answer the following questions based on the information given below:

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.

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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?

Answer & solution

Answer: 21

Solution

Easy

Column-E vacancies are $E1$ and $E2$ . $E1$ is already fixed at $15$ , so push $E2$ to its maximum.

$E2$ cannot have parking ( $E1$ is YY's only parking house), so its base is $10$ . $E2$ sits in Row-2 with adjacency $1$ . Its neighbours in YY are $D2$ , $F2$ , $E1$ .

Maximise $E2$ 's neighbour count. $E1$ is vacant (doesn't count), but $D2$ is occupied and $F2$ can be occupied. So the maximum neighbour count is $2$ .

Plug in the maxima.

\text{price}(E2) = 10 + 5\times 1 + 3\times 2 = 10 + 5 + 6 = 21

This beats $E1=15$ , so $21$ is the column's maximum.

\mathbf{21}\ \text{lakhs}