CAT 2023 Slot 1 — DILR Question 11

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.

How many houses are vacant in Block XX?

Answer & solution

Answer: 3

Solution

Easy

One careful set-up unlocks all five questions. The price formula is the key lever: $\text{price} = \text{base} + 5\times(\text{road adjacency}) + 3\times(\text{neighbour count})$ , with base $=10$ (no parking) or $12$ (parking). Pin down the extreme-price houses first, then propagate the certainties.

The complex has 12 houses in 6 columns (A–F) and 2 rows. Block XX = columns A, B, C; Block YY = columns D, E, F. One road runs in front of Row-2 (the bottom edge) and another runs vertically between the two blocks (between C and D).

Road adjacency (sides touching a road): bottom-row houses touch the front road; D2 and C2 also touch the middle road. So $A2{=}1,\ B2{=}1,\ C2{=}2,\ D2{=}2,\ E2{=}1,\ F2{=}1$ ; in Row-1 only $C1{=}1$ and $D1{=}1$ (middle road), while $A1{=}B1{=}E1{=}F1{=}0$ .

Crack Block XX from its max price 24. The most a house can be worth is base $+5\times(\text{adjacency})+3\times(\text{neighbours})$ . To reach $24$ :

24 = 10 + 5\times 1 + 3\times 3

This is the only combination: base $=10$ , adjacency $=1$ , neighbour count $=3$ . A neighbour count of $3$ needs a house with three same-block neighbours — only $B2$ (touching A2, C2, B1) qualifies, and $B2$ has adjacency $1$ . So $B2$ is vacant, priced 24, and its three neighbours $A2,\ C2,\ B1$ are all occupied.

Fix the rest of XX using Fact 2. Row-1 has exactly two occupied houses, one per block. In XX that occupied Row-1 house is $B1$ (already known occupied), so $A1$ and $C1$ are vacant.

\boxed{\text{XX vacant} = A1,\ B2,\ C1 \qquad \text{XX occupied} = A2,\ B1,\ C2}

Block YY: locate the parking house via E1. Both column-E houses are vacant (Fact 3). $E1$ has adjacency $0$ and neighbour count $1$ (exactly one of $D1,F1$ occupied, by Fact 2). So its price is $10+3=13$ or $12+3=15$ . Since the YY minimum is $15$ and is achieved in column E, $E1=15$ , forcing base $=12$ . Hence $E1$ is the single parking house in YY (Fact 4).

Settle D1 vs F1. Suppose $F1$ were vacant: it has no parking (only $E1$ does), adjacency $0$ , neighbour count $\le 1$ , so price $\le 10+3=13 < 15$ — impossible. So $F1$ is occupied, making $D1$ the vacant Row-1 house in YY. By Fact 3 column D still needs an occupant, so $D2$ is occupied. $F2$ may be either.

\text{YY vacant} = D1,\ E1,\ E2\ (+ \text{maybe } F2);\quad \text{occupied} = D2,\ F1\ (+\text{maybe }F2)

Block XX has exactly the three vacant houses $A1$ , $B2$ , $C1$ .

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