Seating Arrangement - Free MCQ Practice Test with solutions, CAT Logical

[Note: There might be discrepancy in one of the questions of this set. We will update the solution once the objections window closes and answers are updated by IIM Lucknow.]Price of a house = (base price)  + 5 × (road adjacency value) + 3 × (neighbor count)Block XX:
Maximum price of a house in XX is 24
∴ 24 = (10 or 12)  + 5 × (0 or 1 or 2) + 3 × (0 or 1 or 2 or 3)
This is only possible when: 
base price = 10 lakhs, road adjacency = 1 and neighbor count = 3.
⇒ Only house B2 satisfies the given criteria i.e., road adjacency = 1 and neighbor count = 3.
∴ B2 is vacant and its price is 24 lakhs.⇒ A2, B1 and C2 are occupied houses.There is only one occupied house in Row 1 and block XX.
Since B1 is occupied, ⇒ A1 and C1 are vacant.

Block YY:
Both E1 and E2 are vacant and one of them costs 15 lakhs.Let us focus on E1
For E1 neighbor count = 1 (exactly one of D1 or F1 is occupied)
For E1 road adjacency = 0
∴ Cost of E1 = (10 or 12) + 5 × 0 + 3 × 1 = 13 or 15 lakhsSince 15 lakhs is the least cost of a house in block YY, E1 must cost 15 lakhs.
⇒ E1 is the only house in YY which has a parking space.Given, Row-1 has two occupied houses, one in each block.
∴ Exactly one of D1 or F1 is occupied.If F1 is vacant, let’s calculate its price.
It should not have parking space as only 1 house has parking space in block YY. Hence, it’s base price is 10 lakhs. Now even if F2 is occupied, F1’s price will be 10 + 0 + 3 = 13 lakhs.
This is not possible as least price for a house in YY is 15 lakhs.
∴ F1 should be occupied.
⇒ D1 is vacant.Also, at least one house in column D is occupied, hence D2 must be occupied.Now, F2 may be vacant or occupied, both cases are possible.

∴ 3 houses are vacant in block XX.
Hence, 3.

Consider the solution to first question of this set.
B1 and D2 are definitely occupied.
[Two options are correct in this question and both were accepted as correct answer by IIM Lucknow]
Hence, option (a) and (c).

Consider the solution to first question of this set.
In row 2, B2 and E2 are definitely vacant.
Out of D2 and F2, at least one is occupied, hence either 1 is vacant or 0 are vacant.
Case 1: One of D2 or F2 is vacant.
∴ We have B2, F2 and (D2 or F2), i.e., 3 houses vacant in row 2.
Case 2: None of D2 or F2 is vacant.
∴ We have B2 and F2, i.e., 2 houses vacant in row 2.
∴ In row 2, either only 2 houses are vacant or 3 houses are vacant.
Hence, option (d).

Consider the solution to first question of this set.E1 costs 15 laks.
Let us calculate the maximum possible price of E2.
E2's base price is 10 lakh as it cannot have a parking space. (Only 1 house in YY i.e., E1 has a parking space.)
Road adjancecy for E2 = 1 and maximum neighbor count of E2 will be 2 (Both D2 and F2 are occupied and E1 is vacant)
∴ E2's price = 10 + 5 × 1 + 3 × 2 = 21 lakhs
Hence, 21.

Consider the solution to first question of this set.
Only 1 house in YY i.e., E1 has a parking space.
Hence, option (c).