Seating, Positioning and Arrangement Logic Previous Year Questions (PYQs) for CAT: 16+ Solved Questions with Step-by-Step Solutions

    Solve 16+ Seating, Positioning and Arrangement Logic previous year questions for CAT with answers and detailed solutions. Free sample questions below.

    Chapter Roadmap: Seating, Positioning and Arrangement Logic

    Chapter Journey
    1
    Grid and Slot Placement
    Place items in rows and columns using constraints. Foundation topic.
    Weight: 37% | 5 PYQs
    2
    Circular Seating and Passing Movement
    People around a round table passing objects in rounds.
    Weight: 47% | 8 PYQs (Heaviest)
    3
    House Layout and Positional Blocks
    Fixed schematic maps with houses in columns and rows.
    Weight: 37% | 5 PYQs
    End goal: Decode any arrangement set, draw the framework in under 2 minutes, and answer 4 to 5 questions per set with high accuracy.

    What is Grid and Slot Placement?

    The Core Setup

    Every grid placement problem has exactly three parts:

    Element What it is Example
    Grid A fixed structure of rows and columns creating slots A 4 by 4 table = 16 slots
    Items Things to be placed (numbers, people, objects) Numbers 1 to 10
    Conditions Rules restricting where items can go 5 is in Row 2

    The Simple Intuition

    Think of it as a constraint satisfaction puzzle:

    • The grid tells you where things can go.
    • The items tell you what needs to be placed.
    • The conditions tell you how to restrict placement.
    Key Insight: Some slots may be blocked or missing (like a staircase grid). Always count the available slots before you start.

    Seating, Positioning and Arrangement Logic: Solved Questions with Step-by-Step Explanations (5 Problems)

    Question 1 · Data Interpretation and Logical Reasoning MCQ
    Common Description: Seven children, Aarav, Bina, Chirag, Diya, Eshan, Farhan, and Gaurav, are sitting in a circle facing inside (not necessarily in the same order) and playing a game of 'Passing the Buck'.
    The game is played over 10 rounds. In each round, the child holding the Buck must pass it directly to a child sitting in one of the following positions:
    • Immediately to the left;
    • Immediate to the right;
    • Second to the left; or
    • Second to the right.
    The game starts with Bina passing the Buck and ends with Chirag receiving the Buck. The table below provides some information about the pass types and the child receiving the Buck. Some information is missing and labelled as '?'.
    RoundPass TypeReceived by
    1Immediately to the leftAarav
    2Second to the right?
    3Immediately to the rightDiya
    4??
    5?Aarav
    6Second to the left?
    7Immediately to the leftGaurav
    8Immediately to the left?
    9?Farhan
    10?Chirag
    For which of the following children is it possible to determine how many times they received the Buck?
    1. A.

      Farhan

    2. B.

      Eshan

    3. C.

      Bina

    4. D.

      Gaurav

    Correct Answer:

    D

    Step-by-Step Solution

    Key idea: This is a circular passing movement question with modular arithmetic and constraint propagation. We assign numerical positions to the circle and trace the buck's path to find the exact seating arrangement, then check which person's reception count is invariant.

    Step 1: Set up coordinates. Let the 7 positions be 0 to 6 in clockwise order. Since they face inside, Left = +1 (clockwise) and Right = -1 (counter-clockwise). 2L = +2, 2R = -2.

    Step 2: Trace the buck from Bina. Let Bina be at position 0.

    • R1: 1L from 0 position 1. Receiver is Aarav. So Aarav = 1.
    • R2: 2R from 1 1 - 2 = -1 6.
    • R3: 1R from 6 6 - 1 = 5. Receiver is Diya. So Diya = 5.
    • R6: Aarav (1) passes 2L 1 + 2 = 3.
    • R7: 1L from 3 3 + 1 = 4. Receiver is Gaurav. So Gaurav = 4.
    • R8: 1L from 4 4 + 1 = 5. Receiver is Diya.

    Step 3: Determine Farhan and Chirag.

    • R8 receiver is Diya (5). R9 passes to Farhan. Valid targets from 5 are 6, 4, 0, 3. Since 4 is Gaurav and 0 is Bina, Farhan must be 6 or 3.
    • If Farhan = 6: R10 passes to Chirag. Valid targets from 6 are 0, 5, 1, 4. None of these is Chirag (who must be 2 or 3). Contradiction.
    • Thus, Farhan = 3.
    • R10 passes to Chirag. Valid targets from 3 are 4, 2, 5, 1. The only unassigned position is 2. So Chirag = 2.

    Step 4: Eshan is the last unassigned position: 6.

    Step 5: Count receptions and check invariance.

    • Fixed receptions: Aarav (R1, R5), Eshan (R2), Diya (R3, R8), Farhan (R6, R9), Gaurav (R7), Chirag (R10).
    • R4 receiver is from Diya (5). Targets: 6, 4, 0, 3.
    • However, R5 receiver is Aarav (1). The pass from R4 to R5 must reach 1. The positions that can reach 1 are 0, 2, 6, 3.
    • Intersecting Diya's targets {6, 4, 0, 3} with valid passers to 1 {0, 2, 6, 3} eliminates 4 (Gaurav).
    • So R4 receiver is Eshan, Bina, or Farhan. Gaurav cannot receive the buck in R4.
    • Therefore, Gaurav's total reception count is exactly 1. All others have uncertain counts.

    Answer: Gaurav (Option D).

    Question 2 · Data Interpretation and Logical Reasoning MCQ
    Common Description: Seven children, Aarav, Bina, Chirag, Diya, Eshan, Farhan, and Gaurav, are sitting in a circle facing inside (not necessarily in the same order) and playing a game of 'Passing the Buck'.
    The game is played over 10 rounds. In each round, the child holding the Buck must pass it directly to a child sitting in one of the following positions:
    • Immediately to the left;
    • Immediate to the right;
    • Second to the left; or
    • Second to the right.
    The game starts with Bina passing the Buck and ends with Chirag receiving the Buck. The table below provides some information about the pass types and the child receiving the Buck. Some information is missing and labelled as '?'.
    RoundPass TypeReceived by
    1Immediately to the leftAarav
    2Second to the right?
    3Immediately to the rightDiya
    4??
    5?Aarav
    6Second to the left?
    7Immediately to the leftGaurav
    8Immediately to the left?
    9?Farhan
    10?Chirag
    Who is sitting third to the left of Eshan?
    1. A.

      Gaurav

    2. B.

      Divya

    3. C.

      Chirag

    4. D.

      Aarav

    Correct Answer:

    C

    Step-by-Step Solution

    Key idea: This is a circular seating reconstruction question using the passing mechanics from the previous round. We use modular arithmetic to find the exact position of every child, then calculate the relative position.

    Step 1: Set up coordinates. Let the 7 positions be 0 to 6 in clockwise order. Left = +1, Right = -1.

    Step 2: Trace the buck from Bina (position 0).

    • R1: 1L from 0 1. Receiver is Aarav. So Aarav = 1.
    • R2: 2R from 1 1 - 2 = 6.
    • R3: 1R from 6 5. Receiver is Diya. So Diya = 5.
    • R6: Aarav (1) passes 2L 1 + 2 = 3.
    • R7: 1L from 3 4. Receiver is Gaurav. So Gaurav = 4.
    • R8: 1L from 4 5. Receiver is Diya.

    Step 3: Determine Farhan and Chirag.

    • R8 receiver is Diya (5). R9 passes to Farhan. Valid targets from 5: 6, 4, 0, 3.
    • 4 is Gaurav, 0 is Bina. So Farhan is 6 or 3.
    • If Farhan = 6: R10 passes to Chirag. Targets from 6: 0, 5, 1, 4. None is Chirag. Contradiction.
    • Thus, Farhan = 3.
    • R10 passes to Chirag. Targets from 3: 4, 2, 5, 1. Only 2 is unassigned. So Chirag = 2.

    Step 4: Determine Eshan.

    • The only unassigned position is 6. So Eshan = 6.

    Step 5: Calculate the required position.

    • Question: Who is 3rd to the left of Eshan?
    • Eshan is at 6. Left is +1.
    • 3rd to left = 6 + 3 = 9.
    • Modulo 7: .
    • Position 2 is Chirag.

    Answer: Chirag (Option C).

    Question 3 · Data Interpretation and Logical Reasoning NAT
    Common Description: Instructions [30 - 34]
    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.
    BlockColumn-AColumn-BColumn-CRoadColumn-DColumn-EColumn-F
    Row-1A1B1C1ROADD1E1F1
    Row-2A2B2C2ROADD2E2F2
    BlockBlock XXROADBlock YY
    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?
    Correct Answer:

    21

    Step-by-Step Solution

    Key idea: This is a constrained maximization question. We must maximize the quoted price of a vacant house in Column E, but every choice must still satisfy the global minimum-price, occupancy and parking constraints.

    Why: The question asks for the "maximum possible quoted price", which means we need to find the valid configuration that yields the highest value for the target house.

    Step 1: Recall the price formula.

    Price = Base + road adjacency + neighbour count.

    Base is 10 without parking and 12 with parking.

    Step 2: Identify the Column E houses. Both E1 and E2 are vacant.

    Road adjacency values:

    E1 = 0, because it is not adjacent to either road.

    E2 = 1, because it is adjacent to the bottom road.

    Step 3: Use the minimum-price condition. The minimum quoted price in Block YY is 15, and one such house is in Column E.

    Step 4: Check whether E2 can be the 15-price house.

    E2 has road adjacency 1. For price 15 with base 10:

    gives , so neighbour count must be 0.

    E2's neighbours are E1, D2 and F2. Since E1 is vacant, neighbour count 0 would require both D2 and F2 to be vacant.

    But Column D and Column F each must have at least one occupied house. If D2 and F2 are vacant, then D1 and F1 must both be occupied.

    That would give two occupied houses in Row 1 of Block YY, violating the rule that Row 1 has only one occupied house in Block YY.

    Therefore E2 cannot be the 15-price house.

    Step 5: Therefore E1 must be the Column E house with price 15.

    E1 has road adjacency 0. For price 15:

    With base 10, gives , impossible.

    With base 12, gives , so N=1.

    Thus E1 has the parking space, and its price is fixed at 15.

    Step 6: Maximize E2's price.

    E2 is vacant, has no parking (Base 10), and R=1.

    Price = .

    To maximize this, we need to maximize N (neighbour count).

    Neighbours of E2: E1, D2, F2.

    E1 is vacant, so it doesn't count.

    D2 is definitely occupied (derived from D1 being vacant).

    F2 can be occupied or vacant.

    To maximize N, we set F2 to be occupied.

    Then N = 2 (D2 and F2).

    Max Price = .

    Answer: 21

    Question 4 · Data Interpretation and Logical Reasoning MCQ
    Common Description: Instructions [30 - 34]
    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.
    BlockColumn-AColumn-BColumn-CRoadColumn-DColumn-EColumn-F
    Row-1A1B1C1ROADD1E1F1
    Row-2A2B2C2ROADD2E2F2
    BlockBlock XXROADBlock YY
    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. Which of the following options best describes the number of vacant houses in Row-2?
    1. A.

      Exactly 3

    2. B.

      Either 3 or 4

    3. C.

      Exactly 2

    4. D.

      Either 2 or 3

    Correct Answer:

    D

    Step-by-Step Solution

    Key idea: This is a house-layout pricing question. We reverse-engineer the pricing formula to identify occupied and vacant houses, then use row and column occupancy rules to determine the possible number of vacant houses in Row 2.

    Why: The problem gives a formula for the quoted price and states the maximum price in Block XX is 24 and minimum in YY is 15. We must deduce the occupancy of each house to count the vacant ones in Row 2.

    Step 1: Understand the price formula. For a vacant house,

    Price = Base + road adjacency + neighbour count.

    Base is 10 without parking and 12 with parking.

    Step 2: Use Block XX maximum price of 24. In Block XX, the only house that can reach 24 is B2. To reach exactly 24, B2 must have base 10, road adjacency 1, and neighbour count 3. Therefore B2 is vacant and has no parking, and its three neighbours B1, A2 and C2 must be occupied.

    Step 3: Apply Row 1 rule to Block XX. Row 1 has exactly one occupied house in Block XX. Since B1 is already occupied, A1 and C1 must be vacant. In Row 2 of Block XX, A2 and C2 are occupied while B2 is vacant. So Block XX contributes exactly one vacant house in Row 2 (which is B2).

    Step 4: Analyse Block YY. Both E1 and E2 are vacant. The minimum price in Block YY is 15, and one such house is in Column E.

    Step 5: What combinations can give price 15?

    If base is 10, then , which gives road adjacency 1 and neighbour count 0.

    If base is 12, then , which gives road adjacency 0 and neighbour count 1.

    Step 6: Column E possibilities. E2 has road adjacency 1, so it could be a 15-price house only with no parking and neighbour count 0. But E2 is adjacent to D2 and F2. For neighbour count 0, both D2 and F2 must be vacant. But Column D and Column F each must have at least one occupied house. If D2 and F2 are vacant, then D1 and F1 must both be occupied. That would give two occupied houses in Row 1 of Block YY, violating the rule that Row 1 has only one occupied house in Block YY. Therefore E2 cannot be the 15-price house.

    Step 7: Thus E1 must be the Column E house with price 15. E1 has road adjacency 0. For price 15, it must have base 12 (parking) and neighbour count 1. So E1 has the only parking space in Block YY.

    Step 8: Determine Row 1 YY occupancy. Row 1 YY has exactly one occupied house. The candidates are D1 and F1 (since E1 is vacant).

    If F1 were vacant, its price would be Base 10 + . Its only possible occupied neighbor is F2 (since E1 is vacant). So max , giving max price . This violates the minimum price of 15 in Block YY. Thus, F1 cannot be vacant. F1 must be the single occupied house in Row 1 YY.

    Step 9: Determine Row 2 YY occupancy. Since F1 is occupied, D1 must be vacant. Since D1 is vacant, Col D needs an occupied house, so D2 must be occupied. F2 can be either occupied or vacant (both satisfy all constraints).

    Step 10: Count vacant houses in Row 2.

    Row 2 has 6 houses: A2, B2, C2, D2, E2, F2.

    From XX: A2(O), B2(V), C2(O). (1 vacant)

    From YY: D2(O), E2(V), F2(O or V). (1 or 2 vacant)

    Total vacant in Row 2 = 1 + (1 or 2) = 2 or 3.

    Answer: D

    Question 5 · Data Interpretation and Logical Reasoning MCQ
    Common Description: Seven children, Aarav, Bina, Chirag, Diya, Eshan, Farhan, and Gaurav, are sitting in a circle facing inside (not necessarily in the same order) and playing a game of 'Passing the Buck'.
    The game is played over 10 rounds. In each round, the child holding the Buck must pass it directly to a child sitting in one of the following positions:
    • Immediately to the left;
    • Immediate to the right;
    • Second to the left; or
    • Second to the right.
    The game starts with Bina passing the Buck and ends with Chirag receiving the Buck. The table below provides some information about the pass types and the child receiving the Buck. Some information is missing and labelled as '?'.
    RoundPass TypeReceived by
    1Immediately to the leftAarav
    2Second to the right?
    3Immediately to the rightDiya
    4??
    5?Aarav
    6Second to the left?
    7Immediately to the leftGaurav
    8Immediately to the left?
    9?Farhan
    10?Chirag
    Who is sitting immediately to the right of Bina?
    1. A.

      Aarav

    2. B.

      Eshan

    3. C.

      Farhan

    4. D.

      Chirag

    Correct Answer:

    B

    Step-by-Step Solution

    Key idea: This is a relative-position query on a reconstructed circular seating, recognisable because we must determine the exact seating arrangement from a sequence of passes and then answer a directional question.

    Why: The problem provides a table of passes with missing receivers and pass types. By assigning coordinates modulo 7, we can reconstruct the exact positions of all children and then apply the left/right rules.

    Step 1: Fix coordinates. Place Aarav at 0 modulo 7.

    IL = +1, IR = -1, SL = +2, SR = -2.

    Step 2: Reconstruct fixed positions.

    R1: Bina passes IL to Aarav(0). Bina is at 6.

    R2: Aarav(0) passes SR. Receiver is at 5.

    R3: Person at 5 passes IR to Diya. Diya is at 4.

    R6: Aarav(0) passes SL. Receiver is at 2.

    R7: Person at 2 passes IL to Gaurav. Gaurav is at 3.

    Step 3: Locate Farhan, Chirag, Eshan.

    Placed: Aarav(0), Gaurav(3), Diya(4), Bina(6).

    Remaining: 1, 2, 5. People: Chirag, Eshan, Farhan.

    R9: Diya(4) passes to Farhan. Targets from 4: 5, 3, 6, 2. 3 and 6 occupied. Farhan is at 5 or 2.

    R10: Farhan passes to Chirag.

    If Farhan at 5, targets are 6, 4, 0, 3. All occupied. Impossible.

    So Farhan is at 2. Targets from 2: 3, 1, 4, 0. Only 1 is free. Chirag is at 1.

    Eshan is at 5.

    Step 4: Identify Bina's position and direction.

    Bina is at position 6. For a person facing inside, Left is Clockwise (+1) and Right is Counter-Clockwise (-1).

    Immediately to the right of Bina (6) means position .

    Step 5: Identify the person at position 5.

    From the reconstruction, position 5 is occupied by Eshan.

    Answer: B

    More previous year questions (pyqs) in this unit

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    Seating, Positioning and Arrangement Logic Previous Year Questions (PYQs) for CAT: 16+ Solved Questions with Step-by-Step Solutions

    Solve 16+ Seating, Positioning and Arrangement Logic previous year questions for CAT with answers and detailed solutions. Free sample questions below.

    A question from this chapter

    Question 1
    Common Description: Seven children, Aarav, Bina, Chirag, Diya, Eshan, Farhan, and Gaurav, are sitting in a circle facing inside (not necessarily in the same order) and playing a game of 'Passing the Buck'.
    The game is played over 10 rounds. In each round, the child holding the Buck must pass it directly to a child sitting in one of the following positions:
    • Immediately to the left;
    • Immediate to the right;
    • Second to the left; or
    • Second to the right.
    The game starts with Bina passing the Buck and ends with Chirag receiving the Buck. The table below provides some information about the pass types and the child receiving the Buck. Some information is missing and labelled as '?'.
    RoundPass TypeReceived by
    1Immediately to the leftAarav
    2Second to the right?
    3Immediately to the rightDiya
    4??
    5?Aarav
    6Second to the left?
    7Immediately to the leftGaurav
    8Immediately to the left?
    9?Farhan
    10?Chirag
    For which of the following children is it possible to determine how many times they received the Buck?
    Question 2
    Common Description: Seven children, Aarav, Bina, Chirag, Diya, Eshan, Farhan, and Gaurav, are sitting in a circle facing inside (not necessarily in the same order) and playing a game of 'Passing the Buck'.
    The game is played over 10 rounds. In each round, the child holding the Buck must pass it directly to a child sitting in one of the following positions:
    • Immediately to the left;
    • Immediate to the right;
    • Second to the left; or
    • Second to the right.
    The game starts with Bina passing the Buck and ends with Chirag receiving the Buck. The table below provides some information about the pass types and the child receiving the Buck. Some information is missing and labelled as '?'.
    RoundPass TypeReceived by
    1Immediately to the leftAarav
    2Second to the right?
    3Immediately to the rightDiya
    4??
    5?Aarav
    6Second to the left?
    7Immediately to the leftGaurav
    8Immediately to the left?
    9?Farhan
    10?Chirag
    Who is sitting third to the left of Eshan?
    Question 3
    Common Description: Instructions [30 - 34]
    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.
    BlockColumn-AColumn-BColumn-CRoadColumn-DColumn-EColumn-F
    Row-1A1B1C1ROADD1E1F1
    Row-2A2B2C2ROADD2E2F2
    BlockBlock XXROADBlock YY
    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?
    Question 4
    Common Description: Instructions [30 - 34]
    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.
    BlockColumn-AColumn-BColumn-CRoadColumn-DColumn-EColumn-F
    Row-1A1B1C1ROADD1E1F1
    Row-2A2B2C2ROADD2E2F2
    BlockBlock XXROADBlock YY
    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. Which of the following options best describes the number of vacant houses in Row-2?
    Question 5
    Common Description: Seven children, Aarav, Bina, Chirag, Diya, Eshan, Farhan, and Gaurav, are sitting in a circle facing inside (not necessarily in the same order) and playing a game of 'Passing the Buck'.
    The game is played over 10 rounds. In each round, the child holding the Buck must pass it directly to a child sitting in one of the following positions:
    • Immediately to the left;
    • Immediate to the right;
    • Second to the left; or
    • Second to the right.
    The game starts with Bina passing the Buck and ends with Chirag receiving the Buck. The table below provides some information about the pass types and the child receiving the Buck. Some information is missing and labelled as '?'.
    RoundPass TypeReceived by
    1Immediately to the leftAarav
    2Second to the right?
    3Immediately to the rightDiya
    4??
    5?Aarav
    6Second to the left?
    7Immediately to the leftGaurav
    8Immediately to the left?
    9?Farhan
    10?Chirag
    Who is sitting immediately to the right of Bina?
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